Re: how to simulate the situation in DNA evolution for finding the minimum population needed and minimum samples selected to mating in order to no extinction in any one of original species and new spe

[meInvent bbird]

i noticed the faces of human repeated or similar, 
and would like to prove whether evolution a several generations will 
return to the original intelligence of ancester

On Friday, November 18, 2016 at 1:55:31 PM UTC+8, meInvent bbird wrote:
> how to simulate the situation in DNA evolution for finding the minimum 
> population needed and minimum samples selected to mating in order to no 
> extinction in any one of original species and new species
> 
> assume mating are randomly selected to become a couple,
> how to keep species good which means no extinction in original species
> 
> i use i+j to get new species, 
> but i do not know whether there is limit in evolution, i assume
> limit is 7, then extra evolution will go back to past species and 
> cycle again
> 
> import matplotlib.pyplot as plt
> import random
> dict = {}
> dist = {}
> maxnum = 5
> allowedmax = 7
> for i in range(1,allowedmax+1):
> dist[str(i)] = 0
> 
> rr = range (1,maxnum)
> for i in range (1,maxnum):
> for j in range (1,maxnum):
> if i < j:
> print("(" +str(i) + "," + str(j) + ")");
> dict[str(i) + str(j)] = 1;
> dist[str(i+j)] = dist[str(i+j)] + 1
> if i+j > max(rr or [0]):
>  rr = rr + [i+j];
> 
> original = rr;
> for numberofevolutions in range(1,10):
> rr2 = []
> samples = random.sample(original, len(original)-2)
> print("total rr")
> print(str(rr))
> print("samples")
> print(str(samples))
> for i in samples:
> for j in samples:
> if i < j:
> print("(" +str(i) + "," + str(j) + ")");
> if i+j > allowedmax:
> dict[str(i) + str(j)] = (i+j) % allowedmax;
> dist[str((i+j) % allowedmax)] = dist[str((i+j) % 
> allowedmax)] + 1
> if ((i+j) % allowedmax) > max(rr2 or [0]):
> rr2 = rr2 + [((i+j) % allowedmax)];
> else:
> dict[str(i) + str(j)] = i+j;
> dist[str(i+j)] = dist[str(i+j)] + 1
> if i+j > max(rr2 or [0]):
> rr2 = rr2 + [i+j];
> temp = rr
> rr = rr2
> rr2 = temp
> 
> plt.bar(range(len(dist)), dist.values(), align='center')
> plt.xticks(range(len(dist)), dist.keys())
> plt.show()

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how to simulate the situation in DNA evolution for finding the minimum population needed and minimum samples selected to mating in order to no extinction in any one of original species and new species

[meInvent bbird]

how to simulate the situation in DNA evolution for finding the minimum 
population needed and minimum samples selected to mating in order to no 
extinction in any one of original species and new species

assume mating are randomly selected to become a couple,
how to keep species good which means no extinction in original species

i use i+j to get new species, 
but i do not know whether there is limit in evolution, i assume
limit is 7, then extra evolution will go back to past species and 
cycle again

import matplotlib.pyplot as plt
import random
dict = {}
dist = {}
maxnum = 5
allowedmax = 7
for i in range(1,allowedmax+1):
dist[str(i)] = 0

rr = range (1,maxnum)
for i in range (1,maxnum):
for j in range (1,maxnum):
if i < j:
print("(" +str(i) + "," + str(j) + ")");
dict[str(i) + str(j)] = 1;
dist[str(i+j)] = dist[str(i+j)] + 1
if i+j > max(rr or [0]):
 rr = rr + [i+j];

original = rr;
for numberofevolutions in range(1,10):
rr2 = []
samples = random.sample(original, len(original)-2)
print("total rr")
print(str(rr))
print("samples")
print(str(samples))
for i in samples:
for j in samples:
if i < j:
print("(" +str(i) + "," + str(j) + ")");
if i+j > allowedmax:
dict[str(i) + str(j)] = (i+j) % allowedmax;
dist[str((i+j) % allowedmax)] = dist[str((i+j) % 
allowedmax)] + 1
if ((i+j) % allowedmax) > max(rr2 or [0]):
rr2 = rr2 + [((i+j) % allowedmax)];
else:
dict[str(i) + str(j)] = i+j;
dist[str(i+j)] = dist[str(i+j)] + 1
if i+j > max(rr2 or [0]):
rr2 = rr2 + [i+j];
temp = rr
rr = rr2
rr2 = temp

plt.bar(range(len(dist)), dist.values(), align='center')
plt.xticks(range(len(dist)), dist.keys())
plt.show()
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what is the procedure or how to plan how many nodes of dispy need for dsolve differential system in amazon cloud in limited time such as 1 hour, 2 hours.etc

[meInvent bbird]

what is the procedure or how to plan how many nodes of dispy need for dsolve 
differential system in amazon cloud in limited time such as 1 hour, 2 hours.etc 
?


#For Amazon Linux, the user name is ec2-user. For RHEL5, the user name is 
either root or ec2-user. 
#For Ubuntu, the user name is ubuntu. For Fedora, the user name is either 
fedora or ec2-user. 
#For SUSE Linux, the user name is either root or ec2-user. 
#Otherwise, if ec2-user and root don't work, check with your AMI provider.

import random, dispy
import ast 
from __future__ import division 
from sympy import * 
x, y, z, t = symbols('x y z t') 
k, m, n = symbols('k m n', integer=True) 
f, g, h = symbols('f g h', cls=Function) 
import inspect 

def compute(n): # executed on nodes
import random, time, socket
name = socket.gethostname()
cur_best = 1
for ii in range(n[0],n[0]):
for jj in range(n[1],n[1]):
for kk in range(n[2],n[3],100):  
#assume dsolve with sympy for differential system   
dispy_provisional_result((name, r))
cur_best = r
time.sleep(0.1)
# final result
return (name, cur_best)

def job_callback(job): # executed at the client
if job.status == dispy.DispyJob.ProvisionalResult:
#if job.result[1] < 0.005:
# acceptable result; terminate jobs
print('%s computed: %s %s %s %s' % (job.result[0], job.result[1], 
job.result[2], job.result[3], job.result[4]))
# 'jobs' and 'cluster' are created in '__main__' below
for j in jobs:
if j.status in [dispy.DispyJob.Created, dispy.DispyJob.Running,
dispy.DispyJob.ProvisionalResult]:
cluster.cancel(j)

if __name__ == '__main__':
#cluster = dispy.JobCluster(compute, callback=job_callback)
cluster = dispy.JobCluster(compute, 
nodes=['ec2-35-162-137-237.us-west-2.compute.amazonaws.com'], 
ip_addr='127.0.0.1', port=51347, node_port=51348, callback=job_callback, 
keyfile=r"C:\Users\hello\datacenterusekey.ppk")
jobs = []
prevk = 1
count = 0
for ii in range(1,2):
for jj in range(1,2000):
for kk in range(1,2000,100):
if ii < jj and jj < kk:
job = cluster.submit([ii,jj,prevk,kk])
prevk = kk
if job is None:
print('creating job %s failed!' % n)
continue
job.id = count
count = count + 1
jobs.append(job)
cluster.wait()
cluster.print_status()
cluster.close()
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Re: how to compile this code

[meInvent bbird]

>>> class ChangeAddToMultiply(ast.NodeTransformer, ast2.NodeTransformer):
... """Wraps all integers in a call to Integer()"""
... def visit_BinOp(self, node):
... print(dir(node))
... print(dir(node.left))
... if isinstance(node.op, ast.Add):
... ast.Call(Name(id="op2", ctx=ast2.Load()), [node.left, node.right
], [])
... return node
...
Traceback (most recent call last):
  File "", line 1, in 
NameError: name 'ast2' is not defined
>>>
>>> code = inspect.getsourcelines(solve)
>>> tree = ast.parse(code)
Traceback (most recent call last):
  File "", line 1, in 
  File "C:\Python27\lib\ast.py", line 37, in parse
return compile(source, filename, mode, PyCF_ONLY_AST)
TypeError: expected a readable buffer object
>>> tree2 = ast.parse("def op2(a,b): return a*b+a")
>>> tree = ChangeAddToMultiply().visit(tree,tree2)
Traceback (most recent call last):
  File "", line 1, in 
NameError: name 'tree' is not defined
>>> ast.fix_missing_locations(tree)
Traceback (most recent call last):
  File "", line 1, in 
NameError: name 'tree' is not defined
>>> co = compile(tree, '', "exec")
Traceback (most recent call last):
  File "", line 1, in 
NameError: name 'tree' is not defined
>>>
>>> exec(code)
Traceback (most recent call last):
  File "", line 1, in 
TypeError: exec: arg 1 must be a string, file, or code object
>>> exec(co)

*

def op2(a,b): 
return a*b+a 

class ChangeAddToMultiply(ast.NodeTransformer, ast2.NodeTransformer): 
"""Wraps all integers in a call to Integer()""" 
def visit_BinOp(self, node): 
print(dir(node)) 
print(dir(node.left)) 
if isinstance(node.op, ast.Add): 
ast.Call(Name(id="op2", ctx=ast2.Load()), [node.left, node.right], 
[]) 
return node 


code = inspect.getsourcelines(solve) 
tree = ast.parse(code) 
tree2 = ast.parse("def op2(a,b): return a*b+a")
tree = ChangeAddToMultiply().visit(tree,tree2) 
ast.fix_missing_locations(tree) 
co = compile(tree, '', "exec") 

exec(code) 
exec(co) 



On Tuesday, November 1, 2016 at 9:04:43 PM UTC+8, Yann Kaiser wrote:
> You want to replace the `Add` ast with a `Call` ast rather than just
> calling your function.
> 
> Something like:
> 
> if isinstance(node.op, ast.Add):
> return ast.Call(some_ast_expression_that_will_evaluate_to_op,
> [node.left, node.right], [])
> 
> You'll have to replace some_ast_expression_... to something like
> Name(id="op2", ctx=ast.Load()) and inject op2 in the globals when you call
> `exec`.
> 
> On Tue, Nov 1, 2016, 06:36 meInvent bbird <jobmatt...@gmail.com> wrote:
> 
> > would like to change Add operator to custom function op2 in solve function
> > and then this solve([x*y - 1, x + 2], x, y)
> > during solve, the parameters also change Add to custom function op2
> >
> > import ast
> > from __future__ import division
> > from sympy import *
> > x, y, z, t = symbols('x y z t')
> > k, m, n = symbols('k m n', integer=True)
> > f, g, h = symbols('f g h', cls=Function)
> > import inspect
> >
> > def op2(a,b):
> > return a*b+a
> >
> > class ChangeAddToMultiply(ast.NodeTransformer):
> > """Wraps all integers in a call to Integer()"""
> > def visit_BinOp(self, node):
> > print(dir(node))
> > print(dir(node.left))
> > if isinstance(node.op, ast.Add):
> > node.op = op2(node.left, node.right)
> > return node
> >
> > code = inspect.getsourcelines(solve)
> > tree = ast.parse(code)
> > tree = ChangeAddToMultiply().visit(tree)
> > ast.fix_missing_locations(tree)
> > co = compile(tree, '', "exec")
> >
> > exec(code)
> > exec(co)
> > --
> > https://mail.python.org/mailman/listinfo/python-list
> >
> -- 
> Yann Kaiser
> kaiser.y...@gmail.com
> yann.kai...@efrei.net
> +33 6 51 64 01 89
> https://github.com/epsy

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Re: how to debug this distributed program?

[meInvent bbird]

i follow documentation and add keyfile to indicate the path of private key
but still no response from program,

what is the correct way to do?

cluster = dispy.JobCluster(compute, 
nodes=['ec2-35-162-137-237.us-west-2.compute.amazonaws.com'], 
callback=job_callback, 
keyfile=r"C:\Users\martlee2\Downloads\datacenterusekey.ppk")



On Tuesday, November 1, 2016 at 3:58:24 PM UTC+8, Steve D'Aprano wrote:
> On Tue, 1 Nov 2016 06:40 pm, meInvent bbird wrote:
> 
> > how to debug this distributed program?
> 
> 
> The same way you would debug any other program.
> 
> http://sscce.org/
> 
> 
> 
> 
> 
> -- 
> Steve
> “Cheer up,” they said, “things could be worse.” So I cheered up, and sure
> enough, things got worse.

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how to debug this distributed program?

[meInvent bbird]

https://drive.google.com/open?id=0Bxs_ao6uuBDUSEc5S3U3Nko5ZjA


A. i am not sure whether parameter start from 0 or 1
n[0] or n[1] in compute function

B. it run a very long time and nothing to see in amazon linux instance
there is no python program in top command

C. in distributed programming web site, there is no authentication method
using private key in python program,  how do distributed program access
the nodes in amazon cloud when run python program below in window locally?

import random, dispy

def compute(n): # executed on nodes
import random, time, socket
name = socket.gethostname()
cur_best = 1
for ii in range(n[0],n[0]):
for jj in range(n[1],n[1]):
for kk in range(n[2],n[3],100): 
dispy_provisional_result((name, r))
cur_best = r
time.sleep(0.1)
# final result
return (name, cur_best)

def job_callback(job): # executed at the client
if job.status == dispy.DispyJob.ProvisionalResult:
#if job.result[1] < 0.005:
# acceptable result; terminate jobs
print('%s computed: %s %s %s %s' % (job.result[0], job.result[1], 
job.result[2], job.result[3], job.result[4]))
# 'jobs' and 'cluster' are created in '__main__' below
for j in jobs:
if j.status in [dispy.DispyJob.Created, dispy.DispyJob.Running,
dispy.DispyJob.ProvisionalResult]:
cluster.cancel(j)

if __name__ == '__main__':
#cluster = dispy.JobCluster(compute, callback=job_callback)
cluster = dispy.JobCluster(compute, 
nodes=['ec2-100-162-137-237.us-west-2.compute.amazonaws.com'], 
callback=job_callback)
jobs = []
prevk = 1
count = 0
for ii in range(1,2):
for jj in range(1,2000):
for kk in range(1,2000,100):
if ii < jj and jj < kk:
job = cluster.submit([ii,jj,prevk,kk])
prevk = kk
if job is None:
print('creating job %s failed!' % n)
continue
job.id = count
count = count + 1
jobs.append(job)
cluster.wait()
cluster.print_status()
cluster.close()
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how to compile this code

[meInvent bbird]

would like to change Add operator to custom function op2 in solve function
and then this solve([x*y - 1, x + 2], x, y)
during solve, the parameters also change Add to custom function op2

import ast 
from __future__ import division
from sympy import *
x, y, z, t = symbols('x y z t')
k, m, n = symbols('k m n', integer=True)
f, g, h = symbols('f g h', cls=Function)
import inspect

def op2(a,b): 
return a*b+a 

class ChangeAddToMultiply(ast.NodeTransformer): 
"""Wraps all integers in a call to Integer()""" 
def visit_BinOp(self, node): 
print(dir(node))
print(dir(node.left))
if isinstance(node.op, ast.Add): 
node.op = op2(node.left, node.right)
return node 

code = inspect.getsourcelines(solve)
tree = ast.parse(code)
tree = ChangeAddToMultiply().visit(tree) 
ast.fix_missing_locations(tree) 
co = compile(tree, '', "exec") 

exec(code) 
exec(co) 
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Re: how to evaluate a ast tree and change Add to Multiply ?

[meInvent bbird]

Hi Jerry,

how about custom function?

i change to node.op = ast.op2()


import ast 

def op2(a,b):
return a*b+a

class ChangeAddToMultiply(ast.NodeTransformer): 
"""Wraps all integers in a call to Integer()""" 
def visit_BinOp(self, node): 
if isinstance(node.op, ast.Add): 
node.op = ast.op2()
return node 

code = 'print(2+5)' 
tree = ast.parse(code) 
tree = ChangeAddToMultiply().visit(tree) 
ast.fix_missing_locations(tree) 
co = compile(tree, '', "exec") 

exec(code) 
exec(co) 




On Thursday, October 20, 2016 at 12:34:55 AM UTC+8, Jerry Hill wrote:
> On Wed, Oct 12, 2016 at 5:55 AM, meInvent bbird <jobmatt...@gmail.com> wrote:
> > i just expect to
> > rewrite + become multiply
> > by edit the example in the link provided
> 
> This seems to work.  You need to define visit_BinOp instead of
> visit_Num (since you want to mess with the binary operations, not the
> numbers).  Then,in visit_BinOp, we just replace the ast.Add node with
> an ast.Mult node.
> 
> import ast
> 
> class ChangeAddToMultiply(ast.NodeTransformer):
> """Wraps all integers in a call to Integer()"""
> def visit_BinOp(self, node):
> if isinstance(node.op, ast.Add):
> node.op = ast.Mult()
> return node
> 
> code = 'print(2+5)'
> tree = ast.parse(code)
> tree = ChangeAddToMultiply().visit(tree)
> ast.fix_missing_locations(tree)
> co = compile(tree, '', "exec")
> 
> exec(code)
> exec(co)
> 
> -- 
> Jerry

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is it possible use python to run another main point or function in some range of memory in executable file

[meInvent bbird]

is it possible python to run another main point or function in some range of 
memory in executable file

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Re: how to send a json of yield list

[meInvent bbird]

when not to use queue, it is faster now

while True:
for ii in getcombinations():
item = ii
print(item)
sock.send(json.dumps(ii))


On Friday, October 14, 2016 at 5:11:35 PM UTC+8, meInvent bbird wrote:
> succeed to run,
> 
> is it the yield return the whole list 2000 * 2000 * 2000 items?
> 
> as i know that yield is return [1,1,1] etc one by one once it get
> 
> if it return 2000*2000*2000 items, why?
> 
> i have to add a queue get this yield in order to succeed
> 
> but i do not understand the situation when
> using queue and yield at the same time
> 
> will the speed same as the situation that no yield?
> 
> while True:
> #time.sleep(1)
> if q.qsize() > 0:
> print("here1")
> item = q.get()
> print("here2")
> #item = getcombinations()
> sock.send(json.dumps(item))

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Re: how to send a json of yield list

[meInvent bbird]

succeed to run,

is it the yield return the whole list 2000 * 2000 * 2000 items?

as i know that yield is return [1,1,1] etc one by one once it get

if it return 2000*2000*2000 items, why?

i have to add a queue get this yield in order to succeed

but i do not understand the situation when
using queue and yield at the same time

will the speed same as the situation that no yield?

while True:
#time.sleep(1)
if q.qsize() > 0:
print("here1")
item = q.get()
print("here2")
#item = getcombinations()
sock.send(json.dumps(item))
-- 
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Re: how to use pycallgraph in ubuntu and window?

[meInvent bbird]

succeed to draw graph, thanks

originally i think pip install graphviz is enough

then search again

there is an win32 executable file graphviz need to be install

On Friday, October 14, 2016 at 2:58:48 PM UTC+8, dieter wrote:
> meInvent bbird <jobmatt...@gmail.com> writes:
> 
> > i install in ubunbu 14
> >
> > pip install graphviz
> > pip install pycallgraph
> >
> > martin@ubuntu:~/Downloads$ pycallgraph graphviz -- ./pusher.py
> > Traceback (most recent call last):
> > ...
> > pycallgraph.exceptions.PyCallGraphException: The command "dot" is required 
> > to be in your path.
> 
> This tells you that a required (operating system level) utility
> (named "dot") is not available.
> 
> Carefully read the installation instructions, especially any
> remark about dependencies. This should tell you where "dot" is supposed
> to come from. Install all operating system level dependencies.
> 
> If "dot" should have come from "graphviz", then maybe a
> "hash" is necessary to let your shell see new commands.

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Can't pickle : attribute lookup __builtin__.gen erator failed

[meInvent bbird]

Traceback (most recent call last):
  File "C:\Python27\lib\multiprocessing\queues.py", line 262, in _feed
send(obj)
PicklingError: Can't pickle : attribute lookup __builtin__.gen
erator failed


#python pusher.py tcp://*:8080
import sys
import time
import zmq

from multiprocessing import Queue
import threading

import json
from json import dumps, loads, JSONEncoder, JSONDecoder
import pickle
from pycallgraph import PyCallGraph
from pycallgraph.output import GraphvizOutput

class PythonObjectEncoder(JSONEncoder):
def default(self, obj):
if isinstance(obj, (list, dict, str, unicode, int, float, bool, 
type(None))):
return JSONEncoder.default(self, obj)
return {'_python_object': pickle.dumps(obj)}

def as_python_object(dct):
if '_python_object' in dct:
return pickle.loads(str(dct['_python_object']))
return dct

context = zmq.Context()
sock = context.socket(zmq.PUSH)
sock.bind(sys.argv[1])

def getcombinations():
for ii in range(1,2000):
for jj in range(1,2000):
for kk in range(1,2000):
yield [ii,jj,kk]

class StreamArray(list):
def __iter__(self):
return getcombinations()

# according to the comment below
def __len__(self):
return 1

def worker():
while True:
if q.qsize() < 1000:
q.put(getcombinations())

q = Queue(maxsize=1000)
t = threading.Thread(target=worker, args = ())
t.daemon = True
t.start()

while True:
time.sleep(1)
#sock.send(sys.argv[1] + ':' + time.ctime())
combobject = getcombinations()
#sock.send(json.dumps(combobject, cls=PythonObjectEncoder))
if q.qsize() > 0:
item = q.get()
sock.send(json.dumps(item))
#sock.send(json.dumps(combobject, cls=PythonObjectEncoder))
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how to use pycallgraph in ubuntu and window?

[meInvent bbird]

i install in ubunbu 14

pip install graphviz
pip install pycallgraph

martin@ubuntu:~/Downloads$ pycallgraph graphviz -- ./pusher.py
Traceback (most recent call last):
  File "/usr/local/bin/pycallgraph", line 25, in 
with __pycallgraph.PyCallGraph(config=__config):
  File "/usr/local/lib/python2.7/dist-packages/pycallgraph/pycallgraph.py", 
line 32, in __init__
self.reset()
  File "/usr/local/lib/python2.7/dist-packages/pycallgraph/pycallgraph.py", 
line 53, in reset
self.prepare_output(output)
  File "/usr/local/lib/python2.7/dist-packages/pycallgraph/pycallgraph.py", 
line 97, in prepare_output
output.sanity_check()
  File "/usr/local/lib/python2.7/dist-packages/pycallgraph/output/graphviz.py", 
line 63, in sanity_check
self.ensure_binary(self.tool)
  File "/usr/local/lib/python2.7/dist-packages/pycallgraph/output/output.py", 
line 96, in ensure_binary
'The command "{}" is required to be in your path.'.format(cmd))
pycallgraph.exceptions.PyCallGraphException: The command "dot" is required to 
be in your path.


in window 

i had already added path

C:\Python27\Lib\site-packages\graphviz;C:\Python27\Lib\site-packages\pycallgraph

 to environment variable path

but in cmd

'pycallgraph' is not recognized as an internal or external command,
operable program or batch file.

then i use full path

>python "C:\Python27\Lib\site-packages\pycallgraph\py
callgraph.py"  -- "C:\Users\hello\Downloads\pusher.py"
Traceback (most recent call last):
  File "C:\Python27\Lib\site-packages\pycallgraph\pycallgraph.py", line 3, in 
from .output import Output
ValueError: Attempted relative import in non-package


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how to send a json of yield list

[meInvent bbird]

after google a several solutions, 

First method i searched has memory error
sock.send(json.dumps(StreamArray()))
Traceback (most recent call last):
  File "pusher.py", line 43, in 
sock.send(json.dumps(StreamArray()))
  File "C:\Python27\lib\json\__init__.py", line 243, in dumps
return _default_encoder.encode(obj)
  File "C:\Python27\lib\json\encoder.py", line 207, in encode
chunks = self.iterencode(o, _one_shot=True)
  File "C:\Python27\lib\json\encoder.py", line 270, in iterencode
return _iterencode(o, 0)
MemoryError

if use this solution, got another error

combobject = getcombinations()
sock.send(json.dumps(combobject, cls=PythonObjectEncoder))

C:\Users\martlee2\Downloads>python pusher.py tcp://*:8080
Traceback (most recent call last):
  File "pusher.py", line 42, in 
sock.send(json.dumps(combobject, cls=PythonObjectEncoder))
  File "C:\Python27\lib\json\__init__.py", line 250, in dumps
sort_keys=sort_keys, **kw).encode(obj)
  File "C:\Python27\lib\json\encoder.py", line 207, in encode
chunks = self.iterencode(o, _one_shot=True)
  File "C:\Python27\lib\json\encoder.py", line 270, in iterencode
return _iterencode(o, 0)
  File "pusher.py", line 13, in default
return {'_python_object': pickle.dumps(obj)}
  File "C:\Python27\lib\pickle.py", line 1374, in dumps
Pickler(file, protocol).dump(obj)
  File "C:\Python27\lib\pickle.py", line 224, in dump
self.save(obj)
  File "C:\Python27\lib\pickle.py", line 306, in save
rv = reduce(self.proto)
  File "C:\Python27\lib\copy_reg.py", line 70, in _reduce_ex
raise TypeError, "can't pickle %s objects" % base.__name__
TypeError: can't pickle generator objects



#python pusher.py tcp://*:8080
import sys
import time
import zmq
import json
from json import dumps, loads, JSONEncoder, JSONDecoder
import pickle

class PythonObjectEncoder(JSONEncoder):
def default(self, obj):
if isinstance(obj, (list, dict, str, unicode, int, float, bool, 
type(None))):
return JSONEncoder.default(self, obj)
return {'_python_object': pickle.dumps(obj)}

def as_python_object(dct):
if '_python_object' in dct:
return pickle.loads(str(dct['_python_object']))
return dct

context = zmq.Context()
sock = context.socket(zmq.PUSH)
sock.bind(sys.argv[1])

def getcombinations():
for ii in range(1,2000):
for jj in range(1,2000):
for kk in range(1,2000):
yield [ii,jj,kk]

class StreamArray(list):
def __iter__(self):
return getcombinations()

# according to the comment below
def __len__(self):
return 1

while True:
time.sleep(1)
#sock.send(sys.argv[1] + ':' + time.ctime())
combobject = getcombinations()
sock.send(json.dumps(combobject, cls=PythonObjectEncoder))



puller.py

#python puller.py tcp://localhost:8080
import sys
import zmq
import json

context = zmq.Context()
sock = context.socket(zmq.PULL)

for arg in sys.argv[1:]:
sock.connect(arg)

while True:
message = sock.recv()
combinations = json.loads(message)
print(str(combinations))
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Re: how to read linux kernel source with pycparser

[meInvent bbird]

is it possible to git pull a part of directory such as sched
and compile this subdirectory and pycparser it?

i got error when run gcc -E

git submodule add he...@github.com:hello/repo.git kernel/sched
cd kernel/sched
git pull

martin@ubuntu:~/Downloads/kernel/sched$ gcc -E -std=c99 *.c
# 1 "auto_group.c"
# 1 ""
# 1 ""
# 1 "auto_group.c"
# 1 "sched.h" 1
# 1 "/usr/include/linux/sched.h" 1 3 4
# 2 "sched.h" 2
In file included from auto_group.c:1:0:
sched.h:2:32: fatal error: linux/sched/sysctl.h: No such file or directory
compilation terminated.
# 1 "clock.c"
# 1 ""
# 1 ""
# 1 "clock.c"
clock.c:55:28: fatal error: linux/spinlock.h: No such file or directory
compilation terminated.
# 1 "completion.c"
# 1 ""
# 1 ""
# 1 "completion.c"
# 14 "completion.c"
# 1 "/usr/include/linux/sched.h" 1 3 4
# 15 "completion.c" 2
completion.c:15:30: fatal error: linux/completion.h: No such file or directory
compilation terminated.
# 1 "core.c"
# 1 ""
# 1 ""
# 1 "core.c"
core.c:29:25: fatal error: linux/kasan.h: No such file or directory
compilation terminated.
# 1 "cpuacct.c"
# 1 ""
# 1 ""
# 1 "cpuacct.c"
cpuacct.c:1:26: fatal error: linux/cgroup.h: No such file or directory
compilation terminated.
# 1 "cpudeadline.c"
# 1 ""
# 1 ""
# 1 "cpudeadline.c"
cpudeadline.c:14:23: fatal error: linux/gfp.h: No such file or directory
compilation terminated.
# 1 "cpufreq.c"
# 1 ""
# 1 ""
# 1 "cpufreq.c"
# 12 "cpufreq.c"
# 1 "sched.h" 1
# 1 "/usr/include/linux/sched.h" 1 3 4
# 2 "sched.h" 2
In file included from cpufreq.c:12:0:
sched.h:2:32: fatal error: linux/sched/sysctl.h: No such file or directory
compilation terminated.
# 1 "cpufreq_schedutil.c"
# 1 ""
# 1 ""
# 1 "cpufreq_schedutil.c"
cpufreq_schedutil.c:14:27: fatal error: linux/cpufreq.h: No such file or 
directory
compilation terminated.
# 1 "cpupri.c"
# 1 ""
# 1 ""
# 1 "cpupri.c"
cpupri.c:30:23: fatal error: linux/gfp.h: No such file or directory
compilation terminated.
# 1 "cputime.c"
# 1 ""
# 1 ""
# 1 "cputime.c"
cputime.c:1:26: fatal error: linux/export.h: No such file or directory
compilation terminated.
# 1 "deadline.c"
# 1 ""
# 1 ""
# 1 "deadline.c"
# 17 "deadline.c"
# 1 "sched.h" 1
# 1 "/usr/include/linux/sched.h" 1 3 4
# 2 "sched.h" 2
In file included from deadline.c:17:0:
sched.h:2:32: fatal error: linux/sched/sysctl.h: No such file or directory
compilation terminated.
# 1 "debug.c"
# 1 ""
# 1 ""
# 1 "debug.c"
debug.c:13:27: fatal error: linux/proc_fs.h: No such file or directory
compilation terminated.
# 1 "fair.c"
# 1 ""
# 1 ""
# 1 "fair.c"
# 23 "fair.c"
# 1 "/usr/include/linux/sched.h" 1 3 4
# 24 "fair.c" 2
fair.c:24:30: fatal error: linux/latencytop.h: No such file or directory
compilation terminated.
# 1 "idle.c"
# 1 ""
# 1 ""
# 1 "idle.c"



# 1 "/usr/include/linux/sched.h" 1 3 4
# 5 "idle.c" 2
idle.c:5:23: fatal error: linux/cpu.h: No such file or directory
compilation terminated.
# 1 "idle_task.c"
# 1 ""
# 1 ""
# 1 "idle_task.c"
# 1 "sched.h" 1
# 1 "/usr/include/linux/sched.h" 1 3 4
# 2 "sched.h" 2
In file included from idle_task.c:1:0:
sched.h:2:32: fatal error: linux/sched/sysctl.h: No such file or directory
compilation terminated.
# 1 "loadavg.c"
# 1 ""
# 1 ""
# 1 "loadavg.c"
loadavg.c:9:26: fatal error: linux/export.h: No such file or directory
compilation terminated.
# 1 "rt.c"
# 1 ""
# 1 ""
# 1 "rt.c"





# 1 "sched.h" 1
# 1 "/usr/include/linux/sched.h" 1 3 4
# 2 "sched.h" 2
In file included from rt.c:6:0:
sched.h:2:32: fatal error: linux/sched/sysctl.h: No such file or directory
compilation terminated.
# 1 "stats.c"
# 1 ""
# 1 ""
# 1 "stats.c"
stats.c:1:24: fatal error: linux/slab.h: No such file or directory
compilation terminated.
# 1 "stop_task.c"
# 1 ""
# 1 ""
# 1 "stop_task.c"
# 1 "sched.h" 1
# 1 "/usr/include/linux/sched.h" 1 3 4
# 2 "sched.h" 2
In file included from stop_task.c:1:0:
sched.h:2:32: fatal error: linux/sched/sysctl.h: No such file or directory
compilation terminated.
# 1 "swait.c"
# 1 ""
# 1 &q

Re: how to evaluate a ast tree and change Add to Multiply ?

[meInvent bbird]

i just expect to 
rewrite + become multiply
by edit the example in the link provided

but search no examples about this, 
feel unknown about args and keywords etc, 
do not know how to write this 

ast.Call(func=ast.Name(id='Add', ctx=ast.Load()), 
args=[node], keywords=[]) 

On Wednesday, October 12, 2016 at 5:53:42 PM UTC+8, meInvent bbird wrote:
> i use example here
> 
>  http://greentreesnakes.readthedocs.io/en/latest/examples.html
> 
> 
> 
> 
> On Wednesday, October 12, 2016 at 5:47:12 PM UTC+8, Chris Angelico wrote:
> > On Wed, Oct 12, 2016 at 8:32 PM, meInvent bbird <jobmatt...@gmail.com> 
> > wrote:
> > > class ChangeAddtoMultiply(ast.NodeTransformer):
> > > """Wraps all integers in a call to Integer()"""
> > > def visit_Num(self, node):
> > > if isinstance(node.n, int):
> > > return ast.Call(func=ast.Name(id='Add', ctx=ast.Load()),
> > > args=[node], keywords=[])
> > > return node
> > >
> > >
> > 
> > Your class name, docstring, and functionality are all different, and
> > you have given no explanation of what you want us to do. I can't read
> > minds quite that well.
> > 
> > ChrisA

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Re: how to evaluate a ast tree and change Add to Multiply ?

[meInvent bbird]

i use example here

 http://greentreesnakes.readthedocs.io/en/latest/examples.html




On Wednesday, October 12, 2016 at 5:47:12 PM UTC+8, Chris Angelico wrote:
> On Wed, Oct 12, 2016 at 8:32 PM, meInvent bbird <jobmatt...@gmail.com> wrote:
> > class ChangeAddtoMultiply(ast.NodeTransformer):
> > """Wraps all integers in a call to Integer()"""
> > def visit_Num(self, node):
> > if isinstance(node.n, int):
> > return ast.Call(func=ast.Name(id='Add', ctx=ast.Load()),
> > args=[node], keywords=[])
> > return node
> >
> >
> 
> Your class name, docstring, and functionality are all different, and
> you have given no explanation of what you want us to do. I can't read
> minds quite that well.
> 
> ChrisA

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how to faster to know which ast name or id we want to change?

[meInvent bbird]

how to faster to know which ast name or id we want to change?

because i use ast.walk to print all nodes then i know the name + is Add

if using cparser to parse linux kernel, 
it will be a very large
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how to evaluate a ast tree and change Add to Multiply ?

[meInvent bbird]

import sys
import parser
import ast
from collections import deque


class ChangeAddtoMultiply(ast.NodeTransformer):
"""Wraps all integers in a call to Integer()"""
def visit_Num(self, node):
if isinstance(node.n, int):
return ast.Call(func=ast.Name(id='Add', ctx=ast.Load()),
args=[node], keywords=[])
return node


tree = ast.parse('2+5')
for node in ast.walk(tree):
print(str(node))

eval(tree)

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why solveset and dsolve do not have _fields for ast.walk?

[meInvent bbird]

Traceback (most recent call last):
  File "astgraphforsympy.py", line 86, in 
print get_func_calls(tree)
  File "astgraphforsympy.py", line 48, in get_func_calls
for node in ast.walk(tree):
  File "C:\Python27\lib\ast.py", line 213, in walk
todo.extend(iter_child_nodes(node))
  File "C:\Python27\lib\ast.py", line 178, in iter_child_nodes
for name, field in iter_fields(node):
  File "C:\Python27\lib\ast.py", line 166, in iter_fields
for field in node._fields:
AttributeError: Module instance has no attribute '_fields'
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Re: how to refactor nested for loop into smaller for loop assume each of them independent?

[meInvent bbird]

https://social.msdn.microsoft.com/Forums/vstudio/en-US/5f0a9a51-a256-4671-a5fc-e213949e7204/how-to-refactor-3-nested-for-loop-into-smaller-for-loop-assume-each-of-them-independent?forum=csharpgeneral
 

since when i dsolve a differential ideal used near 5GB memory for one ideal
, i feel that i need to use amazon instance, 
i do not know whether sympy use same amount of memory

i prepared to split a for loop which for combination if use i < j and j < k
so 84 Billion /2 = 42 Billions combinations

each script for doing 100 combinations range, it generate around 4XX files

if read the final file it generated,
is

for ii from 1925 to 2025 do
 for jj from 2040 to 2040 do
  for kk from 2040 to 2040 do

it seems missed 2026 to 2040 this range in the final file
i feel that my crafted algorithm is not full combinations

I wrote a algorithm to split for loop to generate maplesoft code for limited 
memory 

Assume I convert this algorithm into Python script and use sympy to do dsolve 

How to distribute to hundreds of Amazon instance to run each smaller for loop 



i am not familiar with using Amazon, 
such as how to create hundreds of EC2 instances

i guess use python script to ssh all remote instance
run a python script with bash command which in python script



On Monday, October 10, 2016 at 6:49:12 AM UTC+8, Chris Angelico wrote:
> On Mon, Oct 10, 2016 at 9:40 AM, Marko Rauhamaa  wrote:
> > Chris Angelico :
> >> Yeah, if it's just for progress status. Of course, that does assume
> >> that the run() function doesn't have anything particularly costly in
> >> it. If it does, well, dis gonna take a while
> >
> > Dealing with multigigabyte data streams is not over the top nowadays.
> 
> This is true, but ultimately, it all comes down to the run() function.
> With no further posts since the drive-by at the beginning, we have
> nothing solid to discuss this on.
> 
> ChrisA

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how to refactor nested for loop into smaller for loop assume each of them independent?

[meInvent bbird]

how to refactor nested for loop into smaller for loop assume each of them 
independent?

because memory is not enough

for ii in range(1,2000):
 for jj in range(1,2000):
  for kk in range(1,2000):
print run(ii,jj,kk)
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Re: 'str' object has no attribute 'intersection' and unhashable set

[meInvent bbird]

after change to frozenset , 

it seems can classify into 7 repeated group

but, in real data

this consecutive lines can be also a group
but i can not find this,
actually i do not understand how the function works
is there any algorithm tutorials or books or external library that
can have a better result for finding repeated lines as group in grouping 
application

1,[(1, 0, 1)]
1,[(1, 0, 1)]
1,[(1, 0, 1)]
1,[(1, 0, 1)]


real data
1,[(1, 0, 1)]
1,[]
1,[(-1, 1, -2), (2, -1/2, 1)]
0,[(1, 0, 0)]
1,[(1, 0, 1)]
1,[]
1,[]
0,[{a: 0}, {b: -1/c, a: 0}, {c: 1, b: -1, a: 0}, {c: -1, b: 1, a: 0}]
0,[(1/2 + sqrt(5)/2, -sqrt(5)/2 + 1/2, -1/2 + sqrt(5)/2), (-sqrt(5)/2 + 1/2, 
1/2 + sqrt(5)/2, -sqrt(5)/2 - 1/2)]
0,[(1, 0, 0)]
1,[(1, 0, 1)]
1,[(1, 0, 1)]
1,[(1, 0, 1)]
1,[(1, 0, 1)]
1,[]
1,[]
0,[(1/2 + sqrt(5)/2, -sqrt(5)/2 + 1/2, -1/2 + sqrt(5)/2), (-sqrt(5)/2 + 1/2, 
1/2 + sqrt(5)/2, -sqrt(5)/2 - 1/2)]
0,[(1, 0, 0)]
1,[]
1,[(-1, 1, -2), (2, -1/2, 1)]
0,[(1, 0, 0)]
1,[(1, 0, 1)]
1,[(1, 0, 1)]
1,[]
1,[(-1, 1, -2), (2, -1/2, 1)]
0,[(-1, -1, 1), (1, 1, -1)]
1,[]
0,[(1/2 + sqrt(5)/2, -sqrt(5)/2 + 1/2, -1/2 + sqrt(5)/2), (-sqrt(5)/2 + 1/2, 
1/2 + sqrt(5)/2, -sqrt(5)/2 - 1/2)]
0,[(-1, -1, 1), (1, 1, -1)]
1,[]
0,[(1/2 + sqrt(5)/2, -sqrt(5)/2 + 1/2, -1/2 + sqrt(5)/2), (-sqrt(5)/2 + 1/2, 
1/2 + sqrt(5)/2, -sqrt(5)/2 - 1/2)]
0,[(-1, -1, 1), (1, 1, -1)]
1,[(-1, 1, -2), (2, -1/2, 1)]
0,[(-1, -1, 1), (1, 1, -1)]
1,[(-1, 1, -2), (2, -1/2, 1)]
0,[(1, 0, 0)]
1,[]
1,[(-1, 1, -2), (2, -1/2, 1)]
0,[(-1, -1, 1), (1, 1, -1)]
1,[]
0,[(1/2 + sqrt(5)/2, -sqrt(5)/2 + 1/2, -1/2 + sqrt(5)/2), (-sqrt(5)/2 + 1/2, 
1/2 + sqrt(5)/2, -sqrt(5)/2 - 1/2)]
0,[(1, 0, 0)]
1,[(1, 0, 1)]
1,[]

def consolidate(sets):
# http://rosettacode.org/wiki/Set_consolidation#Python:_Iterative
setlist = [s for s in sets if s]
for i, s1 in enumerate(setlist):
if s1:
for s2 in setlist[i+1:]:
intersection = s1.intersection(s2)
if intersection:
s2.update(s1)
s1.clear()
s1 = s2
return [s for s in setlist if s]

def wrapper(seqs):
consolidated = consolidate(map(set, seqs))
groupmap = {x: i for i,seq in enumerate(consolidated) for x in seq}
output = {}
for seq in seqs:
target = output.setdefault(groupmap[seq[0]], [])
target.append(seq)
return list(output.values())

with open("testing1.txt", "r") as myfile:
content = myfile.readlines()
gr = [['']]
for ii in range(0,500):
try:
gr = [[frozenset(content[ii].split())]] + gr
except:
print "error" + str(content[ii])
groups = wrapper(gr)
for i, group in enumerate(wrapper(gr)):
print('g{}:'.format(i), group)
print("\n")




On Wednesday, October 5, 2016 at 3:40:25 PM UTC+8, dieter wrote:
> meInvent bbird <jobmatt...@gmail.com> writes:
> ... not looking at the details ...
> 
> "'str' object has not attribute 'intersection'": apparently,
> something is really a string (an 'str') while you expect it to be a set.
> 
> "unhashable set": maybe, you try to put a set into another set (or a dict;
> or somewhere else where hashability is necessary). A "set" itself is
> unhashable (like many mutable standard data types); you may consider to use
> "frozenset" in those cases (of course, a "frozenset" is immutable, i.e.
> cannot be changed after creation).

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Re: how to read linux kernel source with pycparser (Reposting On Python-List Prohibited)

[meInvent bbird]

yes, i searched in google between 2011 and 2012

descending for loop is faster than ascending for loop

On Wednesday, October 5, 2016 at 9:48:41 AM UTC+8, Chris Angelico wrote:
> On Wed, Oct 5, 2016 at 12:41 PM, meInvent bbird <jobmatt...@gmail.com> wrote:
> > so far i do not know which bug i search for
> >
> > i would like to change style of for loop
> >
> > form ascending to descending style
> >
> > for(int i=0; i<3; ++i)
> >
> > to
> >
> > for(int i=3; i>=0; --i)
> >
> > or
> >
> > most crazy to change if else if else into mealy machine
> 
> Are you aware that those loops are significantly different?
> 
> ChrisA

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Re: how to read linux kernel source with pycparser (Reposting On Python-List Prohibited)

[meInvent bbird]

On Wednesday, October 5, 2016 at 2:34:49 AM UTC+8, Lawrence D’Oliveiro wrote:
> On Tuesday, October 4, 2016 at 10:38:14 PM UTC+13, meInvent bbird wrote:
> > how to customize pycparser to search what we want such as bug 
> 
> What kind of bug do you want to search for?

so far i do not know which bug i search for

i would like to change style of for loop

form ascending to descending style

for(int i=0; i<3; ++i)

to

for(int i=3; i>=0; --i)

or 

most crazy to change if else if else into mealy machine
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'str' object has no attribute 'intersection' and unhashable set

[meInvent bbird]

def consolidate(sets):
# http://rosettacode.org/wiki/Set_consolidation#Python:_Iterative
setlist = [s for s in sets if s]
for i, s1 in enumerate(setlist):
if s1:
for s2 in setlist[i+1:]:
intersection = s1.intersection(s2)
if intersection:
s2.update(s1)
s1.clear()
s1 = s2
return [s for s in setlist if s]

def wrapper(seqs):
consolidated = consolidate(map(set, seqs))
groupmap = {x: i for i,seq in enumerate(consolidated) for x in seq}
output = {}
for seq in seqs:
target = output.setdefault(groupmap[seq[0]], [])
target.append(seq)
return list(output.values())

with open("testing1.txt", "r") as myfile:
content = myfile.readlines()
gr = [['']]
for ii in range(0,500):
try:
gr = [[content[ii]]] + gr
except:
print "error" + str(content[ii])
#groups = wrapper(content)
for i, group in enumerate(wrapper(gr)):
print('g{}:'.format(i), group)
print("\n")

Traceback (most recent call last):
  File "", line 10, in 
  File "", line 2, in wrapper
  File "", line 7, in consolidate
AttributeError: 'str' object has no attribute 'intersection'

>>> content[1]
'1,[(-1, 1, -2), (2, -1/2, 1)]\n'
>>> content[2]
'0,[(1, 0, 0)]\n'
>>> content[3]
'1,[(1, 0, 1)]\n'
>>> content[4]
'1,[]\n'
>>> content[5]
'1,[]\n'


then i try to edit to
gr = [[set(content[ii])]] + gr

it return unhashable type 'set'

just would like to search all repeatable pattern for a group of lines
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how to read linux kernel source with pycparser

[meInvent bbird]

i expect to use pycparser to read linux kernel source
and get a AST tree, 

but there are so many directory, 

how to read linux kernel source with pycparser?

how to customize pycparser to search what we want such as bug or fix 
to make a linux patch for linux kernel source with python?

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Re: how to automate java application in window using python

[meInvent bbird]

Hi Matt Wheeler,

can it contorl Maplesoft's maple which is a java executable file?


On Sunday, September 18, 2016 at 5:02:15 PM UTC+8, Matt Wheeler wrote:
> On Thu, 15 Sep 2016, 08:12 meInvent bbird, <jobmatt...@gmail.com> wrote:
> 
> > how to automate java application in window using python
> >
> > 1. scroll up or down of scroll bar
> > 2. click button
> > 3. type text in textbox
> >
> 
> I would recommend having a look at pywinauto
> https://github.com/pywinauto/pywinauto
> 
> It presents a very nice pythonic interface to Windows and the controls
> within them, allowing statements such as (copied from the Readme):
> 
> app.UntitledNotepad.MenuSelect("Help->About Notepad")
> app.AboutNotepad.OK.Click()
> app.UntitledNotepad.Edit.TypeKeys ("pywinauto Works!", with_spaces = True)
> 
> (I found it already quite stable when I first used it a couple of years
> ago, and looking at the Readme now it's clearly still an active project,
> the optional image capture feature using pillow is new)
> 
> >

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Re: is it possible to adjust convex hull to draw blue line instead of green line?

[meInvent bbird]

On Saturday, September 17, 2016 at 2:31:59 PM UTC+8, meInvent bbird wrote:
> On Saturday, September 17, 2016 at 9:35:35 AM UTC+8, Steve D'Aprano wrote:
> > On Sat, 17 Sep 2016 10:20 am, meInvent bbird wrote:
> > 
> > > i succeed to use code to draw green line, but green line not draw the
> > > large area, expect second uploaded picture, the blue line connect
> > > the bottom of red line graph
> > 
> > 
> > Please don't waste our time with dead code that has been commented out or
> > that doesn't do anything.
> > 
> > Here is your code with the commented out dead code removed. You should do
> > this, don't expect us to do it:
> > 
> > 
> > im = img.copy()
> > cntcounter = 0
> > for cnt in contours:
> > print("approx=" + str(approx))
> > cntcounter = cntcounter + 1
> > print("here1")
> > hull = cv2.convexHull(cnt,returnPoints = True)
> > print("here2")
> > while im is None:
> > # WARNING: THIS IS AN INFINITE LOOP
> > time.sleep(1)
> > if im is not None:
> > print("here3")
> > previousx = 0
> > previousy = 0 
> > for c in hull: 
> > if (previousx != 0 and previousy != 0 and c[0][0] != 0
> > and c[0][1] != 0 and abs(previousy - c[0][1]) > 10
> > and abs(c[0][0] - previousx) > 1
> > ):
> > while im is None:
> > # WARNING: THIS IS AN INFINITE LOOP
> > time.sleep(1)
> > cv2.line(im, (previousx, previousy),
> >  (c[0][0], c[0][1]), (0, 255, 0), 2)
> > print("")
> > previousx = c[0][0]
> > previousy = c[0][1]
> > 
> > 
> > 
> > 
> > Now it is much easier to read without the noise.
> > 
> > Problems:
> > 
> > (1) img is not defined;
> > 
> > (2) approx is not defined;
> > 
> > (3) contours is not defined;
> > 
> > (4) cv2 is not defined;
> > 
> > (5) you have TWO possible infinite loops in your code;
> > 
> > (6) time is not defined, but this at least I can guess is the 
> > standard time module;
> > 
> > (7) c is not defined.
> > 
> > 
> > As given to us, we cannot run your code or understand it, because too many
> > things are undefined.
> > 
> > My **guess** is that cv2.line() will take an argument to set the line
> > colour. You should read the documentation for cv2.line().
> > 
> > 
> > Before asking any more questions, please read this:
> > 
> > http://sscce.org/
> > 
> > 
> > 
> > 
> > 
> > 
> > -- 
> > Steve
> > “Cheer up,” they said, “things could be worse.” So I cheered up, and sure
> > enough, things got worse.
> 
> sorry for assuming all have experience in this kind of application
> 
> here is my code
> 
> https://gist.github.com/hoyeunglee/0165b24b10de0baf8538e8866e2dcd60
> 
> Please.


this time, i try to run more times for contour and convex hull

but not blue line expected diagram,

is it possible to adjust the method to draw lines for the bottom of red line

https://gist.github.com/hoyeunglee/9c97fdef1cc32fefde0a454a24a2c2c1
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Re: is it possible to adjust convex hull to draw blue line instead of green line?

[meInvent bbird]

On Saturday, September 17, 2016 at 9:35:35 AM UTC+8, Steve D'Aprano wrote:
> On Sat, 17 Sep 2016 10:20 am, meInvent bbird wrote:
> 
> > i succeed to use code to draw green line, but green line not draw the
> > large area, expect second uploaded picture, the blue line connect
> > the bottom of red line graph
> 
> 
> Please don't waste our time with dead code that has been commented out or
> that doesn't do anything.
> 
> Here is your code with the commented out dead code removed. You should do
> this, don't expect us to do it:
> 
> 
> im = img.copy()
> cntcounter = 0
> for cnt in contours:
> print("approx=" + str(approx))
> cntcounter = cntcounter + 1
> print("here1")
> hull = cv2.convexHull(cnt,returnPoints = True)
> print("here2")
> while im is None:
> # WARNING: THIS IS AN INFINITE LOOP
> time.sleep(1)
> if im is not None:
> print("here3")
> previousx = 0
> previousy = 0 
> for c in hull: 
> if (previousx != 0 and previousy != 0 and c[0][0] != 0
> and c[0][1] != 0 and abs(previousy - c[0][1]) > 10
> and abs(c[0][0] - previousx) > 1
> ):
> while im is None:
> # WARNING: THIS IS AN INFINITE LOOP
> time.sleep(1)
> cv2.line(im, (previousx, previousy),
>  (c[0][0], c[0][1]), (0, 255, 0), 2)
> print("")
> previousx = c[0][0]
> previousy = c[0][1]
> 
> 
> 
> 
> Now it is much easier to read without the noise.
> 
> Problems:
> 
> (1) img is not defined;
> 
> (2) approx is not defined;
> 
> (3) contours is not defined;
> 
> (4) cv2 is not defined;
> 
> (5) you have TWO possible infinite loops in your code;
> 
> (6) time is not defined, but this at least I can guess is the 
> standard time module;
> 
> (7) c is not defined.
> 
> 
> As given to us, we cannot run your code or understand it, because too many
> things are undefined.
> 
> My **guess** is that cv2.line() will take an argument to set the line
> colour. You should read the documentation for cv2.line().
> 
> 
> Before asking any more questions, please read this:
> 
> http://sscce.org/
> 
> 
> 
> 
> 
> 
> -- 
> Steve
> “Cheer up,” they said, “things could be worse.” So I cheered up, and sure
> enough, things got worse.

sorry for assuming all have experience in this kind of application

here is my code

https://gist.github.com/hoyeunglee/0165b24b10de0baf8538e8866e2dcd60

Please.
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is it possible to adjust convex hull to draw blue line instead of green line?

[meInvent bbird]

i succeed to use code to draw green line, but green line not draw the 
large area, expect second uploaded picture, the blue line connect
the bottom of red line graph

im = img.copy()
cntcounter = 0
for cnt in contours:
#epsilon = 0.1*cv2.arcLength(cnt,True)
#approx = cv2.approxPolyDP(cnt,epsilon,True)
#peri = cv2.arcLength(cnt, True)
#approx = cv2.approxPolyDP(cnt, 0.15 * peri, True)
#print("len(approx)="+str(len(approx)))
#if len(approx) == 2:
print("approx=" + str(approx))
cntcounter = cntcounter + 1
print("here1")
#x,y,w,h = cv2.boundingRect(cnt)
hull = cv2.convexHull(cnt,returnPoints = True)
print("here2")
while im is None:
time.sleep(1)
if im is not None:
print("here3")
#cv2.rectangle(im, (x,y), (x+w, y+h), (0,255,0), 2) 

#cv2.imwrite("C:\\Users\\tester\\Documents\\masda"+str(cntcounter)+".png",imi)  
  
#im = cv2.line(im,(x,y),(x+w,y),(0,255,0),2)
#im = cv2.line(im,(x+w,y),(x+w,y+h),(0,255,0),2)
#im = cv2.line(im,(x,y+h),(x+w,y+h),(0,255,0),2)
#im = cv2.line(im,(x,y),(x,y+h),(0,255,0),2)
previousx = 0
previousy = 0 
for c in hull: 
if previousx != 0 and previousy != 0 and c[0][0] != 0 and 
c[0][1] != 0 and abs(previousy - c[0][1]) > 10 and abs(c[0][0] - previousx) > 
1:
while im is None:
time.sleep(1)

cv2.line(im,(previousx,previousy),(c[0][0],c[0][1]),(0,255,0),2)
print("")
#print("previousx=" + str(previousx))
#print("previousy=" + str(previousy))
#print("c[0][0]=" + str(c[0][0]))
#print("c[0][1]=" + str(c[0][1]))
previousx = c[0][0]
previousy = c[0][1]



https://drive.google.com/file/d/0Bxs_ao6uuBDUWVBFZzVIVGotRlk/view?usp=sharing

expected is

https://drive.google.com/file/d/0Bxs_ao6uuBDUNGZFS2F3WnJERzA/view?usp=sharing
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Re: why this code loop forever after a draw a rectangle

[meInvent bbird]

after succeed to draw lines in graph,

however,

https://drive.google.com/file/d/0Bxs_ao6uuBDUWVBFZzVIVGotRlk/view?usp=sharing

expected the blue line connect the bottom of red line graph

https://drive.google.com/file/d/0Bxs_ao6uuBDUNGZFS2F3WnJERzA/view?usp=sharing

how can convex hull be adjusted to do like second graph of sky blue line


On Saturday, September 17, 2016 at 7:48:46 AM UTC+8, meInvent bbird wrote:
> img is the image
> 
> im is a new memory of image using  img.copy()
> 
> 
> On Saturday, September 17, 2016 at 7:46:42 AM UTC+8, meInvent bbird wrote:
> > thank you very much,
> > it out of the loop now.
> > because drawLine function return things
> > 
> > i just change drawLine to rectangle,
> > have not thought that rectangle not return thing, just edit the parameter
> > 
> > 
> > 
> > On Saturday, September 17, 2016 at 5:26:53 AM UTC+8, MRAB wrote:
> > > On 2016-09-16 20:14, Gary Herron wrote:
> > > > On 09/16/2016 04:24 AM, meInvent bbird wrote:
> > > >> im = img.copy()
> > > >> cntcounter = 0
> > > >> for cnt in contours:
> > > >>  epsilon = 0.1*cv2.arcLength(cnt,True)
> > > >>  approx = cv2.approxPolyDP(cnt,epsilon,True)   
> > > >>  #peri = cv2.arcLength(cnt, True)
> > > >>  #approx = cv2.approxPolyDP(c, 0.5 * peri, True)
> > > >>  #print("len(approx)="+str(len(approx)))
> > > >>  if len(approx) == 4:
> > > >>  print("approx=" + str(approx))
> > > >>  cntcounter = cntcounter + 1
> > > >>  print("here1")
> > > >>  x,y,w,h = cv2.boundingRect(cnt)
> > > >>  print("here2")
> > > >>  while im is None:
> > > >>  time.sleep(1)
> > > >>  if im is not None:
> > > >>  print("here3")
> > > >>  im = cv2.rectangle(im.copy(), (x,y), (x+w, y+h), 
> > > >> (0,255,0), 2)
> > > >>  #im = cv2.line(im,(x,y),(x+w,y),(0,255,0),2)
> > > >>  #im = cv2.line(im,(x+w,y),(x+w,y+h),(0,255,0),2)
> > > >>  #im = cv2.line(im,(x,y+h),(x+w,y+h),(0,255,0),2)
> > > >>  #im = cv2.line(im,(x,y),(x,y+h),(0,255,0),2)
> > > >>
> > > >>
> > > >> cv2.imwrite(r'C:\Users\tester\Documents\masda.png',im)
> > > >
> > > >
> > > > These two lines:
> > > >
> > > >while im is None:
> > > >  time.sleep(1)
> > > >
> > > > are an infinite loop if im is None;
> > > >
> > > >
> > > > Since you haven't told us what im (or img, contours, cv2) are, I can't
> > > > tell how im might become None, but it does look like you (confusingly)
> > > > use im for two different things:  an img.copy() and a cv2.rectangle,
> > > > whatever those may be.
> > > >
> > > > Pure guesswork:  if cv2.rectangle draws a rectangle, what does it
> > > > return?  If it doesn't return anything, the line
> > > >  im = cv2.rectangle(...)
> > > > is how im gets the value of None.
> > > >
> > > It looks like the OP is using OpenCV.
> > > 
> > > You're right about cv2.rectangle; it does return None.
> > > 
> > > The line:
> > > 
> > >  im = cv2.rectangle(im.copy(), (x,y), (x+w, y+h), (0,255,0), 2)
> > > 
> > > makes a copy of the image im, draws a rectangle on it, and then binds 
> > > None to im.
> > > 
> > > The copied rectangle is discarded because there's no reference to it, so 
> > > the entire line in pointless.
> > > 
> > > It basically does the same thing as:
> > > 
> > >  im = None
> > > 
> > > only slower!

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Re: why this code loop forever after a draw a rectangle

[meInvent bbird]

img is the image

im is a new memory of image using  img.copy()


On Saturday, September 17, 2016 at 7:46:42 AM UTC+8, meInvent bbird wrote:
> thank you very much,
> it out of the loop now.
> because drawLine function return things
> 
> i just change drawLine to rectangle,
> have not thought that rectangle not return thing, just edit the parameter
> 
> 
> 
> On Saturday, September 17, 2016 at 5:26:53 AM UTC+8, MRAB wrote:
> > On 2016-09-16 20:14, Gary Herron wrote:
> > > On 09/16/2016 04:24 AM, meInvent bbird wrote:
> > >> im = img.copy()
> > >> cntcounter = 0
> > >> for cnt in contours:
> > >>  epsilon = 0.1*cv2.arcLength(cnt,True)
> > >>  approx = cv2.approxPolyDP(cnt,epsilon,True) 
> > >>  #peri = cv2.arcLength(cnt, True)
> > >>  #approx = cv2.approxPolyDP(c, 0.5 * peri, True)
> > >>  #print("len(approx)="+str(len(approx)))
> > >>  if len(approx) == 4:
> > >>  print("approx=" + str(approx))
> > >>  cntcounter = cntcounter + 1
> > >>  print("here1")
> > >>  x,y,w,h = cv2.boundingRect(cnt)
> > >>  print("here2")
> > >>  while im is None:
> > >>  time.sleep(1)
> > >>  if im is not None:
> > >>  print("here3")
> > >>  im = cv2.rectangle(im.copy(), (x,y), (x+w, y+h), 
> > >> (0,255,0), 2)
> > >>  #im = cv2.line(im,(x,y),(x+w,y),(0,255,0),2)
> > >>  #im = cv2.line(im,(x+w,y),(x+w,y+h),(0,255,0),2)
> > >>  #im = cv2.line(im,(x,y+h),(x+w,y+h),(0,255,0),2)
> > >>  #im = cv2.line(im,(x,y),(x,y+h),(0,255,0),2)
> > >>
> > >>
> > >> cv2.imwrite(r'C:\Users\tester\Documents\masda.png',im)
> > >
> > >
> > > These two lines:
> > >
> > >while im is None:
> > >  time.sleep(1)
> > >
> > > are an infinite loop if im is None;
> > >
> > >
> > > Since you haven't told us what im (or img, contours, cv2) are, I can't
> > > tell how im might become None, but it does look like you (confusingly)
> > > use im for two different things:  an img.copy() and a cv2.rectangle,
> > > whatever those may be.
> > >
> > > Pure guesswork:  if cv2.rectangle draws a rectangle, what does it
> > > return?  If it doesn't return anything, the line
> > >  im = cv2.rectangle(...)
> > > is how im gets the value of None.
> > >
> > It looks like the OP is using OpenCV.
> > 
> > You're right about cv2.rectangle; it does return None.
> > 
> > The line:
> > 
> >  im = cv2.rectangle(im.copy(), (x,y), (x+w, y+h), (0,255,0), 2)
> > 
> > makes a copy of the image im, draws a rectangle on it, and then binds 
> > None to im.
> > 
> > The copied rectangle is discarded because there's no reference to it, so 
> > the entire line in pointless.
> > 
> > It basically does the same thing as:
> > 
> >  im = None
> > 
> > only slower!

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Re: why this code loop forever after a draw a rectangle

[meInvent bbird]

thank you very much,
it out of the loop now.
because drawLine function return things

i just change drawLine to rectangle,
have not thought that rectangle not return thing, just edit the parameter



On Saturday, September 17, 2016 at 5:26:53 AM UTC+8, MRAB wrote:
> On 2016-09-16 20:14, Gary Herron wrote:
> > On 09/16/2016 04:24 AM, meInvent bbird wrote:
> >> im = img.copy()
> >> cntcounter = 0
> >> for cnt in contours:
> >>  epsilon = 0.1*cv2.arcLength(cnt,True)
> >>  approx = cv2.approxPolyDP(cnt,epsilon,True)   
> >>  #peri = cv2.arcLength(cnt, True)
> >>  #approx = cv2.approxPolyDP(c, 0.5 * peri, True)
> >>  #print("len(approx)="+str(len(approx)))
> >>  if len(approx) == 4:
> >>  print("approx=" + str(approx))
> >>  cntcounter = cntcounter + 1
> >>  print("here1")
> >>  x,y,w,h = cv2.boundingRect(cnt)
> >>  print("here2")
> >>  while im is None:
> >>  time.sleep(1)
> >>  if im is not None:
> >>  print("here3")
> >>  im = cv2.rectangle(im.copy(), (x,y), (x+w, y+h), 
> >> (0,255,0), 2)
> >>  #im = cv2.line(im,(x,y),(x+w,y),(0,255,0),2)
> >>  #im = cv2.line(im,(x+w,y),(x+w,y+h),(0,255,0),2)
> >>  #im = cv2.line(im,(x,y+h),(x+w,y+h),(0,255,0),2)
> >>  #im = cv2.line(im,(x,y),(x,y+h),(0,255,0),2)
> >>
> >>
> >> cv2.imwrite(r'C:\Users\tester\Documents\masda.png',im)
> >
> >
> > These two lines:
> >
> >while im is None:
> >  time.sleep(1)
> >
> > are an infinite loop if im is None;
> >
> >
> > Since you haven't told us what im (or img, contours, cv2) are, I can't
> > tell how im might become None, but it does look like you (confusingly)
> > use im for two different things:  an img.copy() and a cv2.rectangle,
> > whatever those may be.
> >
> > Pure guesswork:  if cv2.rectangle draws a rectangle, what does it
> > return?  If it doesn't return anything, the line
> >  im = cv2.rectangle(...)
> > is how im gets the value of None.
> >
> It looks like the OP is using OpenCV.
> 
> You're right about cv2.rectangle; it does return None.
> 
> The line:
> 
>  im = cv2.rectangle(im.copy(), (x,y), (x+w, y+h), (0,255,0), 2)
> 
> makes a copy of the image im, draws a rectangle on it, and then binds 
> None to im.
> 
> The copied rectangle is discarded because there's no reference to it, so 
> the entire line in pointless.
> 
> It basically does the same thing as:
> 
>  im = None
> 
> only slower!

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Re: how to automate java application in window using python

[meInvent bbird]

you are right, english is not my first language

i just talk as simple as i can, i do not know previous talking is mean


On Friday, September 16, 2016 at 6:22:34 PM UTC+8, Christian Gollwitzer wrote:
> Am 16.09.16 um 09:01 schrieb Lawrence D’Oliveiro:
> > On Friday, September 16, 2016 at 6:55:07 PM UTC+12, meInvent bbird
> > wrote:
> >> On Thursday, September 15, 2016 at 3:52:41 PM UTC+8, Lawrence
> >> D’Oliveiro wrote:
> >>> On Thursday, September 15, 2016 at 7:13:05 PM UTC+12, meInvent
> >>> bbird wrote:
> >>>> how to automate java application in window using python
> >>>>
> >>>> 1. scroll up or down of scroll bar 2. click button 3. type text
> >>>> in textbox
> >>>
> >>> Well, don’t leave us in suspense! Give us the link to your blog
> >>> post!
> >>
> >> i do not have blog post,
> >
> > Oh, now you make us sad. I thought you were promoting a blog post
> > where you tell us “how to automate java application in window using
> > python” using just 3 steps! But it turns out you have nothing. You
> > are an empty promiser.
> 
> You are being mean. It is quite evident that English is not his first 
> language, and highly probable that this was actually the question "How 
> do I automate a Java application using Python?"
> 
> I don't have a real answer myself, but maybe using Jython one can get 
> access to the Java objects underlying the application (and then, of 
> course, execute methods of the GUI objects). This depends on the GUI 
> toolkit (Swing/SWT) and knowledge of the applications' structure, of 
> course. Another (less robust) way of desktop automation is Sikuli 
> http://www.sikuli.org/ (Windows only, I think)
> 
>   Christian

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Re: why this code loop forever after a draw a rectangle

[meInvent bbird]

i follow this post to give some time it to operate,
wait a long time still looping

http://answers.opencv.org/question/60094/libpng-warning-image-width-is-zero-in-ihdr/


i can not stand this Ninja coding life any more,
i have to open my code for ask this error


import cv2
import numpy as np
#from matplotlib import pyplot as plt
import time

#print("1=" + str(int(sys.argv[1])))
#print("2=" + str(int(sys.argv[2])))
#print("3=" + str(int(sys.argv[3])))

img_rgb = cv2.imread(r'C:\Users\martin\Documents\scree2.png')
img_gray = cv2.cvtColor(img_rgb, cv2.COLOR_BGR2GRAY)
template = cv2.imread(r'C:\Users\martin\Documents\dragob.png',0)
w, h = template.shape[::-1]

res = cv2.matchTemplate(img_gray,template,cv2.TM_CCOEFF_NORMED)
threshold = 0.64

loc = np.where( res >= threshold)
pt = [(0,0)]

while not zip(*loc[::-1]):
threshold = threshold - 0.02
loc = np.where( res >= threshold)

counter = 1
print("threshold="+str(threshold))   
for pt2 in zip(*loc[::-1]):
cv2.rectangle(img_rgb, pt2, (pt2[0] + w, pt2[1] + h), (0,0,255), 2)
pt = pt2
crop_img = img_rgb[pt[1]:(pt[1]+h), pt[0]:(pt[0]+w)]
counter = counter + 1

cv2.imwrite("C:\\Users\\tester\\Documents\\res.png",crop_img)


#import cv2
#winName = "Movement Indicator"
#cv2.namedWindow(winName, cv2.WINDOW_NORMAL)
img = cv2.imread(r'C:\Users\tester\Documents\res.png',1)
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
height, width = gray.shape
edges = cv2.Canny(gray,height,width,apertureSize = 3)
#edges = cv2.Canny(gray,30,200)

#gray = cv2.cvtColor(img, cv2.COLOR_RGB2GRAY)
#ret,thresh = 
cv2.threshold(edges.copy(),cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE,0)
ret,thresh = cv2.threshold(edges,250,150,0)
contours,hierarchy = cv2.findContours(thresh, 1, 2)
#contours = sorted(contours, key = cv2.contourArea, reverse = True)[:10]

im = img.copy()
cntcounter = 0
for cnt in contours:
epsilon = 0.1*cv2.arcLength(cnt,True)
approx = cv2.approxPolyDP(cnt,epsilon,True) 
#peri = cv2.arcLength(cnt, True)
#approx = cv2.approxPolyDP(c, 0.5 * peri, True)
#print("len(approx)="+str(len(approx)))
if len(approx) == 4:
print("approx=" + str(approx))
cntcounter = cntcounter + 1
print("here1")
x,y,w,h = cv2.boundingRect(cnt)
print("here2")
#im = img.copy()
while im is None:
time.sleep(1)
if im is not None:
print("here3")
im = cv2.rectangle(im.copy(), (x,y), (x+w, y+h), (0,255,0), 2) 

#cv2.imwrite("C:\\Users\\martin\\Documents\\masda"+str(cntcounter)+".png",imi)  
  
#im = cv2.line(im,(x,y),(x+w,y),(0,255,0),2)
#im = cv2.line(im,(x+w,y),(x+w,y+h),(0,255,0),2)
#im = cv2.line(im,(x,y+h),(x+w,y+h),(0,255,0),2)
#im = cv2.line(im,(x,y),(x,y+h),(0,255,0),2)








On Friday, September 16, 2016 at 7:34:04 PM UTC+8, Waffle wrote:
> On 16 September 2016 at 14:24, meInvent bbird <jobmatt...@gmail.com> wrote:
> > im = img.copy()
> > cntcounter = 0
> > for cnt in contours:
> > epsilon = 0.1*cv2.arcLength(cnt,True)
> > approx = cv2.approxPolyDP(cnt,epsilon,True)
> > #peri = cv2.arcLength(cnt, True)
> > #approx = cv2.approxPolyDP(c, 0.5 * peri, True)
> > #print("len(approx)="+str(len(approx)))
> > if len(approx) == 4:
> > print("approx=" + str(approx))
> > cntcounter = cntcounter + 1
> > print("here1")
> > x,y,w,h = cv2.boundingRect(cnt)
> > print("here2")
> > while im is None:
> > time.sleep(1)
> > if im is not None:
> > print("here3")
> > im = cv2.rectangle(im.copy(), (x,y), (x+w, y+h), (0,255,0), 
> > 2)
> > #im = cv2.line(im,(x,y),(x+w,y),(0,255,0),2)
> > #im = cv2.line(im,(x+w,y),(x+w,y+h),(0,255,0),2)
> > #im = cv2.line(im,(x,y+h),(x+w,y+h),(0,255,0),2)
> > #im = cv2.line(im,(x,y),(x,y+h),(0,255,0),2)
> >
> >
> > cv2.imwrite(r'C:\Users\tester\Documents\masda.png',im)
> > --
> > https://mail.python.org/mailman/listinfo/python-list
> 
> not sure but..  this bit reads really suspicious:
> 
> while im is None:
> time.sleep(1)
> 
> if im is ever None then how will it ever become not None? unless there
> is some other thread at work i can't really see this happening.
> and if there is some other thread at work then there is probably some
> better solution than sleep()





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why this code loop forever after a draw a rectangle

[meInvent bbird]

im = img.copy()
cntcounter = 0
for cnt in contours:
epsilon = 0.1*cv2.arcLength(cnt,True)
approx = cv2.approxPolyDP(cnt,epsilon,True) 
#peri = cv2.arcLength(cnt, True)
#approx = cv2.approxPolyDP(c, 0.5 * peri, True)
#print("len(approx)="+str(len(approx)))
if len(approx) == 4:
print("approx=" + str(approx))
cntcounter = cntcounter + 1
print("here1")
x,y,w,h = cv2.boundingRect(cnt)
print("here2")
while im is None:
time.sleep(1)
if im is not None:
print("here3")
im = cv2.rectangle(im.copy(), (x,y), (x+w, y+h), (0,255,0), 2)  
   
#im = cv2.line(im,(x,y),(x+w,y),(0,255,0),2)
#im = cv2.line(im,(x+w,y),(x+w,y+h),(0,255,0),2)
#im = cv2.line(im,(x,y+h),(x+w,y+h),(0,255,0),2)
#im = cv2.line(im,(x,y),(x,y+h),(0,255,0),2)


cv2.imwrite(r'C:\Users\tester\Documents\masda.png',im)
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Re: how to automate java application in window using python

[meInvent bbird]

On Friday, September 16, 2016 at 2:26:47 AM UTC+8, bream...@gmail.com wrote:
> On Thursday, September 15, 2016 at 8:13:05 AM UTC+1, meInvent bbird wrote:
> > how to automate java application in window using python
> > 
> > 1. scroll up or down of scroll bar
> > 2. click button
> > 3. type text in textbox
> 
> Please read this http://www.catb.org/esr/faqs/smart-questions.html and then 
> this http://sscce.org/ and then rephrase your post.
> 
> Kindest regards.
> 
> Mark Lawrence.

i find this, but it is not python, it is a product

http://sqa.stackexchange.com/questions/18134/python-based-test-automation-tools-for-desktop-applications
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Re: how to automate java application in window using python

[meInvent bbird]

On Thursday, September 15, 2016 at 3:52:41 PM UTC+8, Lawrence D’Oliveiro wrote:
> On Thursday, September 15, 2016 at 7:13:05 PM UTC+12, meInvent bbird wrote:
> > how to automate java application in window using python
> > 
> > 1. scroll up or down of scroll bar
> > 2. click button
> > 3. type text in textbox
> 
> Well, don’t leave us in suspense! Give us the link to your blog post!

i do not have blog post, 

search nothing about this in google, 

is it possible to automate java application with python?
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how to robot recognize basic geometric object in window or ubuntu window

[meInvent bbird]

i am doing a robot to automate window itself with python 

i give some basic png diagram such as square, circle, triangle
and hope it recognize all kinds of square like things in window 
such as textbox of notepad etc

how a robot recognize basic geometric object in window or ubuntu window

i use cv2 template matching method 
and then draw black square or rectangle as a template

however, it can recognize the size of template only

and recognize not a square diagram from capture screen

then i try a black ground and white line of square, it recognize a File in menu

i feel that it need to try all kinds of color, and size in order to recognize
similar object in screen capture diagram

how do a robot recognize many color and size in seconds?
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how to automate java application in window using python

[meInvent bbird]

how to automate java application in window using python

1. scroll up or down of scroll bar
2. click button
3. type text in textbox
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unexpected rerun program itself when using cv2 and selenium

[meInvent bbird]

unexpected rerun program itself when using cv2 and selenium

when a script, it is expected to run once and stop
but it start a new browser again when using selenium and cv2


d, bbox, filename)
  File "C:\Python27\lib\site-packages\pyscreenshot\procutil.py", line 28, in run
_in_childprocess
p.start()
  File "C:\Python27\lib\multiprocessing\process.py", line 130, in start
self._popen = Popen(self)
  File "C:\Python27\lib\multiprocessing\forking.py", line 258, in __init__
cmd = get_command_line() + [rhandle]
  File "C:\Python27\lib\multiprocessing\forking.py", line 358, in get_command_li
ne
is not going to be frozen to produce a Windows executable.''')
RuntimeError:
Attempt to start a new process before the current process
has finished its bootstrapping phase.

This probably means that you are on Windows and you have
forgotten to use the proper idiom in the main module:

if __name__ == '__main__':
freeze_support()
...

The "freeze_support()" line can be omitted if the program
is not going to be frozen to produce a Windows executable.
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how to set paramiko to use ssh version 1?

[meInvent bbird]

how to set paramiko to use ssh version 1?
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Re: type lookuperror

[meInvent bbird]

On Friday, August 19, 2016 at 5:37:32 AM UTC+8, Lawrence D’Oliveiro wrote:
> On Friday, August 19, 2016 at 3:10:26 AM UTC+12, Chris Angelico wrote:
> > What the OP was looking for was "I want my program to be able to debug
> > itself". That means the program has to be smart enough to figure out
> > its own problems.
> 
> Maybe it is, it just doesn’t agree with you on what those problems are. ;)


a company which write siri in iphone, has already wrote a program
which can write program itself after the program talks with users

it seems possible,

i find many default function when using dir(function) to see
so i guess that these default functions in openstack have its existence reason 
which i guess for program itself.

how to append a function or code in runtime in python?
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Re: type lookuperror

[meInvent bbird]

actually i would like to remove try except code in all function

and i feel that try except code for a large block code can not 
show which function name , which line number error,
if use try except for specified code block to show where it has error
it will have many ugly try except code and need many human force
to craft the code, i hope that code write once for program to program itself 
in future.

actually this method i ask is not good enough since it will need 
to add many function wrapper.

On Thursday, August 18, 2016 at 3:49:04 PM UTC+8, Chris Angelico wrote:
> On Thu, Aug 18, 2016 at 5:44 PM, meInvent bbird <jobmatt...@gmail.com> wrote:
> > if i use
> >
> > result = ""
> > before try
> > and return result at the end
> >
> > as return of function can be any type
> >
> > there will be type mismatch
> >
> > how to return the result of func ?
> 
> If it raises an exception, it *does not have* a return value. That's
> why I let the exception continue to bubble. But fundamentally, you
> can't just carry on regardless. I recommend a complete rethink of what
> you're trying to do here.
> 
> ChrisA

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Re: type lookuperror

[meInvent bbird]

if i use 

result = ""
before try
and return result at the end

as return of function can be any type

there will be type mismatch 

how to return the result of func ?

On Thursday, August 18, 2016 at 3:18:31 PM UTC+8, Chris Angelico wrote:
> On Thu, Aug 18, 2016 at 5:14 PM, Peter Otten <__pete...@web.de> wrote:
> > meInvent bbird wrote:
> >
> >> when try keystone_client.tenants.get
> >> got error,
> >>
> >> isn't this method for all kinds of function?
> >>
> >>>>> m = "4c9a0da00b904422a23341e35be7f8d7"
> >>>>> ten = checkexception(keystone_client.tenants.get,
> >>>>> tenant_id=checkexception(m.encode,encoding='ascii',errors='ignore'))
> >> Unexpected error: 
> >> None
> >
> > That's because your and Chris' version of checkexception() have no explicit
> > return statement and thus always return None. There doesn't seeem to be a
> > tenant_id=None...
> 
> Hmm. Yes, that's another bug I should have fixed as I was rewriting.
> "return result" at the end.
> 
> ChrisA

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Re: type lookuperror

[meInvent bbird]

when try keystone_client.tenants.get
got error,

isn't this method for all kinds of function?

>>> m = "4c9a0da00b904422a23341e35be7f8d7"
>>> ten = checkexception(keystone_client.tenants.get, 
>>> tenant_id=checkexception(m.encode,encoding='ascii',errors='ignore'))
Unexpected error: 
None




On Thursday, August 18, 2016 at 10:22:43 AM UTC+8, Chris Angelico wrote:
> On Thu, Aug 18, 2016 at 12:13 PM, meInvent bbird <jobmatt...@gmail.com> wrote:
> > would like to check errors for every function i run,
> > got error type lookuperror
> >
> > def checkexception(func, **kwargs):
> > try:
> > result = func(*tuple(value for _, value in kwargs.iteritems()))
> > except:
> > print "Unexpected error:", sys.exc_info()[0]
> > try:
> > print(func.__doc__)
> > except:
> > print("no doc error")
> >
> 
> I'm going to be brutally honest, and simply say that this is terrible
> code. I'm not even going to _try_ to fix it. Instead, here's a
> completely rewritten form:
> 
> def checkexception(func, *args, **kwargs):
> try:
> result = func(*args, **kwargs)
> except BaseException as e:
> print("Exception raised: %s" % e)
> try: print(func.__doc__)
> except AttributeError: pass
> raise # Let the exception keep happening.
> 
> But really, there are even better ways to do this. Just let the
> exception happen, and then use something like ipython to help you
> analyze the traceback.
> 
> ChrisA

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type lookuperror

[meInvent bbird]

would like to check errors for every function i run, 
got error type lookuperror

def checkexception(func, **kwargs):
try:
result = func(*tuple(value for _, value in kwargs.iteritems()))
except:
print "Unexpected error:", sys.exc_info()[0]
try: 
print(func.__doc__)
except:
print("no doc error")
   

>>> mm =checkexception("".encode,encoding='ascii',errors='ignore')
Unexpected error: 
S.encode([encoding[,errors]]) -> object

Encodes S using the codec registered for encoding. encoding defaults
to the default encoding. errors may be given to set a different error
handling scheme. Default is 'strict' meaning that encoding errors raise
a UnicodeEncodeError. Other possible values are 'ignore', 'replace' and
'xmlcharrefreplace' as well as any other name registered with
codecs.register_error that is able to handle UnicodeEncodeErrors.


for p in val_list:
   for k, v in p.items():
   if k == "tenant_id":
   print(v)
   m = v
   ten = checkexception(keystone_client.tenants.get, 
checkexception(m.encode,encoding='ascii',errors'ignore'))
   print(ten.name)
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how to python to use virtual memory?

[meInvent bbird]

after set resource hard and soft both out of real memory 70GB to 100GB

and expecting the extra memory setting using virtual memory which is hard disk,

the program still wrongly be killed itself after running in background side

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program return killed after run a very long time

[meInvent bbird]

i suspect it use so much memory,

however i have already assign 30GB memory in openstack cloud

is there any command to allow python to use more memory?

currently

program return killed after run a very long time
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Re: is there a concurrent list for append in parallel programming in python? how to pass parameter in this parallel program with pool?

[meInvent bbird]

how can list be synchronized when multiprocessor working in it?

will one thread updating non-updated version, but another processor updating 
the version?

On Thursday, June 16, 2016 at 4:30:33 PM UTC+8, Steven D'Aprano wrote:
> On Thursday 16 June 2016 17:28, meInvent bbird wrote:
> 
> > is there like c# have concurrent list ?
> 
> What is a concurrent list?
> 
> Can you link to the C# documentation for this?
> 
> To me, "concurrent" describes a style of execution flow, and "list" describes 
> a 
> data structure. I am struggling to understand what "concurrent list" means.
> 
> > i find something these, but how can it pass an initlist list variable
> 
> initlist = ['a', 'b', 'c']
> result = comb(n, initlist)  # pass initlist
> 
> 
> > is it doing the same function as itertools.combinations ?
> 
> It is calling itertools.combinations. So, yes, it is doing the same function 
> as 
> itertools.combinations.
> 
> 
> > def comb(n, initlist): # the argument n is the number of items to select
> > res = list(itertools.combinations(initlist, n)) # create a list from the
> > # iterator 
> > return res
> 
> This does not generate the combinations in parallel. It generates them one at 
> a 
> time, and then creates a list of them.
> 
> This is an interesting question. Somebody could probably write a parallel 
> version of combinations. But I don't know that it would be very useful -- the 
> limiting factor on combinations is more likely to be memory, not time.
> 
> Suppose you generate combinations of 100 items, taken 10 at a time. If I 
> remember the formula for combinations correctly, the total number of 
> combinations is:
> 
> 100!/(10! * 90!) = 17310309456440
> 
> combinations in total. If each generated combination took just *one* byte of 
> memory, that would require over 17 TB of RAM.
> 
> 
> 
> -- 
> Steve

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Re: is there a concurrent list for append in parallel programming in python? how to pass parameter in this parallel program with pool?

[meInvent bbird]

the name in c# is not called concurrent list, it is called
blockingcollection

dictionary called concurrent dictionary

thread safe these kind of things

https://msdn.microsoft.com/en-us/library/dd267312(v=vs.110).aspx

https://msdn.microsoft.com/en-us/library/dd997369(v=vs.110).aspx

https://msdn.microsoft.com/en-us/library/dd997305(v=vs.110).aspx


On Thursday, June 16, 2016 at 4:30:33 PM UTC+8, Steven D'Aprano wrote:
> On Thursday 16 June 2016 17:28, meInvent bbird wrote:
> 
> > is there like c# have concurrent list ?
> 
> What is a concurrent list?
> 
> Can you link to the C# documentation for this?
> 
> To me, "concurrent" describes a style of execution flow, and "list" describes 
> a 
> data structure. I am struggling to understand what "concurrent list" means.
> 
> > i find something these, but how can it pass an initlist list variable
> 
> initlist = ['a', 'b', 'c']
> result = comb(n, initlist)  # pass initlist
> 
> 
> > is it doing the same function as itertools.combinations ?
> 
> It is calling itertools.combinations. So, yes, it is doing the same function 
> as 
> itertools.combinations.
> 
> 
> > def comb(n, initlist): # the argument n is the number of items to select
> > res = list(itertools.combinations(initlist, n)) # create a list from the
> > # iterator 
> > return res
> 
> This does not generate the combinations in parallel. It generates them one at 
> a 
> time, and then creates a list of them.
> 
> This is an interesting question. Somebody could probably write a parallel 
> version of combinations. But I don't know that it would be very useful -- the 
> limiting factor on combinations is more likely to be memory, not time.
> 
> Suppose you generate combinations of 100 items, taken 10 at a time. If I 
> remember the formula for combinations correctly, the total number of 
> combinations is:
> 
> 100!/(10! * 90!) = 17310309456440
> 
> combinations in total. If each generated combination took just *one* byte of 
> memory, that would require over 17 TB of RAM.
> 
> 
> 
> -- 
> Steve

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is there a concurrent list for append in parallel programming in python? how to pass parameter in this parallel program with pool?

[meInvent bbird]

is there like c# have concurrent list ? 

i find something these, but how can it pass an initlist list variable 

is it doing the same function as itertools.combinations ? 

def comb(n, initlist): # the argument n is the number of items to select 
res = list(itertools.combinations(initlist, n)) # create a list from the 
iterator 
return res 

#p = Pool(8) 
#times = range(0, len(initlist)+1)  
   
#values = p.map(comb, times) # pass the range as the sequence of arguments! 
#p.close() 
#p.join() 
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Re: how to for loop append a list [] when using parallel programming

[meInvent bbird]

is there like c# have concurrent list ?

i find something these, but how can it pass an initlist list variable

is it doing the same function as itertools.combinations ?

def comb(n, initlist): # the argument n is the number of items to select
res = list(itertools.combinations(initlist, n)) # create a list from the 
iterator
return res

#p = Pool(8)
#times = range(0, len(initlist)+1)  
  
#values = p.map(comb, times) # pass the range as the sequence of arguments!
#p.close()
#p.join()



On Thursday, June 16, 2016 at 11:17:47 AM UTC+8, John Gordon wrote:
> In <fabcecae-04ae-433c-aad8-43993a21e...@googlegroups.com> meInvent bbird 
> <jobmatt...@gmail.com> writes:
> 
> > how to for loop append a list [] when using parallel programming
> 
> items = []
> for item in parallelized_object_factory():
> items.append(item)
> 
> If you want a more specific answer, ask a more specific question.
> 
> -- 
> John Gordon   A is for Amy, who fell down the stairs
> gor...@panix.com  B is for Basil, assaulted by bears
> -- Edward Gorey, "The Gashlycrumb Tinies"

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how to for loop append a list [] when using parallel programming

[meInvent bbird]

how to for loop append a list [] when using parallel programming
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how to change itertools.combinations(initlist, 2) to parallel for running in multiprocessor environment?

[meInvent bbird]

how to change itertools.combinations(initlist, 2) to parallel for running in 
multiprocessor environment?
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how to search item in list of list

[meInvent bbird]

once a nested list have a word "node" then true else false

def search(current_item):
if isinstance(current_item, list):
if len(current_item)==4: 
if [item for item in current_item if item[4] == "node"] != []:
return True
if True in [search(item) for item in current_item]:
return True
else:
return False

search(mresult)

but it return false


mresult = [[(2, {'11': 1, '10': 1, '00': 0, '01': 1}, ['000', '001', '010', 
'011', '100',
'101', '110', '111'], 'xy', 'start')], [(2, {'11': 1, '10': 1, '00': 0, '01': 1}
, ['000', '001', '010', '011', '100', '101', '110', '111'], 'yz', 'start')], [(2
, {'11': 1, '10': 1, '00': 0, '01': 1}, ['000', '001', '010', '011', '100', '101
', '110', '111'], 'xz', 'start')], [(2, {'11': 1, '10': 0, '00': 0, '01': 0}, ['
000', '001', '010', '011', '100', '101', '110', '111'], 'xy', 'start')], [(2, {'
11': 1, '10': 0, '00': 0, '01': 0}, ['000', '001', '010', '011', '100', '101', '
110', '111'], 'yz', 'start')], [(2, {'11': 1, '10': 0, '00': 0, '01': 0}, ['000'
, '001', '010', '011', '100', '101', '110', '111'], 'xz', 'start')], [(2, {'11':
 1, '10': 0, '00': 1, '01': 1}, ['000', '001', '010', '011', '100', '101', '110'
, '111'], 'xy', 'start')], [(2, {'11': 1, '10': 0, '00': 1, '01': 1}, ['000', '0
01', '010', '011', '100', '101', '110', '111'], 'yz', 'start')], [(2, {'11': 1,
'10': 0, '00': 1, '01': 1}, ['000', '001', '010', '011', '100', '101', '110', '1
11'], 'xz', 'start')], [(1, {'11': 1, '10': 1, '00': 0, '01': 1}, ['00', '01', '
11', '11', '10', '11', '11', '11'], 'xy', 'node')], [(1, {'11': 1, '10': 1, '00'
: 0, '01': 1}, ['00', '01', '10', '11', '11', '11', '11', '11'], 'xy', 'node')],
 [(1, {'11': 1, '10': 1, '00': 0, '01': 1}, ['00', '00', '10', '10', '10', '10',
 '11', '11'], 'xy', 'node')], [(1, {'11': 1, '10': 1, '00': 0, '01': 1}, ['00',
'00', '10', '11', '10', '10', '10', '11'], 'xy', 'node')], [(1, {'11': 1, '10':
1, '00': 0, '01': 1}, ['00', '00', '10', '10', '10', '11', '10', '11'], 'xy', 'n
ode')]]
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yatel install error

[meInvent bbird]

sudo apt-get install python-dev libatlas-base-dev gfortran
pip install yatel


Cleaning up...
Command /usr/bin/python -c "import setuptools, 
tokenize;__file__='/tmp/pip_build_martin/yatel/setup.py';exec(compile(getattr(tokenize,
 'open', open)(__file__).read().replace('\r\n', '\n'), __file__, 'exec'))" 
install --record /tmp/pip-HwGsHy-record/install-record.txt 
--single-version-externally-managed --compile failed with error code 1 in 
/tmp/pip_build_martin/yatel
Traceback (most recent call last):
  File "/usr/bin/pip", line 9, in 
load_entry_point('pip==1.5.4', 'console_scripts', 'pip')()
  File "/usr/lib/python2.7/dist-packages/pip/__init__.py", line 235, in main
return command.main(cmd_args)
  File "/usr/lib/python2.7/dist-packages/pip/basecommand.py", line 161, in main
text = '\n'.join(complete_log)
UnicodeDecodeError: 'ascii' codec can't decode byte 0xe2 in position 72: 
ordinal not in range(128)
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how to use yatel with redis?

[meInvent bbird]

my code is

https://gist.github.com/hoyeunglee/58df4c41a63a2f37e153cbdbc03c16bf


would like to apply itertools.combinations to use redis to use hard disk as
memory rather than using ram(real memory) as memory

def edge_gen(self):
# we combine  haplotypes by two
for hap0, hap1 in itertools.combinations(self.haplotypes_cache.values(), 2):
w = weight.weight("hamming", hap0, hap1)
haps_id = hap0.hap_id, hap1.hap_id
yield dom.Edge(w, haps_id)
-- 
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Re: how to record path in well format and easy to read?

[meInvent bbird]

i forget why i do combination of 2 operators
now i change not to use combination of 2 operators

https://gist.github.com/hoyeunglee/58df4c41a63a2f37e153cbdbc03c16bf


i run 
mresult = DFS(b, 2, 2, mylist, path) 

it do not have deep 2 and deep 1 in the path

they only separated in the list

however, i had already assigned in this format path.append([caseA,path2])
before return path, 

what's wrong with it?


On Sunday, June 12, 2016 at 12:55:17 PM UTC+8, meInvent bbird wrote:
> https://gist.github.com/hoyeunglee/350ac2dd496f7c1c95a428f847e6f2b1
> 
> then can not run deep 3 in python sagecloud
> 
> >>> mresult = DFS(b, 3, 3, mylist, path)
> ('deep=', 3)
> ({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 0, '01': 0})
> ({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 1, '01': 1})
> ({'11': 1, '10': 0, '00': 0, '01': 0}, {'11': 1, '10': 0, '00': 1, '01': 1})
> initlist=
> 18
> path=
> 18
> here is caseA
> ('deep=', 2)
> ({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 0, '01': 0})
> )
> ... retu({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 
> 1, '01': 1})
> ({'11': 1, '10': 0, '00': 0, '01': 0}, {'11': 1, '10': 0, '00': 1, '01': 1})
> initlist=
> 790
> path=
> 936
> ('deep=', 1)
> ({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 0, '01': 0})
> ({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 1, '01': 1})
> ({'11': 1, '10': 0, '00': 0, '01': 0}, {'11': 1, '10': 0, '00': 1, '01': 1})
> initlist=
> 1549682
> path=
> 1870866
> Killed
> 
> 
> On Sunday, June 12, 2016 at 9:38:51 AM UTC+8, meInvent bbird wrote:
> > https://gist.github.com/hoyeunglee/3fea29ed4aadb5dbc11c41f9a36070dc
> > 
> > i discover my code can not record the path
> > 
> > because i do recursive call at the end of function
> > however, i want to do full combination at the end of function before calling
> > recursive call, 
> > 
> > for seeing the result path, choose call recursive call in each else 
> > statement 
> > which means when it is not found yet, it will continue recursive call
> > 
> > this time, i use binary for simple case, 
> > 
> > but final result show only 4 records, first two do not have ok, 
> > 
> > third and fourth can not see ok, it become [...]
> > 
> > how to show this [...] ?
> > 
> > is there any method to record the path in well format and easy to read ?
> > 
> > because in 3 valued case, i do not know name of operators, i can only 
> > record the full column

-- 
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Re: how to record path in well format and easy to read?

[meInvent bbird]

https://gist.github.com/hoyeunglee/350ac2dd496f7c1c95a428f847e6f2b1

then can not run deep 3 in python sagecloud

>>> mresult = DFS(b, 3, 3, mylist, path)
('deep=', 3)
({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 0, '01': 0})
({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 1, '01': 1})
({'11': 1, '10': 0, '00': 0, '01': 0}, {'11': 1, '10': 0, '00': 1, '01': 1})
initlist=
18
path=
18
here is caseA
('deep=', 2)
({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 0, '01': 0})
)
... retu({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 1, 
'01': 1})
({'11': 1, '10': 0, '00': 0, '01': 0}, {'11': 1, '10': 0, '00': 1, '01': 1})
initlist=
790
path=
936
('deep=', 1)
({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 0, '01': 0})
({'11': 1, '10': 1, '00': 0, '01': 1}, {'11': 1, '10': 0, '00': 1, '01': 1})
({'11': 1, '10': 0, '00': 0, '01': 0}, {'11': 1, '10': 0, '00': 1, '01': 1})
initlist=
1549682
path=
1870866
Killed


On Sunday, June 12, 2016 at 9:38:51 AM UTC+8, meInvent bbird wrote:
> https://gist.github.com/hoyeunglee/3fea29ed4aadb5dbc11c41f9a36070dc
> 
> i discover my code can not record the path
> 
> because i do recursive call at the end of function
> however, i want to do full combination at the end of function before calling
> recursive call, 
> 
> for seeing the result path, choose call recursive call in each else statement 
> which means when it is not found yet, it will continue recursive call
> 
> this time, i use binary for simple case, 
> 
> but final result show only 4 records, first two do not have ok, 
> 
> third and fourth can not see ok, it become [...]
> 
> how to show this [...] ?
> 
> is there any method to record the path in well format and easy to read ?
> 
> because in 3 valued case, i do not know name of operators, i can only record 
> the full column

-- 
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how to record path in well format and easy to read?

[meInvent bbird]

https://gist.github.com/hoyeunglee/3fea29ed4aadb5dbc11c41f9a36070dc

i discover my code can not record the path

because i do recursive call at the end of function
however, i want to do full combination at the end of function before calling
recursive call, 

for seeing the result path, choose call recursive call in each else statement 
which means when it is not found yet, it will continue recursive call

this time, i use binary for simple case, 

but final result show only 4 records, first two do not have ok, 

third and fourth can not see ok, it become [...]

how to show this [...] ?

is there any method to record the path in well format and easy to read ?

because in 3 valued case, i do not know name of operators, i can only record 
the full column
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: which library has map reduce and how to use it for this case

[meInvent bbird]

there are six operator, and a logic table initial in b variable 

aa[b[j][i]] 
[aa[b[i][0:1]+b[i][2:3] 

are just like vlookup to find output of each operator 
acting on first column and second column, second column and third column 
, first column and third column 

and searching a output columns which is result sum is 27*2 
and record the path if succeed, 
it record the path  and output any path which has result is 27*2 described 
before 
op1(op2(op3(op1(op2(),),op1(op2(),))), op1(op2(),)) 

there are two cases, first cases b are logic table 
later case are for six operators acting on the result column from previous 
result before recursive call 



def DFS(b, deep, maxx, sourceoperators, path): 
initlist = [] 
if deep > 0: 
print("deep=", deep) 
for aa,bb in itertools.combinations(sourceoperators, 2): 
print(aa,bb) 
if deep == maxx: 
finalresult = [] 
op1xy = [aa[b[i][0:1]+b[i][1:2]] for i in range(len(b))] 
op1yz = [aa[b[i][1:2]+b[i][2:3]] for i in range(len(b))] 
op1xz = [aa[b[i][0:1]+b[i][2:3]] for i in range(len(b))] 
op2xy = [bb[b[i][0:1]+b[i][1:2]] for i in range(len(b))] 
op2yz = [bb[b[i][1:2]+b[i][2:3]] for i in range(len(b))] 
op2xz = [bb[b[i][0:1]+b[i][2:3]] for i in range(len(b))] 
if sum(op1xy) == 54: 
path.append([(deep, aa, b, "xy")]) 
else: 
initlist.append(op1xy) 
if sum(op1yz) == 54: 
path.append([(deep, aa, b, "yz")]) 
else: 
initlist.append(op1yz) 
if sum(op1xz) == 54: 
path.append([(deep, aa, b, "xz")]) 
else: 
initlist.append(op1xz) 
if sum(op2xy) == 54: 
path.append([(deep, bb, b, "xy")]) 
else: 
initlist.append(op2xy) 
if sum(op2yz) == 54: 
path.append([(deep, bb, b, "yz")]) 
else: 
initlist.append(op2yz) 
if sum(op2xz) == 54: 
path.append([(deep, bb, b, "xz")]) 
else: 
initlist.append(op2xz) 
else: 
level = [] 
for j in range(len(b)): 
op1xy = [aa[b[j][i]] for i in range(len(b[j]))] 
op2xy = [bb[b[j][i]] for i in range(len(b[j]))] 
if sum(op1xy) == 54: 
path.append([(deep, aa, b[j], "xy")]) 
else: 
initlist.append(op1xy) 
if sum(op2xy) == 54: 
path.append([(deep, bb, b[j], "xy")]) 
else: 
initlist.append(op2xy) 
level.extend([op1xy, op2xy]) 
if deep == maxx: 
b = [] 
print("initlist=") 
print(len(initlist)) 
for aaa,bbb in itertools.combinations(initlist, 2): 
b.append([str(i)+str(j) for i,j in zip(aaa, bbb)]) 
if deep > 0: 
path2 = DFS(b, deep-1, maxx, sourceoperators, path) 
path.append(path2) 
return path 

path = [] 
mresult = DFS(b, 2, 2, mylist, path) 


On Friday, June 10, 2016 at 10:04:09 PM UTC+8, Michael Selik wrote:
> On Thu, Jun 9, 2016, 9:11 PM Ho Yeung Lee  wrote:
> 
> > input are these six operators, output is finding full column of 27
> > elements add together is 54
> >
> > i got memory error when searching this,
> >
> > i have difficulty in understanding map reduce and transforming my program
> > into map reduce problem
> >
> 
> Why do you think you need map-reduce?
> 
> You'll need to explain the problem in more detail. Instead of talking about
> operators and columns, what is the actual, real-world problem?
> 
> >

-- 
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Re: how to solve memory

[meInvent bbird]

i use sage cloud to run, it killed my program

i checked total memory is large, how much memory it need, 8GB*10, 80GB?

top - 05:55:16 up 10:05,  1 user,  load average: 0.35, 0.63, 0.56
Tasks:  13 total,   1 running,  12 sleeping,   0 stopped,   0 zombie
%Cpu(s):  0.5 us,  0.4 sy,  0.0 ni, 99.1 id,  0.0 wa,  0.0 hi,  0.0 si,  0.0 st
KiB Mem:  26755056 total,  6317696 used, 20437360 free,   611836 buffers
KiB Swap: 67108860 total,   468688 used, 66640172 free.  1223512 cached Mem

def DFS(b, deep, maxx, sourceoperators, path): 
initlist = [] 
if deep > 0: 
print("deep=", deep) 
for aa,bb in itertools.combinations(sourceoperators, 2): 
print(aa,bb) 
if deep == maxx: 
finalresult = [] 
op1xy = [aa[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
op1yz = [aa[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
op1xz = [aa[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
op2xy = [bb[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
op2yz = [bb[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
op2xz = [bb[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
if sum(op1xy) == 54: 
path.append([(deep, aa, b, "xy")]) 
else:
initlist.append(op1xy)
if sum(op1yz) == 54: 
path.append([(deep, aa, b, "yz")]) 
else:
initlist.append(op1yz)
if sum(op1xz) == 54: 
path.append([(deep, aa, b, "xz")]) 
else:
initlist.append(op1xz)
if sum(op2xy) == 54: 
path.append([(deep, bb, b, "xy")]) 
else:
initlist.append(op2xy)
if sum(op2yz) == 54: 
path.append([(deep, bb, b, "yz")]) 
else:
initlist.append(op2yz)
if sum(op2xz) == 54: 
path.append([(deep, bb, b, "xz")]) 
else:
initlist.append(op2xz)
else: 
level = [] 
for j in range(len(b)): 
op1xy = [aa[b[j][i]] for i in range(len(b[j]))] 
op2xy = [bb[b[j][i]] for i in range(len(b[j]))] 
if sum(op1xy) == 54: 
path.append([(deep, aa, b[j], "xy")]) 
else:
initlist.append(op1xy)
if sum(op2xy) == 54: 
path.append([(deep, bb, b[j], "xy")]) 
else:
initlist.append(op2xy)
level.extend([op1xy, op2xy]) 
if deep == maxx: 
b = [] 
print("initlist=")
print(len(initlist))
for aaa,bbb in itertools.combinations(initlist, 2): 
b.append([str(i)+str(j) for i,j in zip(aaa, bbb)]) 
if deep > 0:
path2 = DFS(b, deep-1, maxx, sourceoperators, path) 
path.append(path2)
return path 

path = [] 
mresult = DFS(b, 2, 2, mylist, path) 




On Saturday, June 11, 2016 at 8:34:16 AM UTC+8, MRAB wrote:
> On 2016-06-11 00:31, meInvent bbird wrote:
> > it is quite ridiculous,
> > this time i am sure that i put correct indentation
> > and add a else statement to make sure recursive call inside the
> > if statement , it still memory error,
> >
> > where is the memory error?
> >
> > is itertools.combinations so big for the list?
> >
> [snip]
> 
> How long is initlist?
> 
> When I ran the code, it said over 100_000 items.
> 
> How many combinations would there be?
> 
> Over 10_000_000_000.
> 
> That's how long the list 'b' would be.
> 
> You'll need 10s of gigabytes of memory and a lot of patience!

-- 
https://mail.python.org/mailman/listinfo/python-list

Re: how to solve memory

[meInvent bbird]

i use a version having better indentation,

and then remove redundant which if sum column == 54 , do not add this column
into initlist

and add deep > 0 before recursive call
and print number of initlist is 118,XXX

but it is still running, where it run , and why run a very long time

def DFS(b, deep, maxx, sourceoperators, path): 
initlist = [] 
if deep > 0: 
print("deep=", deep) 
for aa,bb in itertools.combinations(sourceoperators, 2): 
print(aa,bb) 
if deep == maxx: 
finalresult = [] 
op1xy = [aa[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
op1yz = [aa[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
op1xz = [aa[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
op2xy = [bb[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
op2yz = [bb[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
op2xz = [bb[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
if sum(op1xy) == 54: 
path.append([(deep, aa, "xy")]) 
else:
initlist.append(op1xy)
if sum(op1yz) == 54: 
path.append([(deep, aa, "yz")]) 
else:
initlist.append(op1yz)
if sum(op1xz) == 54: 
path.append([(deep, aa, "xz")]) 
else:
initlist.append(op1xz)
if sum(op2xy) == 54: 
path.append([(deep, bb, "xy")]) 
else:
initlist.append(op2xy)
if sum(op2yz) == 54: 
path.append([(deep, bb, "yz")]) 
else:
initlist.append(op2yz)
if sum(op2xz) == 54: 
path.append([(deep, bb, "xz")]) 
else:
initlist.append(op2xz)
else: 
level = [] 
for j in range(len(b)): 
op1xy = [aa[b[j][i]] for i in range(len(b[j]))] 
op2xy = [bb[b[j][i]] for i in range(len(b[j]))] 
if sum(op1xy) == 54: 
path.append([(deep, aa, "xy")]) 
else:
initlist.append(op1xy)
if sum(op2xy) == 54: 
path.append([(deep, bb, "xy")]) 
else:
initlist.append(op2xy)
level.extend([op1xy, op2xy]) 
if deep == maxx: 
b = [] 
print("initlist=")
print(len(initlist))
for aaa,bbb in itertools.combinations(initlist, 2): 
b.append([str(i)+str(j) for i,j in zip(aaa, bbb)]) 
if deep > 0:
path2 = DFS(b, deep-1, maxx, sourceoperators, path) 
path.append(path2)
return path 

path = [] 
mresult = DFS(b, 2, 2, mylist, path) 



On Saturday, June 11, 2016 at 8:34:16 AM UTC+8, MRAB wrote:
> On 2016-06-11 00:31, meInvent bbird wrote:
> > it is quite ridiculous,
> > this time i am sure that i put correct indentation
> > and add a else statement to make sure recursive call inside the
> > if statement , it still memory error,
> >
> > where is the memory error?
> >
> > is itertools.combinations so big for the list?
> >
> [snip]
> 
> How long is initlist?
> 
> When I ran the code, it said over 100_000 items.
> 
> How many combinations would there be?
> 
> Over 10_000_000_000.
> 
> That's how long the list 'b' would be.
> 
> You'll need 10s of gigabytes of memory and a lot of patience!

-- 
https://mail.python.org/mailman/listinfo/python-list

Re: how to solve memory

[meInvent bbird]

this time i remove redundant which not add full column sum already has beeb 27*2

but it always has syntax error, indentation error, 

where is wrong?

def DFS(b, deep, maxx, sourceoperators, path):
 initlist = []
 if deep > 0:
  print("deep=", deep)
  for aa,bb in itertools.combinations(sourceoperators, 2):
   print(aa,bb)
   if deep == maxx:
 finalresult = []
 op1xy = [aa[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
 op1yz = [aa[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
 op1xz = [aa[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
 op2xy = [bb[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
 op2yz = [bb[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
 op2xz = [bb[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
 if sum(op1xy) == 54:
  path.append([(deep, aa, b, "xy")])
 else:
  initlist.append(op1xy)
 if sum(op1yz) == 54:
  path.append([(deep, aa, b, "yz")])
 else:
  initlist.append(op1yz)  
 if sum(op1xz) == 54:
  path.append([(deep, aa, b, "xz")])
 else:
  initlist.append(op1xz)
 if sum(op2xy) == 54:
  path.append([(deep, bb, b, "xy")])
 else:
  initlist.append(op2xy)
 if sum(op2yz) == 54:
  path.append([(deep, bb, b, "yz")])
 else:
  initlist.append(op2yz)
 if sum(op2xz) == 54:  
  path.append([(deep, bb, b, "xz")])
 else:
  initlist.append(op2xz)
   else:
 level = []
 for j in range(len(b)):
  op1xy = [aa[b[j][i]] for i in range(len(b[j]))]
  op2xy = [bb[b[j][i]] for i in range(len(b[j]))]
  if sum(op1xy) == 54:
   path.append([(deep, aa, b[j], "xy")])
  else:
   initlist.append(op1xy)
  if sum(op2xy) == 54:
   path.append([(deep, bb, b[j], "xy")])
  else:
   initlist.append(op2xy)
  level.append(op1xy)
  level.append(op2xy)
   print("initlist=")
   print(len(initlist)
 if deep == maxx:
   if deep > 0:
 b = []
 for aaa,bbb in itertools.combinations(initlist, 2): 
  b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
 print("deep=")
 print(deep)
 path2 = DFS(b, deep-1, maxx, sourceoperators, path)
 path.append(path2)
   else:
 print("")
   print("path=")
   print(len(path))
 else:
   if deep > 0:
 for aaa,bbb in itertools.combinations(initlist, 2):
  b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
 print("deep=")
 print(deep)
 path2 = DFS(b, deep-1, maxx, sourceoperators, path)
 path.append(path2)
   else:
     print("")
   print("path=")
   print(len(path))
 return path



On Saturday, June 11, 2016 at 8:34:16 AM UTC+8, MRAB wrote:
> On 2016-06-11 00:31, meInvent bbird wrote:
> > it is quite ridiculous,
> > this time i am sure that i put correct indentation
> > and add a else statement to make sure recursive call inside the
> > if statement , it still memory error,
> >
> > where is the memory error?
> >
> > is itertools.combinations so big for the list?
> >
> [snip]
> 
> How long is initlist?
> 
> When I ran the code, it said over 100_000 items.
> 
> How many combinations would there be?
> 
> Over 10_000_000_000.
> 
> That's how long the list 'b' would be.
> 
> You'll need 10s of gigabytes of memory and a lot of patience!

-- 
https://mail.python.org/mailman/listinfo/python-list

Re: how to solve memory

[meInvent bbird]

it is quite ridiculous, 
this time i am sure that i put correct indentation
and add a else statement to make sure recursive call inside the 
if statement , it still memory error, 

where is the memory error?

is itertools.combinations so big for the list?


  if deep == maxx:
   if deep > 0:
b = []
for aaa,bbb in itertools.combinations(initlist, 2): 
 b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
print("deep=")
print(deep)
path2 = DFS(b, deep-1, maxx, sourceoperators, path)
path.append(path2)
   else:
print ""
  else:
   if deep > 0:
for aaa,bbb in itertools.combinations(initlist, 2):
 b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
print("deep=")
print(deep)
path2 = DFS(b, deep-1, maxx, sourceoperators, path)
path.append(path2)
   else:
print ""
 return path


On Saturday, June 11, 2016 at 12:26:55 AM UTC+8, John Gordon wrote:
> In <8a1c372e-bd6c-4923-8ae1-8f129ec74...@googlegroups.com> meInvent bbird 
> <jobmatt...@gmail.com> writes:
> 
> > > I already responded to your earlier post about this program.  Did you
> > > read it?
> 
> > sorry i missed your reply post, 
> > i find post by me and search your name, no this name
> > is your name changed in previous post?
> 
> My comment was that the recursive calls weren't indented in the
> "if deep > 0" block, therefore DFS was being called infinitely with smaller
> and smaller values of deep.  But it appears you have fixed that issue.
> 
> -- 
> John Gordon   A is for Amy, who fell down the stairs
> gor...@panix.com  B is for Basil, assaulted by bears
> -- Edward Gorey, "The Gashlycrumb Tinies"
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: how to solve memory

[meInvent bbird]

i put final part of code inside one more space 
and add constraint deep > 0 
but still memory error 

and print the deep number to see, it do not run infinity again 


V6 is just my superstitution, because V is disable 6 is separation 

i run your version directly, it has key = '0' error 

M1 = {}
M2 = {}
M3 = {}
M4 = {}
M5 = {}
V6 = {}
M1['00']=0
M1['01']=1
M1['02']=1
M1['10']=2
M1['11']=2
M1['12']=1
M1['20']=1
M1['21']=1
M1['22']=1
M2['00']=0
M2['01']=1
M2['02']=1
M2['10']=1
M2['11']=2
M2['12']=2
M2['20']=1
M2['21']=1
M2['22']=1
M3['00']=0
M3['01']=1
M3['02']=1
M3['10']=0
M3['11']=1
M3['12']=2
M3['20']=0
M3['21']=2
M3['22']=2
M4['00']=0
M4['01']=1
M4['02']=0
M4['10']=1
M4['11']=2
M4['12']=2
M4['20']=0
M4['21']=1
M4['22']=1
M5['00']=0
M5['01']=1
M5['02']=1
M5['10']=2
M5['11']=1
M5['12']=2
M5['20']=0
M5['21']=2
M5['22']=1
V6['00']=1
V6['01']=1
V6['02']=2
V6['10']=1
V6['11']=1
V6['12']=1
V6['20']=2
V6['21']=2
V6['22']=2
MM = {}
MM[0] = M1
MM[1] = M2
MM[2] = M3
MM[3] = M4
MM[4] = M5
MM[5] = V6
m = 3
b = [str(i)+str(j)+str(k) for i in range(m) for j in range(m) for k in range(m)]
import itertools
deep = 3
final = []
mylist = [MM[i] for i in range(0,7-1)]
b = [str(i)+str(j)+str(k) for i in range(m) for j in range(m) for k in range(m)]
def DFS(b, deep, maxx, sourceoperators, path):
 initlist = []
 if deep > 0:
  print("deep=", deep)
  for aa,bb in itertools.combinations(sourceoperators, 2):
   print(aa,bb)
   if deep == maxx:
finalresult = []
op1xy = [aa[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
op1yz = [aa[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
op1xz = [aa[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
op2xy = [bb[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
op2yz = [bb[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
op2xz = [bb[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
if sum(op1xy) == 54:
  path.append([(deep, aa, "xy")])
if sum(op1yz) == 54:
  path.append([(deep, aa, "yz")])
if sum(op1xz) == 54:
  path.append([(deep, aa, "xz")])
if sum(op2xy) == 54:
  path.append([(deep, bb, "xy")])
if sum(op2yz) == 54:
  path.append([(deep, bb, "yz")])
if sum(op2xz) == 54:  
  path.append([(deep, bb, "xz")])
initlist.append(op1xy)
initlist.append(op1yz)
initlist.append(op1xz)
initlist.append(op2xy)
initlist.append(op2yz)
initlist.append(op2xz)
   else:
level = []
for j in range(len(b)):
 op1xy = [aa[b[j][i]] for i in range(len(b[j]))]
 op2xy = [bb[b[j][i]] for i in range(len(b[j]))]
 if sum(op1xy) == 54:
  path.append([(deep, aa, "xy")])
 if sum(op2xy) == 54:
  path.append([(deep, bb, "xy")])
 level.append(op1xy)
 level.append(op2xy)
 initlist.append(op1xy)
 initlist.append(op2xy)
  if deep == maxx:
   if deep > 0:
b = []
for aaa,bbb in itertools.combinations(initlist, 2): 
 b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
print("deep=")
print(deep)
path2 = DFS(b, deep-1, maxx, sourceoperators, path)
path.append(path2)
  else:
   if deep > 0:
for aaa,bbb in itertools.combinations(initlist, 2):
 b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
print("deep=")
print(deep)
path2 = DFS(b, deep-1, maxx, sourceoperators, path)
path.append(path2)
 return path

path = []
mresult = DFS(b, 2, 2, mylist, path)



On Friday, June 10, 2016 at 10:08:44 PM UTC+8, Steven D'Aprano wrote:
> On Fri, 10 Jun 2016 10:58 am, meInvent bbird wrote:
> 
> [snip unreadable code]
> 
> I just ran your code, and it almost crashed my computer. I think it is quite
> rude to post code without an explanation of what the problem is.
> 
> 
> I'm afraid that your code is virtually unreadable to me. It is too verbose,
> the indentation is awful (single space indents is impossible for me to
> track by eye) and it makes too much work out of simple operations. And none
> of the variable names mean anything.
> 
> I've started to simplify and clean the code, and got to the following, which
> I hope you will agree is easier to read and more compact:
> 
> 
> 
> import itertools
> M1 = {'00': 0, '01': 2, '02': 1, '10': 1, '11': 1, 
>   '12': 1, '20': 1, '21': 1, '22': 2}
> M2 = {'00': 0, '01': 1, '02': 1, '10': 1, '11': 1, 
>   '12': 1, '20': 1, '21': 1, '22': 1}
> M3 = {'00': 2, '01': 2, '02': 2, '10': 0, '11': 2, 
>   '12': 1, '20': 0, '21': 1, '22': 2}
> M4 = {'00': 1, '01': 2, '02': 1, '10': 2, '11': 2, 
>   '12': 2, '20': 0, '21': 1, '22': 2}
> M5 = {'00': 0, '01': 1, '02': 1, '10': 0, '11': 2, 
>   '12': 1, '20': 0, '21': 1, '22': 1}
> # Why V instead of M?
> V6 = {'00': 1, '01': 1, '02': 2, '10': 1, '11': 2, 
>   '12': 1, '20': 1, '21': 2, '22': 2}
> 
> MM = {0: M1, 1: M2, 2: M3, 3: M4, 4: M5, 5: V6}
&

Re: how to solve memory

[meInvent bbird]

On Friday, June 10, 2016 at 10:08:44 PM UTC+8, Steven D'Aprano wrote:
> On Fri, 10 Jun 2016 10:58 am, meInvent bbird wrote:
> 
> [snip unreadable code]
> 
> I just ran your code, and it almost crashed my computer. I think it is quite
> rude to post code without an explanation of what the problem is.
> 
> 
> I'm afraid that your code is virtually unreadable to me. It is too verbose,
> the indentation is awful (single space indents is impossible for me to
> track by eye) and it makes too much work out of simple operations. And none
> of the variable names mean anything.
> 
> I've started to simplify and clean the code, and got to the following, which
> I hope you will agree is easier to read and more compact:
> 
> 
> 
> import itertools
> M1 = {'00': 0, '01': 2, '02': 1, '10': 1, '11': 1, 
>   '12': 1, '20': 1, '21': 1, '22': 2}
> M2 = {'00': 0, '01': 1, '02': 1, '10': 1, '11': 1, 
>   '12': 1, '20': 1, '21': 1, '22': 1}
> M3 = {'00': 2, '01': 2, '02': 2, '10': 0, '11': 2, 
>   '12': 1, '20': 0, '21': 1, '22': 2}
> M4 = {'00': 1, '01': 2, '02': 1, '10': 2, '11': 2, 
>   '12': 2, '20': 0, '21': 1, '22': 2}
> M5 = {'00': 0, '01': 1, '02': 1, '10': 0, '11': 2, 
>   '12': 1, '20': 0, '21': 1, '22': 1}
> # Why V instead of M?
> V6 = {'00': 1, '01': 1, '02': 2, '10': 1, '11': 2, 
>   '12': 1, '20': 1, '21': 2, '22': 2}
> 
> MM = {0: M1, 1: M2, 2: M3, 3: M4, 4: M5, 5: V6}
> m = 3
> b = [str(i)+str(j)+str(k) for i in range(m) 
>  for j in range(m) for k in range(m)]
> mylist = [MM[i] for i in range(6)]
> 
> def DFS(b, deep, maxx, sourceoperators, path):
> initlist = []
> if deep > 0:
> print("deep=", deep)
> for aa,bb in itertools.combinations(sourceoperators, 2):
> print(aa,bb)
> if deep == maxx:
> finalresult = []
> op1xy = [aa[b[i][0:1]] for i in range(len(b))]
> op1yz = [aa[b[i][1:2]] for i in range(len(b))]
> op1xz = [aa[b[i][0]+b[i][2]] for i in range(len(b))]
> op2xy = [bb[b[i][0:1]] for i in range(len(b))]
> op2yz = [bb[b[i][1:2]] for i in range(len(b))]
> op2xz = [bb[b[i][0]+b[i][2]] for i in range(len(b))]
> if sum(op1xy) == 54:
> path.append([(deep, aa, "xy")])
> if sum(op1yz) == 54:
> path.append([(deep, aa, "yz")])
> if sum(op1xz) == 54:
> path.append([(deep, aa, "xz")])
> if sum(op2xy) == 54:
> path.append([(deep, bb, "xy")])
> if sum(op2yz) == 54:
> path.append([(deep, bb, "yz")])
> if sum(op2xz) == 54:
> path.append([(deep, bb, "xz")])
> initlist.extend([op1xy, op1yz, op1xz, op2xy, op2yz, op2xz])
> else:
> level = []
> for j in range(len(b)):
> op1xy = [aa[b[j][i]] for i in range(len(b[j]))]
> op2xy = [bb[b[j][i]] for i in range(len(b[j]))]
> if sum(op1xy) == 54:
> path.append([(deep, aa, "xy")])
> if sum(op2xy) == 54:
> path.append([(deep, bb, "xy")])
> level.extend([op1xy, op2xy])
> initlist.extend([op1xy, op2xy])
> if deep == maxx:
> b = []
> for aaa,bbb in itertools.combinations(initlist, 2):
> b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
> path = DFS(b, deep-1, maxx, sourceoperators, path)
> return path
> 
> path = []
> mresult = DFS(b, 2, 2, mylist, path)
> 
> 
> 
> Unfortunate, in cleaning up your code, I have changed something, because my
> version and your version do not do the same thing.
> 
> After nearly crashing my computer running your version, I am not going to
> spend the time trying to debug this. If you want our help, I suggest that
> you clean up the code. Currently it is an incomprehensible mess to me. Even
> after the cleanup, I have no idea what this piece of code is supposed to do
> or what it is calculating.
> 
> Can you explain the purpose of the code? What is it calculating? Can you
> show the expected results? What does "DFS" mean? Why do you have M1 - M5
> but then a mysterious V6?
> 
> I can see you are calculating *something* to do with permutations of 0 1 2,
> but I don't understand what.
> 
> I think I can see one problem: DFS is a recursive function, but it never
> ends the recursion. It ALW

Re: how to solve memory

[meInvent bbird]

On Friday, June 10, 2016 at 10:46:43 PM UTC+8, John Gordon wrote:
> In <4f853aa2-cc00-480b-9fd7-79b05cbd4...@googlegroups.com> meInvent bbird 
> <jobmatt...@gmail.com> writes:
> 
> > https://drive.google.com/file/d/0Bxs_ao6uuBDULVNsRjZSVjdPYlE/view?usp=sharing
> 
> I already responded to your earlier post about this program.  Did you
> read it?
> 
> -- 
> John Gordon   A is for Amy, who fell down the stairs
> gor...@panix.com  B is for Basil, assaulted by bears
> -- Edward Gorey, "The Gashlycrumb Tinies"



sorry i missed your reply post, 
i find post by me and search your name, no this name
is your name changed in previous post?

could you post the link here?
is it an email or google group post?
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: how to solve memory

[meInvent bbird]

i put final part of code inside one more space
and add constraint deep > 0
but still memory error

and print the deep number to see, it do not run infinity again


V6 is just my superstitution, because V is disable 6 is separation

i run your version directly, it has key = '0' error

M1 = {}
M2 = {}
M3 = {}
M4 = {}
M5 = {}
V6 = {}
M1['00']=0
M1['01']=1
M1['02']=2
M1['10']=1
M1['11']=1
M1['12']=2
M1['20']=2
M1['21']=2
M1['22']=2
M2['00']=0
M2['01']=0
M2['02']=2
M2['10']=0
M2['11']=1
M2['12']=2
M2['20']=2
M2['21']=2
M2['22']=2
M3['00']=0
M3['01']=0
M3['02']=0
M3['10']=0
M3['11']=1
M3['12']=2
M3['20']=0
M3['21']=2
M3['22']=2
M4['00']=0
M4['01']=1
M4['02']=0
M4['10']=1
M4['11']=1
M4['12']=1
M4['20']=0
M4['21']=1
M4['22']=2
M5['00']=0
M5['01']=0
M5['02']=0
M5['10']=0
M5['11']=1
M5['12']=2
M5['20']=0
M5['21']=2
M5['22']=2
V6['00']=2
V6['01']=2
V6['02']=2
V6['10']=1
V6['11']=2
V6['12']=2
V6['20']=0
V6['21']=1
V6['22']=2
MM = {}
MM[0] = M1
MM[1] = M2
MM[2] = M3
MM[3] = M4
MM[4] = M5
MM[5] = V6
m = 3
b = [str(i)+str(j)+str(k) for i in range(m) for j in range(m) for k in range(m)]
import itertools
deep = 3
final = []
mylist = [MM[i] for i in range(0,7-1)]
b = [str(i)+str(j)+str(k) for i in range(m) for j in range(m) for k in range(m)]
def DFS(b, deep, maxx, sourceoperators, path):
 initlist = []
 if deep > 0:
  print("deep=", deep)
  for aa,bb in itertools.combinations(sourceoperators, 2):
   print(aa,bb)
   if deep == maxx:
finalresult = []
op1xy = [aa[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
op1yz = [aa[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
op1xz = [aa[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
op2xy = [bb[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
op2yz = [bb[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
op2xz = [bb[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
if sum(op1xy) == 54:
  path.append([(deep, aa, "xy")])
if sum(op1yz) == 54:
  path.append([(deep, aa, "yz")])
if sum(op1xz) == 54:
  path.append([(deep, aa, "xz")])
if sum(op2xy) == 54:
  path.append([(deep, bb, "xy")])
if sum(op2yz) == 54:
  path.append([(deep, bb, "yz")])
if sum(op2xz) == 54:  
  path.append([(deep, bb, "xz")])
initlist.append(op1xy)
initlist.append(op1yz)
initlist.append(op1xz)
initlist.append(op2xy)
initlist.append(op2yz)
initlist.append(op2xz)
   else:
level = []
for j in range(len(b)):
 op1xy = [aa[b[j][i]] for i in range(len(b[j]))]
 op2xy = [bb[b[j][i]] for i in range(len(b[j]))]
 if sum(op1xy) == 54:
  path.append([(deep, aa, "xy")])
 if sum(op2xy) == 54:
  path.append([(deep, bb, "xy")])
 level.append(op1xy)
 level.append(op2xy)
 initlist.append(op1xy)
 initlist.append(op2xy)
  if deep == maxx:
   if deep > 0:
b = []
for aaa,bbb in itertools.combinations(initlist, 2): 
 b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
print("deep=")
print(deep)
path = DFS(b, deep-1, maxx, sourceoperators, path)
  else:
   if deep > 0:
for aaa,bbb in itertools.combinations(initlist, 2):
 b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
print("deep=")
print(deep)
path = DFS(b, deep-1, maxx, sourceoperators, path)
 return path

path = []
mresult = DFS(b, 2, 2, mylist, path)


On Friday, June 10, 2016 at 10:08:44 PM UTC+8, Steven D'Aprano wrote:
> On Fri, 10 Jun 2016 10:58 am, meInvent bbird wrote:
> 
> [snip unreadable code]
> 
> I just ran your code, and it almost crashed my computer. I think it is quite
> rude to post code without an explanation of what the problem is.
> 
> 
> I'm afraid that your code is virtually unreadable to me. It is too verbose,
> the indentation is awful (single space indents is impossible for me to
> track by eye) and it makes too much work out of simple operations. And none
> of the variable names mean anything.
> 
> I've started to simplify and clean the code, and got to the following, which
> I hope you will agree is easier to read and more compact:
> 
> 
> 
> import itertools
> M1 = {'00': 0, '01': 2, '02': 1, '10': 1, '11': 1, 
>   '12': 1, '20': 1, '21': 1, '22': 2}
> M2 = {'00': 0, '01': 1, '02': 1, '10': 1, '11': 1, 
>   '12': 1, '20': 1, '21': 1, '22': 1}
> M3 = {'00': 2, '01': 2, '02': 2, '10': 0, '11': 2, 
>   '12': 1, '20': 0, '21': 1, '22': 2}
> M4 = {'00': 1, '01': 2, '02': 1, '10': 2, '11': 2, 
>   '12': 2, '20': 0, '21': 1, '22': 2}
> M5 = {'00': 0, '01': 1, '02': 1, '10': 0, '11': 2, 
>   '12': 1, '20': 0, '21': 1, '22': 1}
> # Why V instead of M?
> V6 = {'00': 1, '01': 1, '02': 2, '10': 1, '11': 2, 
>   '12': 1, '20': 1, '21': 2, '22': 2}
> 
> MM = {0: M1, 1: M2, 2: M3, 3: M4, 4: M5, 5: V6}
> m = 3
> b = [str(i)+str(j)+str(k) for i in ra

Re: run code error with numpy and scipy

[meInvent bbird]

On Thursday, June 9, 2016 at 9:52:43 PM UTC+8, MRAB wrote:
> On 2016-06-09 07:02, meInvent bbird wrote:
> > when i run code in window has an error not a valid win32 dll
> >
> > then i install scipy first in this site, then install numpy-mkl
> > all python 2.7 and win32
> >
> > http://www.lfd.uci.edu/~gohlke/pythonlibs/
> >
> > then i run code below, got error , and captured screen in this link
> >
> > https://drive.google.com/file/d/0Bxs_ao6uuBDUQXROd2VqSURGa00/view?usp=sharing
> >
> [snip]
> You're entering it at the interactive Python prompt.
> 
> When it sees the blank line, it thinks you've finished entering the 
> function, so it complains when the next line is indented. (Things are 
> slightly different when working interactively than when running from a 
> file.)
> 
> You should put the code into a file and then run that.

thank you i will try it.
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: how to solve memory

[meInvent bbird]

On Friday, June 10, 2016 at 9:13:13 AM UTC+8, Michael Torrie wrote:
> On 06/09/2016 06:58 PM, meInvent bbird wrote:
> > 
> 
> Do you have a question for the list? If so, please state what it is, and
> describe what you are doing and what isn't working.  If you can boil it
> down to a dozen lines of run-able, self-contained code that illustrates
> the problem, that is helpful too.


there are six operator, and a logic table initial in b variable

aa[b[j][i]]
[aa[b[i][0:1]+b[i][2:3]

are just like vlookup to find output of each operator 
acting on first column and second column, second column and third column
, first column and third column

and searching a output columns which is result sum is 27*2
and record the path if succeed, 
it record the path  and output any path which has result is 27*2 described 
before
op1(op2(op3(op1(op2(),),op1(op2(),))), op1(op2(),))

there are two cases, first cases b are logic table 
later case are for six operators acting on the result column from previous 
result before recursive call

-- 
https://mail.python.org/mailman/listinfo/python-list

how to solve memory

[meInvent bbird]

https://drive.google.com/file/d/0Bxs_ao6uuBDULVNsRjZSVjdPYlE/view?usp=sharing

M1 = {}
M2 = {}
M3 = {}
M4 = {}
M5 = {}
V6 = {}
M1['00']=0
M1['01']=2
M1['02']=1
M1['10']=1
M1['11']=1
M1['12']=1
M1['20']=1
M1['21']=1
M1['22']=2
M2['00']=0
M2['01']=1
M2['02']=1
M2['10']=1
M2['11']=1
M2['12']=1
M2['20']=1
M2['21']=1
M2['22']=1
M3['00']=2
M3['01']=2
M3['02']=2
M3['10']=0
M3['11']=2
M3['12']=1
M3['20']=0
M3['21']=1
M3['22']=2
M4['00']=1
M4['01']=2
M4['02']=1
M4['10']=2
M4['11']=2
M4['12']=2
M4['20']=0
M4['21']=1
M4['22']=2
M5['00']=0
M5['01']=1
M5['02']=1
M5['10']=0
M5['11']=2
M5['12']=1
M5['20']=0
M5['21']=1
M5['22']=1
V6['00']=1
V6['01']=1
V6['02']=2
V6['10']=1
V6['11']=2
V6['12']=1
V6['20']=1
V6['21']=2
V6['22']=2
MM = {}
MM[0] = M1
MM[1] = M2
MM[2] = M3
MM[3] = M4
MM[4] = M5
MM[5] = V6
m = 3
b = [str(i)+str(j)+str(k) for i in range(m) for j in range(m) for k in range(m)]
import itertools
deep = 3
final = []
mylist = [MM[i] for i in range(0,7-1)]
b = [str(i)+str(j)+str(k) for i in range(m) for j in range(m) for k in range(m)]
def DFS(b, deep, maxx, sourceoperators, path):
 initlist = []
 if deep > 0:
  print("deep=", deep)
  for aa,bb in itertools.combinations(sourceoperators, 2):
   print(aa,bb)
   if deep == maxx:
finalresult = []
op1xy = [aa[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
op1yz = [aa[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
op1xz = [aa[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
op2xy = [bb[b[i][0:1]+b[i][1:2]] for i in range(len(b))]
op2yz = [bb[b[i][1:2]+b[i][2:3]] for i in range(len(b))]
op2xz = [bb[b[i][0:1]+b[i][2:3]] for i in range(len(b))]
if sum(op1xy) == 54:
  path.append([(deep, aa, "xy")])
if sum(op1yz) == 54:
  path.append([(deep, aa, "yz")])
if sum(op1xz) == 54:
  path.append([(deep, aa, "xz")])
if sum(op2xy) == 54:
  path.append([(deep, bb, "xy")])
if sum(op2yz) == 54:
  path.append([(deep, bb, "yz")])
if sum(op2xz) == 54:  
  path.append([(deep, bb, "xz")])
initlist.append(op1xy)
initlist.append(op1yz)
initlist.append(op1xz)
initlist.append(op2xy)
initlist.append(op2yz)
initlist.append(op2xz)
   else:
level = []
for j in range(len(b)):
 op1xy = [aa[b[j][i]] for i in range(len(b[j]))]
 op2xy = [bb[b[j][i]] for i in range(len(b[j]))]
 if sum(op1xy) == 54:
  path.append([(deep, aa, "xy")])
 if sum(op2xy) == 54:
  path.append([(deep, bb, "xy")])
 level.append(op1xy)
 level.append(op2xy)
 initlist.append(op1xy)
 initlist.append(op2xy)
 if deep == maxx:
  b = []
  #print(len(list(itertools.combinations(initlist, 2
  for aaa,bbb in itertools.combinations(initlist, 2): 
   b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
  path = DFS(b, deep-1, maxx, sourceoperators, path)
 else:
  #print(len(list(itertools.combinations(initlist, 2
  for aaa,bbb in itertools.combinations(initlist, 2):
   b.append([str(i)+str(j) for i,j in zip(aaa, bbb)])
  path = DFS(b, deep-1, maxx, sourceoperators, path)
 return path

path = []
mresult = DFS(b, 2, 2, mylist, path)
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run code error with numpy and scipy

[meInvent bbird]

when i run code in window has an error not a valid win32 dll

then i install scipy first in this site, then install numpy-mkl
all python 2.7 and win32

http://www.lfd.uci.edu/~gohlke/pythonlibs/

then i run code below, got error , and captured screen in this link

https://drive.google.com/file/d/0Bxs_ao6uuBDUQXROd2VqSURGa00/view?usp=sharing

import numpy as np
from scipy.sparse.linalg import svds
from functools import partial
from scipy.sparse import csr_matrix

def emsvd(Y, k=None, tol=1E-3, maxiter=None):
 if k is None:
  svdmethod = partial(np.linalg.svd, full_matrices=False)
 else:
  svdmethod = partial(svds, k=k)
 if maxiter is None:
  maxiter = np.inf

 muhat = np.nanmean(Y, axis=0, keepdims=1)
 valid = np.isfinite(Y)
 Y_hat = np.where(valid, Y, muhat)

 halt = False
 ii = 1
 v_prev = 0

 while not halt:
  U, s, Vt = svdmethod(Y_hat - muhat)
  Y_hat[~valid] = (U.dot(np.diag(s)).dot(Vt) + muhat)[~valid]
  muhat = Y_hat.mean(axis=0, keepdims=1)
  v = s.sum()
  if ii >= maxiter or ((v - v_prev) / v_prev) < tol:
   halt = True
   ii += 1
   v_prev = v
 return Y_hat, muhat, U, s, Vt


A = csr_matrix([[1, 2, 0], [0, 0, 3], [4, 0, 5]])
emsvd(A)

np.nanmean(A, axis=0, keepdims=1)
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Re: how to extract a variable as parameter which has index using by a for loop?

[meInvent bbird]

just extract b[i][0:1] and b[i][1:2] out of for loop

but i depend on for loop

def node(mmm, A, B): 
 H2 = [MM[mmm][A+B] for i in range(len(b))] 
 return H2 
 
node(5, b[i][0:1], b[i][1:2]) 

it is not convenient to disclose in detail
just expect to discuss from the view of programming

On Wednesday, June 8, 2016 at 4:56:56 PM UTC+8, Steven D'Aprano wrote:
> On Wednesday 08 June 2016 17:31, meInvent bbird wrote:
> 
> > b = [str(i)+str(j)+str(k) for i in range(m) for j in range(m) for k in
> > range(m)]
> > b[21][0:1]+b[21][1:2]
> > b[21][1:2]+b[21][2:3]
> > b[21][0:1]+b[21][2:3]
> > 
> > 
> > originally,
> > 
> > mmm = 5
> > H2 = [MM[mmm][b[i][0:1]+b[i][1:2]] for i in range(len(b))]
> 
> This is a mess. I don't understand what you are trying to do. You have these 
> variable names that don't mean anything, like "b" and "H2", and others which 
> aren't defined, like MM. I don't understand what you are trying to 
> accomplish, 
> or the purpose of your code.
> 
> 
> > how to extract b[i][0:1] and b[i][1:2] as parameters?
> 
> I don't understand the question.
> 
> 
> > def node(mmm, b[i][0:1], b[i][1:2]):
> >  H2 = [MM[mmm][A+B] for i in range(len(b))]
> >  return H2
> > 
> > node(5, b[i][0:1], b[i][1:2])
> 
> Explain what node() is supposed to do, in English. Don't write any code yet. 
> What is its purpose? What does it return? What arguments does it need to take 
> in order to perform its purpose?
> 
> 
> -- 
> Steve
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how to extract a variable as parameter which has index using by a for loop?

[meInvent bbird]

b = [str(i)+str(j)+str(k) for i in range(m) for j in range(m) for k in range(m)]
b[21][0:1]+b[21][1:2]
b[21][1:2]+b[21][2:3]
b[21][0:1]+b[21][2:3]


originally,

mmm = 5
H2 = [MM[mmm][b[i][0:1]+b[i][1:2]] for i in range(len(b))]

how to extract b[i][0:1] and b[i][1:2] as parameters?

def node(mmm, b[i][0:1], b[i][1:2]):
 H2 = [MM[mmm][A+B] for i in range(len(b))]
 return H2

node(5, b[i][0:1], b[i][1:2])
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Re: why for loop print only once after add if statement

[meInvent bbird]

when read my code again, i discover this error i fixed before

but forget to do for another branch of if statement

now fixed

On Saturday, May 28, 2016 at 6:19:23 PM UTC+8, meInvent bbird wrote:
> for item, i in enumerate(aa)
>   print item
> 
> this writing, it can print all values
> 
> for item, i in enumerate(aa)
>   if item == findit:
> print item
> 
> this only print the first value, means it only print once then not print 
> again,
> 
> where is wrong?

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Re: why for loop print only once after add if statement

[meInvent bbird]

thanks, i discover that i misunderstand i and item,

they should be swapped

On Saturday, May 28, 2016 at 6:19:23 PM UTC+8, meInvent bbird wrote:
> for item, i in enumerate(aa)
>   print item
> 
> this writing, it can print all values
> 
> for item, i in enumerate(aa)
>   if item == findit:
> print item
> 
> this only print the first value, means it only print once then not print 
> again,
> 
> where is wrong?

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how to write code generator for Isabelle by using pygments?

[meInvent bbird]

how to write code generator for Isabelle by using pygments?

i am thinking to write a machine learning code to generate code 
by learning example from Isabelle code

however, after google, not much information about this.
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