Re: Get Count of function arguments passed in
On Wednesday, 11 September 2019 20:25:32 UTC+10, Sayth Renshaw wrote:
> On Wednesday, 11 September 2019 20:11:21 UTC+10, Sayth Renshaw wrote:
> > Hi
> >
> > I want to allow as many lists as needed to be passed into a function.
> > But how can I determine how many lists have been passed in?
> >
> > I expected this to return 3 but it only returned 1.
> >
> > matrix1 = [[1, -2], [-3, 4],]
> > matrix2 = [[2, -1], [0, -1]]
> > matrix3 = [[2, -1], [0, -1]]
> > # print(add(matrix1, matrix2))
> >
> > def add(*matrix):
> > print(len(locals()))
> >
> > add(matrix1,matrix2,matrix3)
> >
> > Cheers
> >
> > Sayth
>
> Tried creating a list of the arguments, however I am creating too many
> positional arguments.
>
> matrix1 = [[1, -2], [-3, 4],]
> matrix2 = [[2, -1], [0, -1]]
> matrix3 = [[2, -1], [0, -1]]
>
> matrix = []
> def add_many(a = list(*matrix)):
> for num in a:
> for i in range(len(matrix[num])):
> for j in range(len(matrix[num])):
> print(matrix[num][i][j] + matrix[num][i][j])
>
> add_many(matrix1,matrix2)
Last failure for the moment
matrix1 = [[1, -2], [-3, 4],]
matrix2 = [[2, -1], [0, -1]]
matrix3 = [[2, -1], [0, -1]]
matrix = []
def add_many(*matrix):
for num in range(add_many().func_code.co_argcount):
for i in range(len(matrix[num])):
for j in range(len(matrix[num])):
print(matrix[i][j] + matrix[i][j])
add_many(matrix1,matrix2)
Sayth
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Re: Get Count of function arguments passed in
On Wednesday, 11 September 2019 20:11:21 UTC+10, Sayth Renshaw wrote: > Hi > > I want to allow as many lists as needed to be passed into a function. > But how can I determine how many lists have been passed in? > > I expected this to return 3 but it only returned 1. > > matrix1 = [[1, -2], [-3, 4],] > matrix2 = [[2, -1], [0, -1]] > matrix3 = [[2, -1], [0, -1]] > # print(add(matrix1, matrix2)) > > def add(*matrix): > print(len(locals())) > > add(matrix1,matrix2,matrix3) > > Cheers > > Sayth Tried creating a list of the arguments, however I am creating too many positional arguments. matrix1 = [[1, -2], [-3, 4],] matrix2 = [[2, -1], [0, -1]] matrix3 = [[2, -1], [0, -1]] matrix = [] def add_many(a = list(*matrix)): for num in a: for i in range(len(matrix[num])): for j in range(len(matrix[num])): print(matrix[num][i][j] + matrix[num][i][j]) add_many(matrix1,matrix2) -- https://mail.python.org/mailman/listinfo/python-list
Get Count of function arguments passed in
Hi I want to allow as many lists as needed to be passed into a function. But how can I determine how many lists have been passed in? I expected this to return 3 but it only returned 1. matrix1 = [[1, -2], [-3, 4],] matrix2 = [[2, -1], [0, -1]] matrix3 = [[2, -1], [0, -1]] # print(add(matrix1, matrix2)) def add(*matrix): print(len(locals())) add(matrix1,matrix2,matrix3) Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: pandas loc on str lower for column comparison
On Tuesday, 10 September 2019 12:56:36 UTC+10, Sayth Renshaw wrote: > On Friday, 6 September 2019 07:52:56 UTC+10, Piet van Oostrum wrote: > > Piet van Oostrum <> writes: > > > > > That would select ROWS 0,1,5,6,7, not columns. > > > To select columns 0,1,5,6,7, use two-dimensional indexes > > > > > > df1 = df.iloc[:, [0,1,5,6,7]] > > > > > > : selects all rows. > > > > And that also solves your original problem. > > > > This statement: > > > > df1['Difference'] = df1.loc['Current Team'].str.lower().str.strip() == > > df1.loc['New Team'].str.lower().str.strip() > > > > should not use .loc, because then you are selecting rows, not columns, but: > > > > df1['Difference'] = df1['Current Team'].str.lower().str.strip() == df1['New > > Team'].str.lower().str.strip() > > -- > > Piet van Oostrum <> > > WWW: http://piet.vanoostrum.org/ > > PGP key: [8DAE142BE17999C4] > > That actually creates another error. > > A value is trying to be set on a copy of a slice from a DataFrame. > Try using .loc[row_indexer,col_indexer] = value instead > > See the caveats in the documentation: > http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy > > So tried this > df['c'] = df.apply(lambda df1: df1['Current Team'].str.lower().str.strip() == > df1['New Team'].str.lower().str.strip(), axis=1) > > Based on this SO answer https://stackoverflow.com/a/46570641 > > Thoughts? > > Sayth This works on an individual row df2 = df1.loc[(df1['Current Team'] == df1['New Team']),'D'] = 'Wow' But how do I apply it to the whole new column and return the new dataset? Trying to use lambda but it cannot contain assigment df2 = df1.apply(lambda df1: [ (df1['Current Team'] == df1['New Team']) ]['D'] = 'succeed') df2 Confused Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: pandas loc on str lower for column comparison
On Friday, 6 September 2019 07:52:56 UTC+10, Piet van Oostrum wrote: > Piet van Oostrum <> writes: > > > That would select ROWS 0,1,5,6,7, not columns. > > To select columns 0,1,5,6,7, use two-dimensional indexes > > > > df1 = df.iloc[:, [0,1,5,6,7]] > > > > : selects all rows. > > And that also solves your original problem. > > This statement: > > df1['Difference'] = df1.loc['Current Team'].str.lower().str.strip() == > df1.loc['New Team'].str.lower().str.strip() > > should not use .loc, because then you are selecting rows, not columns, but: > > df1['Difference'] = df1['Current Team'].str.lower().str.strip() == df1['New > Team'].str.lower().str.strip() > -- > Piet van Oostrum <> > WWW: http://piet.vanoostrum.org/ > PGP key: [8DAE142BE17999C4] That actually creates another error. A value is trying to be set on a copy of a slice from a DataFrame. Try using .loc[row_indexer,col_indexer] = value instead See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy So tried this df['c'] = df.apply(lambda df1: df1['Current Team'].str.lower().str.strip() == df1['New Team'].str.lower().str.strip(), axis=1) Based on this SO answer https://stackoverflow.com/a/46570641 Thoughts? Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: pandas loc on str lower for column comparison
That is actually consistent with Excel row, column. Can see why it works that way then. Thanks -- https://mail.python.org/mailman/listinfo/python-list
Re: pandas loc on str lower for column comparison
On Sunday, 1 September 2019 10:48:54 UTC+10, Sayth Renshaw wrote: > I've created a share doc same structure anon data from my google drive. > > https://drive.google.com/file/d/0B28JfFTPNr_lckxQRnFTRF9UTEFYRUVqRWxCNVd1VEZhcVNr/view?usp=sharing > > Sayth I tried creating the df1 dataframe by using iloc instead of loc to avoid any column naming issues. So i created a list of integers for iloc representing the columns in current example. df1 = df.iloc[[0,1,5,6,7]] However, I ust be misunderstanding the docs https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.iloc.html#pandas.DataFrame.iloc Allowed inputs are: An integer, e.g. 5. A list or array of integers, e.g. [4, 3, 0]. Because while it works I appear to grab all columns 13 when I requested 5. UID NameFTE Agent IDCurrent Leader New Leader Current TeamNew TeamCurrent SiteNew SiteUnnamed: 10 Unnamed: 11 Unnamed: 12 How do I misunderstand iloc? Thanks, Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Need help: integrating unittest with setuptools
On Monday, 2 September 2019 04:44:29 UTC+10, YuXuan Dong wrote: > Hi, everybody: > > I have met a problem while I ran `python setup.py test`: > > unittest.case.SkipTest: No module named 'winreg' > > I ran the command in MacOS and my project is written for only UNIX-like > systems. I don't use any Windows-specified API. How dose `winreg` come here? > > In my `setup.py`: > > test_suite="test" > > In my `test/test.py`: > > import unittest > > class TestAll(unittest.TestCase): > def testall(self): > return None > > It works if I ran `python -m uniittest test.py` alone but raises the above > exception if I ran `python setup.py test`. > > I'm working on this for the whole day, searching for every keywords I can > think of with Google but can't find why or how. Could you help me? Thanks. > > -- > YX. D. Does this help? https://stackoverflow.com/questions/4320761/importerror-no-module-named-winreg-python3 Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: How to create a list and append to list inside a mqtt and GUI program?
me__ == "__main__":
> app = QApplication(sys.argv)
> mainWin = MainWindow()
> mainWin.show()
> publishedMessage = mainWin.getGUIFilename()
> sys.exit(app.exec_())
>
> MQTT.py
> import logging
> from datetime import timedelta
> import time
> from thespian.actors import *
> from transitions import Machine
> import paho.mqtt.client as mqtt
> import importlib
> import os.path
> import sys
> from PyQt5.QtWidgets import *
> from PyQt5.QtCore import *
> from PyQt5 import QtWidgets, uic
>
> class device(mqtt.Client):
> def on_connect(self, mqttc, obj, flags, rc):
> if rc == 0:
> print("Connected to broker")
> else:
> print("Connection failed")
>
> # mqttc.subscribe("microscope/light_sheet_microscope/UI")
>
> def on_message(self, mqttc, userdata, message):
> msg = str(message.payload.decode("utf-8"))
> print("message recieved= " + msg)
> # print("File which you want to import(with .py extension)")
> print("message topic=", message.topic)
> print("message qos=", message.qos)
> print("message retain flag=", message.retain)
>
> def run(self):
> self.connect("broker.hivemq.com", 1883, 60)
>
> test1.py
>
> from PyQt5 import QtCore, QtGui, QtWidgets
> import sys
> from PyQt5.QtWidgets import *
> from PyQt5.QtCore import *
> from PyQt5 import QtWidgets, uic
> from mqtt import *
>
> class SubWindow(QWidget):
> def __init__(self, parent = None):
> super(SubWindow, self).__init__(parent)
> self.setMinimumSize(QSize(300, 200))
> label = QLabel("Laser", self)
>
> self.modeButton = QtWidgets.QPushButton("Click me",self)
> self.modeButton.setGeometry(QtCore.QRect(10, 40, 81, 23))
> self.modeButton.setObjectName("Turn on")
> self.modeButton.clicked.connect(self.modFun)
>
> def modFun(self):
> print("creating new instance " + "Laser")
> client = device("Laser")
> client.run()
>
> client.loop_start() # start the loop
> device_message = "ON"
> time.sleep(2)
> print("Subscribing to topic",
> "microscope/light_sheet_microscope/UI")
> client.subscribe("microscope/light_sheet_microscope/UI")
> print("Publishing message to topic",
> "microscope/light_sheet_microscope/UI")
> client.publish("microscope/light_sheet_microscope/UI",
> device_message)
> time.sleep(2) # wait
> client.loop_stop() # stop the loop
>
> def closeEvent(self, event):
> self.close()
>
> test2.py
>
> from PyQt5 import QtCore, QtGui, QtWidgets
> import sys
> from PyQt5.QtWidgets import *
> from PyQt5.QtCore import *
> from PyQt5 import QtWidgets, uic
>
> class SubWindow(QWidget):
> def __init__(self, parent = None):
> super(SubWindow, self).__init__(parent)
> self.setMinimumSize(QSize(300, 250))
> label = QLabel("Sub Window", self)
>
> self.modeButton = QtWidgets.QPushButton("Click me",self)
> self.modeButton.setGeometry(QtCore.QRect(10, 40, 81, 23))
>
> self.modeButton.clicked.connect(self.modFun)
>
>
>
> def modFun(self):
> print("Hello there i'm Click me")
>
> def closeEvent(self, event):
> self.close()
>
> Thanks.
list_of_file_name = []
my_file = getGUIFilename()
list.append(my_file)
Sayth
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Re: pandas loc on str lower for column comparison
I've created a share doc same structure anon data from my google drive. https://drive.google.com/file/d/0B28JfFTPNr_lckxQRnFTRF9UTEFYRUVqRWxCNVd1VEZhcVNr/view?usp=sharing Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: pandas loc on str lower for column comparison
On Sunday, 1 September 2019 05:19:34 UTC+10, Piet van Oostrum wrote:
> Sayth Renshaw writes:
>
> > But on both occasions I receive this error.
> >
> > # KeyError: 'the label [Current Team] is not in the [index]'
> >
> > if I test df1 before trying to create the new column it works just fine.
> >
> What do you mean by testing df1?
>
> And could it be that 'Current Team' is spelled differently in the assignment
> than in the construction of df1? For example a difference in spaces, like a
> triling space or a breaking vs. non-breaking space? Please check that both
> are exactly the same.
>
> --
> Piet van Oostrum
> WWW: http://piet.vanoostrum.org/
> PGP key: [8DAE142BE17999C4]
Hi
Version info
1.1.0 #scipy
0.23.4 #pandas
RangeIndex: 35 entries, 0 to 34
Data columns (total 5 columns):
UID 35 non-null object
Name35 non-null object
New Leader 35 non-null object
Current Team35 non-null object
New Team35 non-null object
dtypes: object(5)
memory usage: 1.4+ KB
I had to anonymise the sheet. But same structure. tested produces same errors.
import pandas as pd
import scipy
print(scipy.version.version)
print(pd.__version__)
xls = pd.ExcelFile("Melbourne-anon.xlsx")
df = xls.parse('Sheet1', skiprows= 4)
# df1 = df.loc[:, ('UID','Name','New Leader','Current Team', 'New Team')]
# print(df['Current Team'] == df1['Current Team'])
df1 = df.loc[:, ('UID','Name','New Leader','Current Team', 'New Team')]
# df1['Difference'] = df1.loc['Current Team'].str.lower().str.strip() ==
df1.loc['New Team'].str.lower().str.strip()
# df1 = df[['UID','Name','New Leader','Current Team', 'New Team']].copy()
df1['Difference'] = df1.loc['Current Team'].str.lower().str.strip() ==
df1.loc['New Team'].str.lower().str.strip()
df1.info()
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Re: pandas loc on str lower for column comparison
On Friday, 30 August 2019 00:49:32 UTC+10, Piet van Oostrum wrote:
> Piet van Oostrum writes:
>
> > So the correct way to do this is to make df1 a copy rather than a view.
> >
> > df1 = df.loc[:, ('UID','Name','New Leader','Current Team', 'New Team')]
>
> Or maybe even make an explicit copy:
>
> df1 = df[['UID','Name','New Leader','Current Team', 'New Team']].copy()
> --
> Piet van Oostrum
> WWW: http://piet.vanoostrum.org/
> PGP key: [8DAE142BE17999C4]
I have tried both
df1 = df.loc[:, ('UID','Name','New Leader','Current Team', 'New Team')]
df1['Difference'] = df1.loc['Current Team'].str.lower().str.strip() ==
df1.loc['New Team'].str.lower().str.strip()
and
df1 = df[['UID','Name','New Leader','Current Team', 'New Team']].copy()
df1['Difference'] = df1.loc['Current Team'].str.lower().str.strip() ==
df1.loc['New Team'].str.lower().str.strip()
But on both occasions I receive this error.
# KeyError: 'the label [Current Team] is not in the [index]'
if I test df1 before trying to create the new column it works just fine.
Sayth
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Re: [SOLVED] Re: Compare zip lists where order is important
On Thursday, 29 August 2019 20:33:46 UTC+10, Peter Otten wrote:
> Sayth Renshaw wrote:
>
> > will find the added
> > pairs, but ignore the removed ones. Is that what you want?
> >
> > Yes, I think. I want to find the changed pairs. The people that moved team
> > numbers.
>
> To find the people that moved team numbers I would tear the pairs apart.
> Like:
>
> >>> people = ["Tim","Bill","Sally","Ally","Fred","Fredricka"]
> >>> team_number = [1,1,2,2,3,3]
> >>> shuffle_people = ["Fredricka","Bill","Sally","Tim","Ally","Fred"]
> >>> shuffle_team_number = [1,1,2,2,3,3]
> >>> old = dict(zip(people, team_number))
> >>> new = dict(zip(shuffle_people, shuffle_team_number))
> >>> for name in old.keys() & new.keys():
> ... old_team = old[name]
> ... new_team = new[name]
> ... if old_team != new_team:
> ... print(name, "went from", old_team, "to", new_team)
> ...
> Tim went from 1 to 2
> Fredricka went from 3 to 1
> Ally went from 2 to 3
The reason I opted away from Dictionaries is if there was a team with people
with same name. Then the keys would be the same.
So if Sally left and team 2 had one Tim move in and a new Tim start.
shuffle_people = ["Fredricka","Bill","Tim","Tim","Ally","Fred"]
shuffle_team_number = [1,1,2,2,3,3]
becomes
{'Fredricka': 1, 'Bill': 1, 'Tim': 2, 'Ally': 3, 'Fred': 3}
This still appears to work but is wrong.
for name in old.keys()& new.keys():
old_team = old[name]
new_team = new[name]
if old_team != new_team:
print(name, "went from", old_team, "to", new_team)
Ally went from 2 to 3
Tim went from 1 to 2
Fredricka went from 3 to 1
But I guess in reality I would use a UID and then look up the UID in a list or
database.
Cheers
Sayth
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Re: pandas loc on str lower for column comparison
On Friday, 30 August 2019 00:49:32 UTC+10, Piet van Oostrum wrote:
> Piet van Oostrum
>
> > So the correct way to do this is to make df1 a copy rather than a view.
> >
> > df1 = df.loc[:, ('UID','Name','New Leader','Current Team', 'New Team')]
>
> Or maybe even make an explicit copy:
>
> df1 = df[['UID','Name','New Leader','Current Team', 'New Team']].copy()
Thank you so much.
Sayth
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Re: [SOLVED] Re: Compare zip lists where order is important
will find the added pairs, but ignore the removed ones. Is that what you want? Yes, I think. I want to find the changed pairs. The people that moved team numbers. Sayth -- https://mail.python.org/mailman/listinfo/python-list
pandas loc on str lower for column comparison
Hi
I am importing 4 columns into a dataframe from a spreadsheet.
My goal is to create a 5th column with TRUE or False if column 4 (str) matches
column 3.
Trying to leverage this answer https://stackoverflow.com/a/35940955/461887
This is my code
import pandas as pd
xls = pd.ExcelFile("Melbourne.xlsx")
df = xls.parse('Sheet1', skiprows= 4)
df1 = df[['UID','Name','New Leader','Current Team', 'New Team']]
df1['Difference'] = df1['Current Team'].str.lower().str.replace('s/+',"") ==
df1['New Team'].str.lower().str.replace('s/+',"")
Which gives this error
C:\Users\u369811\AppData\Local\Continuum\anaconda3\lib\site-packages\ipykernel_launcher.py:6:
SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
See the caveats in the documentation:
http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
If I update the line to use loc as this I still receive a long error.
df1['Difference'] = df1.loc['Current Team'].str.lower().str.replace('s/+',"")
== df1.loc['New Team'].str.lower().str.replace('s/+',"")
What am I missing?
Sayth
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[SOLVED] Re: Compare zip lists where order is important
On Thursday, 29 August 2019 14:03:44 UTC+10, Sayth Renshaw wrote:
> On Thursday, 29 August 2019 13:53:43 UTC+10, Sayth Renshaw wrote:
> > On Thursday, 29 August 2019 13:25:01 UTC+10, Sayth Renshaw wrote:
> > > Hi
> > >
> > > Trying to find whats changed in this example. Based around work and team
> > > reschuffles.
> > >
> > > So first I created my current teams and then my shuffled teams.
> > >
> > > people = ["Tim","Bill","Sally","Ally","Fred","Fredricka"]
> > > team_number = [1,1,2,2,3,3]
> > >
> > > shuffle_people = ["Fredricka","Bill","Sally","Tim","Ally","Fred"]
> > > shuffle_team_number = [1,1,2,2,3,3]
> > >
> > > Then combine.
> > >
> > > teams = list(zip(people,team_number))
> > > shuffle_teams = list(zip(shuffle_people, shuffle_team_number))
> > >
> > > Then I am attempting to compare for change.
> > >
> > > [i for i, j in zip(teams, shuffle_teams) if i != j]
> > >
> > > #Result
> > > [('Tim', 1), ('Ally', 2), ('Fred', 3), ('Fredricka', 3)]
> > >
> > > #Expecting to see
> > >
> > > [('Fredricka', 1),('Tim', 2)]
> > >
> > > What's a working way to go about this?
> > >
> > > Sayth
> >
> > It looks like Tuples are comparing by position changes not content changes.
> >
> > So this fails too
> >
> > set(shuffle_teams) & set(teams)
> > # {('Bill', 1), ('Fred', 3), ('Sally', 2)}
>
> Well this works although its not clear which line is the change.
>
> set(teams).symmetric_difference(set(shuffle_teams))
>
> {('Ally', 2),
> ('Ally', 3), # This is the change Ally changed from 2 to 3
> ('Fredricka', 1), # However here this first line is the change.
> ('Fredricka', 3),
> ('Tim', 1),
> ('Tim', 2)}
>
> Hints?
set(shuffle_teams).difference(set(teams))
{('Ally', 3), ('Fredricka', 1), ('Tim', 2)}
Sayth
--
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Re: Compare zip lists where order is important
On Thursday, 29 August 2019 13:53:43 UTC+10, Sayth Renshaw wrote:
> On Thursday, 29 August 2019 13:25:01 UTC+10, Sayth Renshaw wrote:
> > Hi
> >
> > Trying to find whats changed in this example. Based around work and team
> > reschuffles.
> >
> > So first I created my current teams and then my shuffled teams.
> >
> > people = ["Tim","Bill","Sally","Ally","Fred","Fredricka"]
> > team_number = [1,1,2,2,3,3]
> >
> > shuffle_people = ["Fredricka","Bill","Sally","Tim","Ally","Fred"]
> > shuffle_team_number = [1,1,2,2,3,3]
> >
> > Then combine.
> >
> > teams = list(zip(people,team_number))
> > shuffle_teams = list(zip(shuffle_people, shuffle_team_number))
> >
> > Then I am attempting to compare for change.
> >
> > [i for i, j in zip(teams, shuffle_teams) if i != j]
> >
> > #Result
> > [('Tim', 1), ('Ally', 2), ('Fred', 3), ('Fredricka', 3)]
> >
> > #Expecting to see
> >
> > [('Fredricka', 1),('Tim', 2)]
> >
> > What's a working way to go about this?
> >
> > Sayth
>
> It looks like Tuples are comparing by position changes not content changes.
>
> So this fails too
>
> set(shuffle_teams) & set(teams)
> # {('Bill', 1), ('Fred', 3), ('Sally', 2)}
Well this works although its not clear which line is the change.
set(teams).symmetric_difference(set(shuffle_teams))
{('Ally', 2),
('Ally', 3), # This is the change Ally changed from 2 to 3
('Fredricka', 1), # However here this first line is the change.
('Fredricka', 3),
('Tim', 1),
('Tim', 2)}
Hints?
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Re: Compare zip lists where order is important
On Thursday, 29 August 2019 13:25:01 UTC+10, Sayth Renshaw wrote:
> Hi
>
> Trying to find whats changed in this example. Based around work and team
> reschuffles.
>
> So first I created my current teams and then my shuffled teams.
>
> people = ["Tim","Bill","Sally","Ally","Fred","Fredricka"]
> team_number = [1,1,2,2,3,3]
>
> shuffle_people = ["Fredricka","Bill","Sally","Tim","Ally","Fred"]
> shuffle_team_number = [1,1,2,2,3,3]
>
> Then combine.
>
> teams = list(zip(people,team_number))
> shuffle_teams = list(zip(shuffle_people, shuffle_team_number))
>
> Then I am attempting to compare for change.
>
> [i for i, j in zip(teams, shuffle_teams) if i != j]
>
> #Result
> [('Tim', 1), ('Ally', 2), ('Fred', 3), ('Fredricka', 3)]
>
> #Expecting to see
>
> [('Fredricka', 1),('Tim', 2)]
>
> What's a working way to go about this?
>
> Sayth
It looks like Tuples are comparing by position changes not content changes.
So this fails too
set(shuffle_teams) & set(teams)
# {('Bill', 1), ('Fred', 3), ('Sally', 2)}
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Re: How to use regex to search string between {}?
> > A site like http://www.pyregex.com/ allows you to check your regex with > slightly fewer clicks and keystrokes than editing your program. Thanks Jason -- https://mail.python.org/mailman/listinfo/python-list
Compare zip lists where order is important
Hi
Trying to find whats changed in this example. Based around work and team
reschuffles.
So first I created my current teams and then my shuffled teams.
people = ["Tim","Bill","Sally","Ally","Fred","Fredricka"]
team_number = [1,1,2,2,3,3]
shuffle_people = ["Fredricka","Bill","Sally","Tim","Ally","Fred"]
shuffle_team_number = [1,1,2,2,3,3]
Then combine.
teams = list(zip(people,team_number))
shuffle_teams = list(zip(shuffle_people, shuffle_team_number))
Then I am attempting to compare for change.
[i for i, j in zip(teams, shuffle_teams) if i != j]
#Result
[('Tim', 1), ('Ally', 2), ('Fred', 3), ('Fredricka', 3)]
#Expecting to see
[('Fredricka', 1),('Tim', 2)]
What's a working way to go about this?
Sayth
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Re: Enumerate - int object not subscriptable
On Wednesday, 21 August 2019 03:16:01 UTC+10, Ian wrote: > Or use the "pairwise" recipe from the itertools docs: > > from itertools import tee > > def pairwise(iterable): > "s -> (s0,s1), (s1,s2), (s2, s3), ..." > a, b = tee(iterable) > next(b, None) > return zip(a, b) > > for num1, num2 in pairwise(a): > print(num1, num2) > > > Stanley C. Kitching > > Human Being > > Phoenix, Arizona > > > > -- This definitely ended up being the ticket. def pairwise(iterable): "s -> (s0,s1), (s1,s2), (s2, s3), ..." a, b = tee(iterable) next(b, None) return zip(a, b) def output_data(s): serie = fibo(input_length) x = [] y = [] for num1, num2 in pairwise(serie): y.append( num2 / num1) for item in y: x.append(y.index(item)) draw_graph(x, y) Kept trying with the enumerate style for i, num in enumerate(serie): while i < len(serie)-1: x.append(int(serie[i + 1])/ int(serie[i])) i += 1 but just couldn't quite get it to work. Itertools definitely made it easier. for i, num in enumerate(serie): while i < len(serie)-1: x.append(int(serie[i + 1])/ int(serie[i])) i += 1 -- https://mail.python.org/mailman/listinfo/python-list
Enumerate - int object not subscriptable
Hi I want to do basic math with a list. a = [1, 2, 3, 4, 5, 6, 7, 8] for idx, num in enumerate(a): print(idx, num) This works, but say I want to print the item value at the next index as well as the current. for idx, num in enumerate(a): print(num[idx + 1], num) I am expecting 2, 1. But am receiving TypeError: 'int' object is not subscriptable Why? Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Copying a row from a range of Excel files to another
Cecil Westerhof wrote:
> I was asked to copy a certain line from about 300 Excel lines to a new
> Excel file. That is not something I would like to do by hand and I
> immediately thought: that should be possible with Python.
>
> And it is. I was surprised how fast I could write that with openpyxl.
> My first try was not very neat, but a proof of concept. Then by
> looking better at the possibilities I could get much cleaner code. But
> I am still not completely happy. At the moment I have the following
> code:
> wb_out = Workbook()
> for filepath in filepathArr:
> current_row = []
> wb_in = load_workbook(filepath)
> for cell in wb_in.active[src_row]:
> current_row.append(cell.value)
> wb_out.active.append(current_row)
> wb_in.close()
> wb_out.save(report_start + datetime.now().strftime('%Y-%m-%d') +
> report_end)
> wb_out.close()
>
> I could not find a way to copy a row from one workbook to another.
> That is why I put the row in current_row and do an append. Am I
> overlooking something, or is that really the way to do this?
>
>
>
> --
> Cecil Westerhof
Pandas may work out better.
import pandas as pd
df = pd.read_excel('spreadsheet')
# find the row by filtering with regex
df2 = df.'column'.str.contains('criteria as regex')
df2.pd.save_excel('output.xlsx')
Cheers
Sayth
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Re: for line3 in myips matching too longer matches.
Chris Roberts wrote:
> ###
> CODE:
>elif line1.rstrip(‘\n’) in line2.strip(‘\n’):
>for line3 in myips:
>print “###”
>print “line1 is %s” % line1.rstrip(‘\n’)
>print “line2 is %s” % line2.strip(‘\n’)
> ###
> OUTPUT:
> line1 is 10.10.168.2
> line2 is - address: 10.10.168.27 # myhost
> ###
>
> I think the problem is here:
> line1.rstrip(‘\n’) in line2.strip(‘\n’):
>
> I want it to
> match only 10.10.168.2 AND 10.10.168.2:
> NOT match 10.10.168.2[0-9]
>
> If someone out there knows a simple solution. I would love to see it.
> Thanks in advance.
> crzzy1
Not sure exactly what the input is but a comprehension would do this.
[x for x in input_line.split(' ') if == '10.10.168.2']
Cheers
Sayth
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Re: pandas split and melt()
>
> Since I didn't find a cool shortcut I decided to use brute force:
>
> $ cat pandas_explode_column.py
> import pandas as pd
>
> df = pd.DataFrame(
> [
> [
> "2019-06-21 11:15:00",
> "WNEWSKI, Joan;#17226;#BALIN, Jock;#18139;#DUNE, Colem;#17230;"
> ],
> [
> "2019-06-22 10:00:00", "Doe, John;#42;Robbins, Rita;"
> ]
> ],
> columns=["Session date", "Consultant"]
> )
>
> def explode_consultants(consultants):
> consultants = (c.lstrip("#") for c in consultants.split(";"))
> return (c for c in consultants if c.strip("0123456789"))
>
> def explode_column(df, column, split):
> for _index, row in df.iterrows():
> for part in split(row[column]):
> yield [part if c == column else row[c] for c in df.columns]
>
> def explode(df, column, split):
> return pd.DataFrame(
> explode_column(df, "Consultant", split), columns=df.columns
> )
>
> df2 = explode(df, "Consultant", explode_consultants)
>
> print(df)
> print(df2)
> $ python3 pandas_explode_column.py
> Session date Consultant
> 0 2019-06-21 11:15:00 WNEWSKI, Joan;#17226;#BALIN, Jock;#18139;#DUNE...
> 1 2019-06-22 10:00:00 Doe, John;#42;Robbins, Rita;
>
> [2 rows x 2 columns]
> Session date Consultant
> 0 2019-06-21 11:15:00 WNEWSKI, Joan
> 1 2019-06-21 11:15:00BALIN, Jock
> 2 2019-06-21 11:15:00DUNE, Colem
> 3 2019-06-22 10:00:00 Doe, John
> 4 2019-06-22 10:00:00 Robbins, Rita
>
> [5 rows x 2 columns]
> $
Mind a little blown :-). Going to have to play and break this several times to
fully get it.
Thanks
Sayth
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Re: pandas split and melt()
Update.
Option 1. - This annihilates all text in the column leaving nothing.
completed_tasks['Consultant'] =
completed_tasks['Consultant'].str.rstrip('.#123')
Option 2. - returns unhashable series list.
output =
completed_tasks[completed_tasks['Consultant']].str.contains(r'/\b[^\d\W]+\b/g')
neither option works as the column is a list.
Thanks
Sayth
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Re: pandas split and melt()
> > NB. There are varied amounts of consultants so splitting across columns is > > uneven. if it was even melt seems like it would be good > > https://dfrieds.com/data-analysis/melt-unpivot-python-pandas > > Keep it simple: https://docs.python.org/3.6/library/string.html > > -- > Regards =dn I have managed to split but not filter the numbers. All other code being the same. ... completed_tasks = df.loc[(df['Completed'] == 'Yes') & (df['Motor/Property'] == 'Motor') & (df['Delivery Method'] == 'Group Coaching')] completed_tasks['Consultant'] = completed_tasks['Consultant'].str.split(pat = ";") pattern = "\\#d" completed_tasks['Consultant'] = completed_tasks['Consultant'].str.split(pat = ";") completed_tasks['Consultant'] = completed_tasks['Consultant'].str.contains(pattern) ... Works without the regex which causes this error AttributeError: Can only use .str accessor with string values, which use np.object_ dtype pattern = "\\#d" completed_tasks['Consultant'] = completed_tasks['Consultant'].str.split(pat = ";") Sayth -- https://mail.python.org/mailman/listinfo/python-list
pandas split and melt()
Hi
Having fun with pandas filtering a work excel file.
My current script opens selected and filters the data and saves as excel.
import pandas as pd
import numpy as np
log = pd.read_excel("log_dump_py.xlsx")
df = log.filter(items=['Completed', 'Priority', 'Session date', 'Consultant',
'Coach',
'Delivery Method', 'Focus of Coaching', 'Leader', 'Site',
'Coaching Description','Motor/Property',
],)
completed_tasks = df.loc[(df['Completed'] == 'Yes') & (df['Motor/Property'] ==
'Motor') & (df['Delivery Method'] == 'Group Coaching')]
print(completed_tasks.head(n=5))
completed_tasks.to_excel("filtered_logs.xlsx")
This leaves me with a set of several columns. The main column of concern for
this example is a consultant
Session dateConsultant
2019-06-21 11:15:00 WNEWSKI, Joan;#17226;#BALIN, Jock;#18139;#DUNE,
Colem;#17230;
How can I split the consultant column, keep only names and drop the numbers and
for every session date create a line with data and consultants name?
NB. There are varied amounts of consultants so splitting across columns is
uneven. if it was even melt seems like it would be good
https://dfrieds.com/data-analysis/melt-unpivot-python-pandas
Thanks
Sayth
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Re: generator to write N lines to file
> To get the individual lines you have to yield them > > for line in lines_gen: > yield line > > This can be rewritten with some syntactic sugar as > > yield from lines_gen > > > for line in getWord(fileName, 5): > > with open(dumpName, 'a') as f: > > f.write(line) > > Note that getWord as written does not close the infile when you break out of > the loop, e. g. > > for line in getWord(filename, 5): > break > > To avoid that you can use a context manager: > > from itertools import islice > from contextlib import contextmanager > > infile = ... > outfile = ... > > @contextmanager > def head(infile, numlines): > with open(infile) as f: > yield islice(f, numlines) > > with open(outfile, "w") as f: > with head(infile, 5) as lines: > f.writelines(lines) Thank You Peter great advice as always. Sayth -- https://mail.python.org/mailman/listinfo/python-list
generator to write N lines to file
Afternoon Trying to create a generator to write the first N lines of text to a file. However, I keep receiving the islice object not the text, why? from itertools import islice fileName = dir / "7oldsamr.txt" dumpName = dir / "dump.txt" def getWord(infile, lineNum): with open(infile, 'r+') as f: lines_gen = islice(f, lineNum) yield lines_gen for line in getWord(fileName, 5): with open(dumpName, 'a') as f: f.write(line) Thanks, Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Function to determine list max without itertools
> Now, what happens when the code is tested with various (different) sets
> of test-data?
> (remember the last question from my previous msg!?)
>
It fails on any list entry that isn't a float or an int, giving a TypeError.
> Once you are happy with the various test-runs, do you have any ideas for
> making the code more efficient?
> --
> Regards =dn
Efficient No, but resistant to error I can buy using try except to pass through
TypeErrors.
def max_try(listarg):
myMax = listarg[0]
try:
for item in listarg:
if item > myMax:
myMax = item
except TypeError:
print(f'Only numbers are supported, this entry "{item}" was not')
pass
return myMax
Thanks.
PS Since I am going through the list fully the only optimisation I can think of
a generator to feed it seems sort of redundant. Unless of course it was a list
of 10,000 numbers then maybe its useful.
Maybe a generator with a condition that if list length is greater than 500 to
chunk it into 250 lengths.
But then If we go along this path then you could say if every other task in an
application waited on it could be very important so then async await might be
an option.
Sayth
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Re: Function to determine list max without itertools
On Friday, 19 April 2019 17:01:33 UTC+10, Sayth Renshaw wrote: > Set the first item in the list as the current largest. > Compare each subsequent integer to the first. > if this element is larger, set integer. def maxitwo(listarg): myMax = listarg[0] for item in listarg: if item > myMax: myMax = item return myMax Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Function to determine list max without itertools
Set the first item in the list as the current largest. Compare each subsequent integer to the first. if this element is larger, set integer. -- https://mail.python.org/mailman/listinfo/python-list
Re: Function to determine list max without itertools
> > In English rather than Python, how do you find the maximum element in a > list? > > -- > Rob Gaddi, Highland Technology Get first 1 item in the list and compare it to the rest. If it is larger than rest its the max. However if another list member is larger it replaces the first item and comparison continues. Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Function to determine list max without itertools
> > > It's still overly complicated. > This is where I have ended up. Without itertools and max its what I got currently. def maximum(listarg): myMax = listarg[0] for item in listarg: for i in listarg[listarg.index(item)+1:len(listarg)]: if myMax < i: myMax = i return myMax How would you simplify it? -- https://mail.python.org/mailman/listinfo/python-list
Re: Function to determine list max without itertools
Thank you for the advice everyone. > > The first thing to try is find every place where you update myMax, and This was actually where I was going wrong. I was setting max but then overwriting it with item. Then kept checking item only to return myMax. I went looking for other solutions as I thought I must be well off the path in the shrubs but I was actually close. This is how I ended up. There may be better solutions but this works. def maximum(listarg): items = list(listarg) myMax = items[0] for item in items: for i in items[items.index(item)+1:len(items)]: if myMax < i: myMax = i else: pass return myMax if __name__ == "__main__": print(maximum([4,3,6,2,1,4])) Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Function to determine list max without itertools
wrote: > > > > > > I have created a function that takes a list as an argument. > > Without using itertools I want to compare each item in the list to find the > > max. > > > > However instead of the max I keep getting the last item in the list. Where > > is my logic wrong here? > > > > def maximum(listarg): > > items = list(listarg) > > myMax = 0 > > for index, item in enumerate(items): > > for otheritem in items[index + 1 :]: > > if item < otheritem: > > myMax = otheritem > > elif item > otheritem: > > myMax = item > > else: > > myMax = myMax > > > > Seems like it should work but doesn't. > > > > I'd recommend rethinking your algorithm first, and then thinking about > debugging your actual code. You should be able to find the largest in > a collection without making many many passes over the items - a single > pass should be sufficient. > > ChrisA Most I am finding either use the max function or itertools. Both of which I am trying not to use so that I actually do it myself. This one on SO is where I was going https://stackoverflow.com/a/3990826/461887 def maxelements(seq): ''' Return list of position(s) of largest element ''' max_indices = [] if seq: max_val = seq[0] for i,val in ((i,val) for i,val in enumerate(seq) if val >= max_val): if val == max_val: max_indices.append(i) else: max_val = val max_indices = [i] return max_indices This is probably the nicest one but still uses Max. >>> a=[5,4,3,2,1] >>> def eleMax(items, start=0, end=None): ... return max(items[start:end]) Thanks Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Importing module from another subdirectory
On Thursday, 18 April 2019 06:59:43 UTC+10, Rich Shepard wrote: > What is the proper syntax to import the model class in the model/ > subdirectory into a tkinter view module, e.g., activities.py? The syntax, > 'import model as m' fails because it is not in the same subdirectory as the > importing module. > > The program directory tree is: > > bustrac/ > README.rst > bustrac.py* > controller/ > model/ > scripts/ > views/ > > Running pdb in python3-3.7.3 produces the same error: > $ python3 > Python 3.7.3 (default, Mar 26 2019, 06:40:28) > [GCC 5.5.0] on linux > Type "help", "copyright", "credits" or "license" for more information. > >>> import pdb > >>> import activities_data_entry > Traceback (most recent call last): >File "", line 1, in >File > "/home/rshepard/development/business_tracker/views/activities_data_entry.py", > line 1, in > import model.activities as act > ModuleNotFoundError: No module named 'model' > > My previous use of python has had all files in the same directory so I've > not before had to learn how to address this issue. Pointers appreciated. > > Regards, > > Rich Morning Apologies I don't know the answer but went looking. This guide should answer the question. Didn't know it was so difficult to be honest. https://chrisyeh96.github.io/2017/08/08/definitive-guide-python-imports.html#example-directory-structure Then there was this rather long list of traps. http://python-notes.curiousefficiency.org/en/latest/python_concepts/import_traps.html Most guides say it was fixed in 3.3 but still seems quite confusing post 3.3 Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Function to determine list max without itertools
I have created a function that takes a list as an argument. Without using itertools I want to compare each item in the list to find the max. However instead of the max I keep getting the last item in the list. Where is my logic wrong here? def maximum(listarg): items = list(listarg) myMax = 0 for index, item in enumerate(items): for otheritem in items[index + 1 :]: if item < otheritem: myMax = otheritem elif item > otheritem: myMax = item else: myMax = myMax Seems like it should work but doesn't. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: hello this ali .. i want some question about python
On Saturday, 6 April 2019 08:21:51 UTC+11, maak khan wrote: > i need your help guys .. plz With? -- https://mail.python.org/mailman/listinfo/python-list
Re: I really liked this Javscript FizzBuzz can it be as nice in Python?
Wow in both of these examples I have no idea what is happening. Enjoying trying
to figure it out :-)
> print(*[[n,"Fizz","Buzz","Fizzbuzz"][int("300102100120100"[n%15])] for
> n in range(1,101)], sep="\n")
> This is not good code, and if anyone asks, I didn't say you were
> allowed to do this in an interview.
Sorry too late that was yesterday, they loved it :-)
> Here's some itertools gymnastics ;)
>
> from itertools import *
>
> def make(s, n):
> return cycle(chain(repeat("", n-1), (s,)))
>
> fizzbuzz = map("".join, zip(make("Fizz", 3), make("Buzz", 5)))
>
> for i, fb in enumerate(islice(fizzbuzz, 100), 1):
> print(fb or i)
Thanks
Sayth
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I really liked this Javscript FizzBuzz can it be as nice in Python?
I saw this fizzbuzz in Eloquent Javascript and thought its really nice. Not all
the usual if else version, just if.
for (let n = 1; n <= 100; n++) {
let output = "";
if (n % 3 == 0) output += "Fizz";
if (n % 5 == 0) output += "Buzz";
console.log(output || n);
}
I can't quite get a nice version like this out.
This was my attempt to mimick it. I am sure Python can get it cleaner, but I
had never thought about not doing if else for Fizzbuzz its just the way I did
it.
n = range(100)
output = ""
for num in n:
if (num % 3 == 0): output += "Fizz"
if (num % 5 == 0): output += "Buzz"
print(output or num)
Haven't quite got it. But is possible nice and succinct like the javascript
version. Maybe lambda will do it, gonna try that.
Any unique mindlowing style FizzBuzz I would never have considered and can
learn off?
Cheers
Sayth
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Re: scalable bottleneck
On Thursday, 4 April 2019 10:51:35 UTC+11, Paul Rubin wrote: > Sayth Renshaw writes: > > for x in range ( max_root ): > > 1) Do you see a memory bottleneck here? If so, what is it? > > 2) Can you think of a way to fix the memory bottleneck? > > In Python 2, range(n) creates a list in memory. To make an iterator you > would use xrange instead of range. In Python 3, range(n) is an iterator. > > Either way, fetch_squares creates a list of squares and not an iterator. > > > for square in fetch_squares (MAX ): > > do_something_with ( square ) > > Here's you're creating that list and throwing it away. So you have the > right idea with: > > def supply_squares(max_root): > for x in max_root: > yield x > > But supply_squares returns a generator which means you have to iterate > through it. > > print(supply_squares(item)) > > will print something like > > > > which is what you saw. You want: > >for square in supply_squares(MAX): > print square Ah thanks for your help. Happy I identified the correct bottleneck too. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
scalable bottleneck
In an email, I received this question as part of a newsletter. def fetch_squares ( max_root ): squares = [] for x in range ( max_root ): squares . append (x **2) return squares MAX = 5 for square in fetch_squares (MAX ): do_something_with ( square ) 1) Do you see a memory bottleneck here? If so, what is it? 2) Can you think of a way to fix the memory bottleneck? Want to know if I am trying to solve the correct bottleneck. I am thinking that the bottleneck is to create a list only to iterate the same list you created sort of doubling the time through. Is that the correct problem to solve? If it is then I thought the best way is just to supply the numbers on the fly, a generator. def supply_squares(max_root): for x in max_root: yield x MAX = 5 So then I set up a loop and do whatever is needed. At this time I am creating generator objects. But is this the correct way to go? More of a am I thinking correctly questino. item = 0 while item < MAX: print(supply_squares(item)) item += 1 Thanks Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Implement C's Switch in Python 3
On Sunday, 3 February 2019 14:31:14 UTC+11, Avi Gross wrote:
> Yes, you caught the usual flaw in the often irregular English language.
>
> The 11th, 12th and 13th do all begin with 1 so there is a simple fix in the
> function version by checking if day//10 is 1.
>
> Revised example:
>
> """ Use last digit to determine suffix handling teens well """
>
> def nthSuffix(day):
> if (day // 10 == 1):
> suffix = "th"
> else:
> nth = day % 10
> suffix = "st" if nth == 1 else ("nd" if nth == 2 else ("rd" if nth
> == 3 else "th"))
>
> return str(day) + suffix
>
> >>> [ nthSuffix(day) for day in range(1,32)]
> ['1st', '2nd', '3rd', '4th', '5th', '6th', '7th', '8th', '9th', '10th',
> '11th', '12th', '13th', '14th', '15th', '16th', '17th', '18th', '19th',
> '20th', '21st', '22nd', '23rd', '24th', '25th', '26th', '27th', '28th',
> '29th', '30th', '31st']
>
> Want a dictionary version? Use the above to make a full dictionary:
>
> >>> chooseFrom = { day : nthSuffix(day) for day in range(1,32)}
> >>> chooseFrom
> {1: '1st', 2: '2nd', 3: '3rd', 4: '4th', 5: '5th', 6: '6th', 7: '7th', 8:
> '8th', 9: '9th', 10: '10th', 11: '11th', 12: '12th', 13: '13th', 14: '14th',
> 15: '15th', 16: '16th', 17: '17th', 18: '18th', 19: '19th', 20: '20th', 21:
> '21st', 22: '22nd', 23: '23rd', 24: '24th', 25: '25th', 26: '26th', 27:
> '27th', 28: '28th', 29: '29th', 30: '30th', 31: '31st'}
> >>> chooseFrom[1]
> '1st'
> >>> chooseFrom[11]
> '11th'
> >>> chooseFrom[21]
> '21st'
>
> Subject: Re: Implement C's Switch in Python 3
>
> On 2019-02-03 02:51, Avi Gross wrote:
> > I may be missing something, but the focus seems to be only on the
> > rightmost digit. You can get that with
> >
> I had the same thought, but came across a problem. "11st", "12nd", "13rd"?
>
> [snip]
> >
> > Output:
> >
> >>>> for day in range(1, 32):
> > print( nthSuffix(day))
> >
> > 1st
> > 2nd
> > 3rd
> > 4th
> > 5th
> > 6th
> > 7th
> > 8th
> > 9th
> > 10th
> > 11st
> > 12nd
> > 13rd
> > 14th
> > 15th
> > 16th
> > 17th
> > 18th
> > 19th
> > 20th
> > 21st
> > 22nd
> > 23rd
> > 24th
> > 25th
> > 26th
> > 27th
> > 28th
> > 29th
> > 30th
> > 31st
> >
> [snip]
> --
I do like this.
Want a dictionary version? Use the above to make a full dictionary:
>>> chooseFrom = { day : nthSuffix(day) for day in range(1,32)}
>>> chooseFrom
{1: '1st', 2: '2nd', 3: '3rd', 4: '4th', 5: '5th', 6: '6th', 7: '7th', 8:
'8th', 9: '9th', 10: '10th', 11: '11th', 12: '12th', 13: '13th', 14: '14th',
15: '15th', 16: '16th', 17: '17th', 18: '18th', 19: '19th', 20: '20th', 21:
'21st', 22: '22nd', 23: '23rd', 24: '24th', 25: '25th', 26: '26th', 27:
'27th', 28: '28th', 29: '29th', 30: '30th', 31: '31st'}
>>> chooseFrom[1]
'1st'
>>> chooseFrom[11]
'11th'
>>> chooseFrom[21]
'21st'
Not having a default case as in switch forced you to write out all possible
combinations.
I think the intent and readbility of switch statements is a bit nicer.
Cheers
Sayth
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Re: Implement C's Switch in Python 3
> >I am trying to convert a switch statement from C into Python. (why?
> >practising).
> >
> >This is the C code.
> >
> >printf("Dated this %d", day);
> > switch (day) {
> >case 1: case 21: case 31:
> >printf("st"); break;
> >case 2: case 22:
> >printf("nd"); break;
> >case 3: case 23:
> >printf("rd"); break;
> >default: printf("th"); break;
> >
> > }
> > printf(" day of ");
> >
> >#Premise if the use enter an int as the date 21 for example it would print
> >21st. It appends the correct suffix onto a date.
> >Reading and trying to implement a function that uses a dictionary. Not
> >sure how to supply list into it to keep it brief and with default case
> >of 'th'.
> >
> >This is my current code.
> >
> >def f(x):
> >return {
> >[1, 21, 31]: "st",
> >[2, 22]: "nd",
> >[3, 23]: "rd",
> >}.get(x, "th")
> >
> >
> >print(f(21))
> >
> >I have an unhashable type list. Whats the best way to go?
>
> Skip has commented on lists being unhashable. We can elaborate on that
> if you like.
>
> However, even if you went to tuples (which would let you construct the
> dict you lay out above), there is another problem.
>
> You're looking up "x" in the dict. But the keys of the dict are not
> integers, they are lists (or tuples) or integers, so they won't match.
>
> You _could_ do this:
>
> return {
> 1: "st", 21: "st", 31: "st",
> 2: "nd", 22: "nd",
> 3: "rd", 23: "rd",
> }.get(x, "th")
>
> which makes distinct entries for each integer value of interest.
>
> The conventional approach would normally be:
>
> if x in (1, 21, 31):
> return "st"
> if x in (2, 22):
> return "nd"
> if x in (3, 23):
> return "rd"
> return "th"
>
> While this works for a small number of choices, if you had a huge dict
> with lots of possible values you could return to your
> dict-keyed-on-tuples approach. You would need to try each tuple in turn:
>
> mapping = {
> (1, 21, 31): "st",
> (2, 22): "nd",
> (3, 23): "rd",
> }
> for key, suffix in mapping.items():
> if x in key:
> return suffix
> return "th"
>
> However, for big dictionaries (with many keys) you loose a key strength
> of dicts: constant time lookup. You can see the above code (and the
> earlier "if" code) are rather linear, with run time going up linearly
> with the number of keys. You're better with the int->string single value
> dict version.
>
> Cheers,
> Cameron Simpson
It seems odd with C having switch that its cleaner and more efficient than
python where we are having to implement our own functions to recreate switch
everytime.
Or perhaps use a 3rd party library like
https://github.com/mikeckennedy/python-switch
You have both given good options, it seems there are no standard approaches in
this case.
Cheers
Sayth
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Implement C's Switch in Python 3
Hi
I am trying to convert a switch statement from C into Python. (why? practising).
This is the C code.
printf("Dated this %d", day);
switch (day) {
case 1: case 21: case 31:
printf("st"); break;
case 2: case 22:
printf("nd"); break;
case 3: case 23:
printf("rd"); break;
default: printf("th"); break;
}
printf(" day of ");
#Premise if the use enter an int as the date 21 for example it would print
21st. It appends the correct suffix onto a date.
Reading and trying to implement a function that uses a dictionary. Not sure how
to supply list into it to keep it brief and with default case of 'th'.
This is my current code.
def f(x):
return {
[1, 21, 31]: "st",
[2, 22]: "nd",
[3, 23]: "rd",
}.get(x, "th")
print(f(21))
I have an unhashable type list. Whats the best way to go?
Cheers
Sayth
--
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Re: Use a function arg in soup
Thanks Peter ### (2) attrs is a dict, so iterating over it will lose the values. Are you sure you want that? ### Yes initially I want just the keys as a list so I can choose to filter them out to only the ones I want. # getAttr Thanks very much will get my function up and working. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Use a function arg in soup
Hi.
I want to use a function argument as an argument to a bs4 search for attributes.
I had this working not as a function
# noms = soup.findAll('nomination')
# nom_attrs = []
# for attr in soup.nomination.attrs:
# nom_attrs.append(attr)
But as I wanted to keep finding other elements I thought I would turn it into a
generator function.
def attrKey(element):
for attr in soup.element.attrs:
yield attr
results in a Nonetype as soup.element.attrs is passed and the attribute isn't
substituted.
# Attempt 2
def attrKey(element):
for attr in "soup.{}.attrs".format(element):
yield attr
# Attempt 3
def attrKey(element):
search_attr = "soup.{}.attrs".format(element)
for attr in search_attr:
yield attr
so I would call it like
attrKey('nomination')
Any ideas on how the function argument can be used as the search attribute?
Thanks
Sayth
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Re: Format list of list sub elements keeping structure.
> > Well, your code was close. All you needed was a little tweak > to make it work like you requested. So keep working at it, > and if you have a specific question, feel free to ask on the > list. > > Here's a tip. Try to simplify the problem. Instead of > looping over a list of lists, and then attempting to do a > format in the middle of an iteration, a format that you > really don't know how to do in a vacuum (no pressure, > right???), pull one of the sublists out and try to format it > _first_. IOWs: isolate the problem. > > And, when you can format _one_ list the way you want -- > spoiler alert! -- you can format an infinite number of lists > the way you want. Loops are cool like that. Well, most of > the time. > > The key to solving most complex problems is to (1) break > them down into small parts, (2) solve each small part, and > (3) assemble the whole puzzle. This is a skill you must > master. And it's really not difficult. It just requires a > different way of thinking about tasks. Thank you Rick, good advice. I really am enjoying coding at the moment, got myself and life in a good headspace. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Format list of list sub elements keeping structure.
> > Then using this cool answer on SO [...] > > Oh. I thought you wanted to learn how to solve problems. I had no idea you > were auditioning for the James Dean part. My bad. Awesome response burn lol. I am trying to solve problems. Getting tired of dealing with JSON and having to figure out the structure each time. Just want to automate that part so I can move through the munging part and spend more time on higher value tasks. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Format list of list sub elements keeping structure.
> myjson = ...
> path = "['foo']['bar'][42]"
> print(eval("myjson" + path))
>
> ?
>
> Wouldn't it be better to keep 'data' as is and use a helper function like
>
> def get_value(myjson, path):
> for key_or_index in path:
> myjson = myjson[key_or_index]
> return myjson
>
> path = ['foo', 'bar', 42]
> print(get_value(myjson, path))
>
> ?
Currently I do leave the data I extract the keys out as a full path.
If I use pprint as suggested I get close.
['glossary'],
['glossary', 'title'],
['glossary', 'GlossDiv'],
['glossary', 'GlossDiv', 'title'],
['glossary', 'GlossDiv', 'GlossList'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'ID'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'SortAs'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossTerm'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'Acronym'],
...]
But to select elements from the json I need the format json['elem1']['elem2] .
I want to be able to get any json in future and parse it into my function and
return a list of all json key elements.
Then using this cool answer on SO https://stackoverflow.com/a/14692747/461887
from functools import reduce # forward compatibility for Python 3
import operator
def getFromDict(dataDict, mapList):
return reduce(operator.getitem, mapList, dataDict)
def setInDict(dataDict, mapList, value):
getFromDict(dataDict, mapList[:-1])[mapList[-1]] = value
Then get the values from the keys
>>> getFromDict(dataDict, ["a", "r"])
1
That would mean I could using my function if I get it write be able to feed it
any json, get all the full paths nicely printed and then feed it back to the SO
formula and get the values.
It would essentially self process itself and let me get a summary of all keys
and their data.
Thanks
Sayth
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Re: Format list of list sub elements keeping structure.
On Tuesday, 24 July 2018 14:25:48 UTC+10, Rick Johnson wrote:
> Sayth Renshaw wrote:
>
> > elements = [['[{0}]'.format(element) for element in elements]for elements
> > in data]
>
> I would suggest you avoid list comprehensions until you master long-form
> loops.
I actually have the answer except for a glitch where on list element is an int.
My code
for item in data:
out = '[{0}]'.format("][".join(item))
print(out)
which prints out
[glossary]
[glossary][title]
[glossary][GlossDiv]
[glossary][GlossDiv][title]
[glossary][GlossDiv][GlossList]
However, in my source I have two lines like this
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef',
'GlossSeeAlso', 0],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef',
'GlossSeeAlso', 1],
when it hits these lines I get
TypeError: sequence item 6: expected str instance, int found
Do I need to do an explicit check for these 2 cases or is there a simpler way?
Cheers
Sayth
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Re: Format list of list sub elements keeping structure.
On Tuesday, 24 July 2018 14:25:48 UTC+10, Rick Johnson wrote:
> Sayth Renshaw wrote:
>
> > elements = [['[{0}]'.format(element) for element in elements]for elements
> > in data]
>
> I would suggest you avoid list comprehensions until you master long-form
> loops.
My general issue is that I want to pick up all the elements in each sublist and
operate on them to concatenate them together.
However, using for loops I get each list then each element of each list.
When in my head I want each list then all elements of each list.
Sayth
--
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Re: Format list of list sub elements keeping structure.
I am very close to the end result. I now have it as
Output
[ ['[glossary]'],
['[glossary]', '[title]'],
['[glossary]', '[GlossDiv]'],
['[glossary]', '[GlossDiv]', '[title]'],
['[glossary]', '[GlossDiv]', '[GlossList]'],
['[glossary]', '[GlossDiv]', '[GlossList]', '[GlossEntry]'],
.]
I used.
elements = [['[{0}]'.format(element) for element in elements]for elements in
data]
Is there a good way to strip the ', ' and make the list a list of 1 element
lists? Thoughts on achieving this?
So
[ ['[glossary]'],
['[glossary][title]'],
]
Cheers
Sayth
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Re: Format list of list sub elements keeping structure.
> >
> > for item in data:
> > for elem in item:
> > out = ("[{0}]").format(elem)
> > print(out)
>
> Hint: print implicitly adds a newline to the output string. So collect all
> the values of each sublist and print a line-at-time to output, or use the
> end= argument of Py3's print, or find another solution. Also remember that
> indention is significant in Python.
Thanks
Getting closer.
answer = []
for item in data:
for elem in item:
out = ("[{0}]").format(elem)
answer.append(out)
print(answer)
Think I need to bring it in a list not an element of a list and process it.
Cheers
Sayth
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Format list of list sub elements keeping structure.
I have data which is a list of lists of all the full paths in a json document.
How can I change the format to be usable when selecting elements?
data = [['glossary'],
['glossary', 'title'],
['glossary', 'GlossDiv'],
['glossary', 'GlossDiv', 'title'],
['glossary', 'GlossDiv', 'GlossList'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'ID'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'SortAs'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossTerm'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'Acronym'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'Abbrev'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef', 'para'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef',
'GlossSeeAlso'],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef',
'GlossSeeAlso', 0],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef',
'GlossSeeAlso', 1],
['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossSee']]
I am trying to change it to be.
[['glossary'],
['glossary']['title'],
['glossary']['GlossDiv'],
]
Currently when I am formatting I am flattening the structure(accidentally).
for item in data:
for elem in item:
out = ("[{0}]").format(elem)
print(out)
Which gives
[glossary]
[title]
[GlossDiv]
[title]
[GlossList]
[GlossEntry]
[ID]
[SortAs]
[GlossTerm]
[Acronym]
[Abbrev]
[GlossDef]
[para]
[GlossSeeAlso]
[0]
[1]
[GlossSee]
Cheers
Sayth
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Re: Data Integrity Parsing json
On Thursday, 26 April 2018 07:57:28 UTC+10, Paul Rubin wrote: > Sayth Renshaw writes: > > What I am trying to figure out is how I give myself surety that the > > data I parse out is correct or will fail in an expected way. > > JSON is messier than people think. Here's an article with some > explanation and test code: > > http://seriot.ch/parsing_json.php Thanks Paul there is a lot of good information in that article. Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Looping on a list in json
no doubt tho after playing with this is that enumerate value ends up in the
output which is a dictionary. The enumerate has no key which makes it invalid
json if dumped.
Not massive issue but getting the effect of enumerate without polluting output
would be the winner.
>runner_lists = {}
>for n, item in enumerate(result):
># if this one is interested / not -filtered:
>print(n, item)
>runner_lists[n] = result[n]["RacingFormGuide"]["Event"]["Runners"]
Sayth
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Re: Looping on a list in json
Sorry
figured it. Needed to use n to iterate when creating.
runner_lists = {}
for n, item in enumerate(result):
# if this one is interested / not -filtered:
print(n, item)
runner_lists[n] = result[n]["RacingFormGuide"]["Event"]["Runners"]
Sayth
--
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Re: Looping on a list in json
> I'd just keep the interesting runners, along with their race numbers, in a
> dict. The enumerate function is handy here. Something like (untested):
>
> runner_lists = {}
> for n, item in enumerate(result):
> if this one is interested/not-filtered:
> runner_lists[n] = result["RacingFormGuide"]["Event"]["Runners"]
>
> and just return runner_lists. That way you know what the race numbers were
> for
> each list of runners.
>
>
> Cheers,
> Cameron Simpson
The main issue is that enumerate doesn't enumerate on the lists when trying to
filter.
result = meeting_id["Races"]
so yes enumerating this works, showing just printing n and item.
runner_lists = {}
for n, item in enumerate(result):
# if this one is interested / not -filtered:
print(n, item)
0 {'FeatureRaceBonusActive': 'Disabled', 'FixedPriceSummary': {'FixedPrices':
[{'SportId': 8, 'LeagueId': 102, 'MeetingId': 1218, 'MainEventId': 650350,
'SubEventId': 3601361, 'Status': 'F', 'StatusDescription': 'FINALISED',
'BetTypeName': 'Win', 'EnablePlaceBetting': True}]}, 'RacingFormGuide':
{'Copyright': .
and so on it goes through the 7 items in this file.
but including
runner_lists = {}
for n, item in enumerate(result):
# if this one is interested / not -filtered:
print(n, item)
runner_lists[n] = result["RacingFormGuide"]["Event"]["Runners"]
## Produces
Traceback (most recent call last):
dict_keys(['RaceDay', 'ErrorInfo', 'Success'])
File "/home/sayth/PycharmProjects/ubet_api_mongo/parse_json.py", line 31, in
runner_lists[n] = result["RacingFormGuide"]["Event"]["Runners"]
TypeError: list indices must be integers or slices, not str
Cheers
Sayth
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Re: Read Firefox sqlite files with Python
On Sunday, 5 November 2017 04:32:26 UTC+11, Steve D'Aprano wrote:
> I'm trying to dump a Firefox IndexDB sqlite file to text using Python 3.5.
>
>
> import sqlite3
> con = sqlite3.connect('foo.sqlite')
> with open('dump.sql', 'w') as f:
> for line in con.iterdump():
> f.write(line + '\n')
>
>
> The error I get is:
>
> Traceback (most recent call last):
> File "", line 2, in
> File "/usr/local/lib/python3.5/sqlite3/dump.py", line 30, in _iterdump
> schema_res = cu.execute(q)
> sqlite3.DatabaseError: file is encrypted or is not a database
>
>
> If I open the file in a hex editor, it starts with:
>
> SQLite format 3
>
> and although I can see a few human readable words, the bulk of the file looks
> like noise.
>
>
>
>
> --
> Steve
> “Cheer up,” they said, “things could be worse.” So I cheered up, and sure
> enough, things got worse.
https://stackoverflow.com/a/18601429
Version mismatch between sqlite CLI and python sqlite API? I created again my
db from the script instead of the CLI. Now insert and select work from the
script, but not from the CLI. $sqlite -version returns 2.8.17, while the python
version is 2.7.3.
Sayth
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Re: Looping on a list in json
On Sunday, 5 November 2017 09:53:37 UTC+11, Cameron Simpson wrote:
> >I want to get a result from a largish json api. One section of the json
> >structure returns lists of data. I am wanting to get each resulting list
> >returned.
> >
> >This is my code.
> >import json
> >from pprint import pprint
> >
> >with open(r'/home/sayth/Projects/results/Canterbury_2017-01-20.json', 'rb')
> >as f, open('socks3.json','w') as outfile:
> >to_read = json.load(f)
> [...]
> >meeting_id = to_read["RaceDay"]["Meetings"][0]
> >result = meeting_id["Races"]
> >#failing
> >for item in result:
> >pprint(["RacingFormGuide"]["Event"]["Runners"])
>
> I'd just keep the interesting runners, along with their race numbers, in a
> dict. The enumerate function is handy here. Something like (untested):
>
> runner_lists = {}
> for n, item in enumerate(result):
> if this one is interested/not-filtered:
> runner_lists[n] = result["RacingFormGuide"]["Event"]["Runners"]
>
> and just return runner_lists. That way you know what the race numbers were
> for
> each list of runners.
>
> >What is the best way to and return the data?
>
> The basic idea is to make a small data structure of your own (just the
> dictionary runner_lists in the example above) and fill it in with the
> infomation you care about in a convenient and useful shape. Then just return
> the data structure.
>
> The actual data structure will depend on what you need to do with this later.
>
> Cheers,
Thank you. That does seem a good approach.
I was intending to merge other dictionary data from other dicts within the json
structure and that's where the trouble starts i guess trying to get too much
from json.
Thanks
Sayth
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Looping on a list in json
Hi
I want to get a result from a largish json api. One section of the json
structure returns lists of data. I am wanting to get each resulting list
returned.
This is my code.
import json
from pprint import pprint
with open(r'/home/sayth/Projects/results/Canterbury_2017-01-20.json', 'rb') as
f, open('socks3.json','w') as outfile:
to_read = json.load(f)
print(to_read.keys())
# pprint(to_read)
meet = to_read["RaceDay"]["Meetings"]
meeting_id = to_read["RaceDay"]["Meetings"][0]
pprint(meeting_id.keys())
# result = meeting_id["Races"][1]["RacingFormGuide"]["Event"]["Runners"]
result = meeting_id["Races"]
#failing
for item in result:
pprint(["RacingFormGuide"]["Event"]["Runners"])
The key to the issue is that
result = meeting_id["Races"][0]["RacingFormGuide"]["Event"]["Runners"]
result = meeting_id["Races"][1]["RacingFormGuide"]["Event"]["Runners"]
result = meeting_id["Races"][2]["RacingFormGuide"]["Event"]["Runners"]
the numbers though in the above results could go from 0 to 10.
What is the best way to and return the data?
would just save meeting_id["Races"] to my result however there are a lot of
other junk dictionaries and lists I am filtering.
Cheers
Sayth
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Re: pathlib PurePosixPath
> > Hi > > > > How do I create a valid file name and directory with pathlib? > > > > When I create it using PurePosixPath I end up with an OSError due to an > > obvously invlaid path being created. > > You're on Windows. The rules for POSIX paths don't apply to your file > system, and... > > > OSError: [Errno 22] Invalid argument: > > 'C:\\Users\\Sayth\\Projects\\results/Warwick Farm2017-09-06T00:00:00.json' > > ... the colon is invalid on Windows file systems. You'll have to > replace those with something else. > > ChrisA Thanks. Updated the script. But shouldn't it create the file if it doesn't exist? Which none of them will. for dates in fullUrl: # print(dates) time.sleep(0.3) r = requests.get(dates) data = r.json() if data["RaceDay"] is not None: a = data["RaceDay"]["MeetingDate"] b = a[:7] file_name = data["RaceDay"]["Meetings"][0]["VenueName"] + '_' + b + '.json' result_path = pathlib.PurePath(r'C:\Users\Sayth\Projects\results', file_name) with open(result_path, 'a') as f: f.write(data) ##Output File "C:/Users/Sayth/PycharmProjects/ubet_api_mongo/json_download.py", line 42, in with open(result_path, 'a') as f: FileNotFoundError: [Errno 2] No such file or directory: 'C:\\Users\\Sayth\\Projects\\results\\Warwick Farm_2017-09.json' Process finished with exit code 1 Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
pathlib PurePosixPath
Hi How do I create a valid file name and directory with pathlib? When I create it using PurePosixPath I end up with an OSError due to an obvously invlaid path being created. import pathlib for dates in fullUrl: # print(dates) time.sleep(0.3) r = requests.get(dates) data = r.json() if data["RaceDay"] is not None: file_name = data["RaceDay"]["Meetings"][0]["VenueName"] + data["RaceDay"]["MeetingDate"] + '.json' result_path = pathlib.PurePosixPath(r'C:\Users\Sayth\Projects\results', file_name) with open(result_path, 'a') as f: f.write(data) ##Output C:\Users\Sayth\Anaconda3\envs\json\python.exe C:/Users/Sayth/PycharmProjects/ubet_api_mongo/json_download.py Traceback (most recent call last): File "C:/Users/Sayth/PycharmProjects/ubet_api_mongo/json_download.py", line 40, in with open(result_path, 'a') as f: OSError: [Errno 22] Invalid argument: 'C:\\Users\\Sayth\\Projects\\results/Warwick Farm2017-09-06T00:00:00.json' Process finished with exit code 1 Not sure exactly which way to fix it. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Suggestions on storing, caching, querying json
On Thursday, 5 October 2017 15:13:43 UTC+11, Sayth Renshaw wrote: > HI > > Looking for suggestions around json libraries. with Python. I am looking for > suggestions around a long term solution to store and query json documents > across many files. > > I will be accessing an api and downloading approx 20 json files from an api a > week. Having downloaded this year I have over 200 files already. So it will > grow at a reasonable rate. > > What I have initially done is store them into a mongo db. Now I am wondering > if this is useful or prudent since other than querying the json I wont have > much use of other mongo features. > > When querying the json files though queries will utilise multiple json files > at once, not just retrieving a single record. The usage is for data analysis. > > Is there a good json storage option, with caching and optimal querying etc. > > Regarding querying I did find a library for json searching called ObjectPath > written in Python http://objectpath.org/reference.html > > Looking to leverage your experience. > > Cheers > > Sayth There is a new extension for redis ReJson and redis-py for using redis and python as a json store. http://rejson.io/ and https://github.com/andymccurdy/redis-py. Not sure if this has much more upside than mongo other than having a more fmailiar query language like JsonPath http://rejson.io/path/ Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Suggestions on storing, caching, querying json
HI Looking for suggestions around json libraries. with Python. I am looking for suggestions around a long term solution to store and query json documents across many files. I will be accessing an api and downloading approx 20 json files from an api a week. Having downloaded this year I have over 200 files already. So it will grow at a reasonable rate. What I have initially done is store them into a mongo db. Now I am wondering if this is useful or prudent since other than querying the json I wont have much use of other mongo features. When querying the json files though queries will utilise multiple json files at once, not just retrieving a single record. The usage is for data analysis. Is there a good json storage option, with caching and optimal querying etc. Regarding querying I did find a library for json searching called ObjectPath written in Python http://objectpath.org/reference.html Looking to leverage your experience. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: None is None but not working
Thank you it was data["RaceDay"] that was needed.
ata = r.json()
if data["RaceDay"] is None:
print("Nothing here")
else:
print(data["RaceDay"])
Nothing here
Nothing here
Nothing here
{'MeetingDate': '2017-01-11T00:00:00', .
Thanks
Sayth
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None is None but not working
Hi
I have got a successful script setup to rotate through dates and download json
data from the url.
As the api returns 200 whether successful I want to check if the file returned
is not successful.
when a file doesn't exist the api returns
{'RaceDay': None, 'ErrorInfo': {'SystemId': 200, 'ErrorNo': 55013,
'DisplayMessage': 'File Not Found.', 'ContactSupport': False,
'SupportErrorReference': '200-55013'}, 'Success': False}
When I call data = r.json() it says its type is None if it is not successful so
I thought it easier to check that.
However checking for None does not work the flow in my if else falls straight
to else.
for dates in fullUrl:
r = requests.get(dates)
data = r.json()
if data is None:
print("Nothing here")
else:
print(data["RaceDay"])
and I get output of
None
None
{'MeetingDate': '2017-01- ... and so on.
How can I actually get this to check?
If i use type(data) I also get None.
Cheers
Sayth
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Re: Easy way to get a list of tuples.
> >
> > Thanks Thomas yes you are right with append. I have tried it but just
> > can't get it yet as append takes only 1 argument and I wish to give it 3.
> >
> You have not showed us what you tried, but you are probably missing a pair
> of brackets.
>
> C:\Users\User>python
> Python 3.6.0 (v3.6.0:41df79263a11, Dec 23 2016, 08:06:12) [MSC v.1900 64 bit
> (AMD64)] on win32
> Type "help", "copyright", "credits" or "license" for more information.
> >>> x = []
> >>> x.append(('a', 'b', 'c'))
> >>> x.append(('p', 'q', 'r'))
> >>> x
> [('a', 'b', 'c'), ('p', 'q', 'r')]
> >>>
>
> Does this help?
>
> Frank Millman
Oh yes I just had one set of brackets with my append.
Thanks
Frank
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Re: Easy way to get a list of tuples.
On Thursday, 21 September 2017 20:31:28 UTC+10, Thomas Jollans wrote: > On 2017-09-21 12:18, Sayth Renshaw wrote: > > This is my closest code > > > > data = r.json() > > > > raceData = [] > > > > for item in data["RaceDay"]['Meetings'][0]['Races']: > > raceDetails = item['RacingFormGuide']['Event']['Race'] > > raceData += > > (raceDetails['Name'],raceDetails['Number'],raceDetails['Distance']) > > > > print(raceDetails) > > > > You're close! > > The operator += extends a list with the items of another sequence (or > iterable). What you're looking for is the method .append(), which adds a > single element. > > Observe: > > Python 3.6.0 |Continuum Analytics, Inc.| (default, Dec 23 2016, 12:22:00) > [GCC 4.4.7 20120313 (Red Hat 4.4.7-1)] on linux > Type "help", "copyright", "credits" or "license" for more information. > py> a_list = [] > py> a_list += 1,2,3 > py> a_list > [1, 2, 3] > py> a_list.append(4) > py> a_list > [1, 2, 3, 4] > py> a_list += 4 > Traceback (most recent call last): > File "", line 1, in > TypeError: 'int' object is not iterable > py> a_list.append((5,6,7)) > py> a_list > [1, 2, 3, 4, (5, 6, 7)] > py> > > > -- > Thomas Jollans Thanks Thomas yes you are right with append. I have tried it but just can't get it yet as append takes only 1 argument and I wish to give it 3. I am really having trouble creating the groups of 3, since I am getting one consistent stream. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Easy way to get a list of tuples.
Hi
I have been toying with json and I particular area where I cannot get the
desired result a list of tuples as my return. The json from the API is way to
long but I don't think it will matter.
.. hitting url
data = r.json()
for item in data["RaceDay"]['Meetings'][0]['Races']:
raceDetails = item['RacingFormGuide']['Event']['Race']
print(raceDetails)
This returns
{'Number': 1, 'NumberDisplay': '01', 'Distance': 1000, 'DistanceDisplay': '1000
METRES', 'Name': 'CLASS 3 HANDICAP', 'NameForm': 'HIGHWAY-C3'}
{'Number': 2, 'NumberDisplay': '02', 'Distance': 1600, 'DistanceDisplay': '1600
METRES', 'Name': 'BM 90 HANDICAP', 'NameForm': 'BM90'}
{'Number': 3, 'NumberDisplay': '03', 'Distance': 1100, 'DistanceDisplay': '1100
METRES', 'Name': 'HERITAGE STAKES', 'NameForm': 'HERITAGE'}
{'Number': 4, 'NumberDisplay': '04', 'Distance': 1400, 'DistanceDisplay': '1400
METRES', 'Name': 'BILL RITCHIE HANDICAP', 'NameForm': 'RITCHIE'}
{'Number': 5, 'NumberDisplay': '05', 'Distance': 1400, 'DistanceDisplay': '1400
METRES', 'Name': 'TEA ROSE STAKES', 'NameForm': 'TEA ROSE'}
{'Number': 6, 'NumberDisplay': '06', 'Distance': 1600, 'DistanceDisplay': '1600
METRES', 'Name': 'GEORGE MAIN STAKES', 'NameForm': 'GEO MAIN'}
{'Number': 7, 'NumberDisplay': '07', 'Distance': 1100, 'DistanceDisplay': '1100
METRES', 'Name': 'THE SHORTS', 'NameForm': 'THE SHORTS'}
{'Number': 8, 'NumberDisplay': '08', 'Distance': 2000, 'DistanceDisplay': '2000
METRES', 'Name': 'KINGTON TOWN STAKES', 'NameForm': 'KING TOWN'}
{'Number': 9, 'NumberDisplay': '09', 'Distance': 1200, 'DistanceDisplay': '1200
METRES', 'Name': 'BM 84 HANDICAP', 'NameForm': 'BM84'}
My goal is to select a few elements and create a list of 3 element tuples
like this
[('CLASS 3 HANDICAP', 1, 1000), ('BM 90 HANDICAP', 2, 1600), ('HERITAGE
STAKES', 3, 1100), ('BILL RITCHIE HANDICAP', 4, 1400), ('TEA ROSE STAKES', 5,
1400), ('GEORGE MAIN STAKES', 6, 1600), ('THE SHORTS', 7, 1100), ('KINGTON TOWN
STAKES', 8, 2000), ('BM 84 HANDICAP', 9, 1200)]
I get close creating a list of elements but each attempt I try to create the
list of tuples fails.
This is my closest code
data = r.json()
raceData = []
for item in data["RaceDay"]['Meetings'][0]['Races']:
raceDetails = item['RacingFormGuide']['Event']['Race']
raceData +=
(raceDetails['Name'],raceDetails['Number'],raceDetails['Distance'])
print(raceDetails)
which returns
['CLASS 3 HANDICAP', 1, 1000, 'BM 90 HANDICAP', 2, 1600, 'HERITAGE STAKES', 3,
1100, 'BILL RITCHIE HANDICAP', 4, 1400, 'TEA ROSE STAKES', 5, 1400, 'GEORGE
MAIN STAKES', 6, 1600, 'THE SHORTS', 7, 1100, 'KINGTON TOWN STAKES', 8, 2000,
'BM 84 HANDICAP', 9, 1200]
How do I get the tuples?
Cheers
Sayth
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Re: Why is my class undefined?
On Thursday, 17 August 2017 09:03:59 UTC+10, Ian wrote:
wrote:
> > Morning
> >
> > I haven't ventured into classes much before. When trying to follow some
> > examples and create my own classes in a jupyter notebook I receive an error
> > that the class is undefined.
> >
> > So I created for practise a frog class
> >
> > class frog(object):
> >
> > def __init__(self, ftype, word):
> > self.ftype = ftype
> > self.word = ftype
> >
> > def jump(self):
> > """
> > Make frog jump
> > """
> > return "I am jumping"
> >
> > if __name__ == "__main__":
> > tree_frog = frog("Tree Frog", "Ribbitt")
> > print(frog.ftype)
> > print(frog.word)
> >
> >
> > I receive this error
> >
> > NameError Traceback (most recent call last)
> > in ()
> > > 1 class frog(object):
> > 2
> > 3 def __init__(self, ftype, word):
> > 4 self.ftype = ftype
> > 5 self.word = ftype
> >
> > in frog()
> > 12
> > 13 if __name__ == "__main__":
> > ---> 14 tree_frog = frog("Tree Frog", "Ribbitt")
> > 15 print(frog.ftype)
> > 16 print(frog.word)
> >
> > NameError: name 'frog' is not defined
> >
> > what exactly am I doing wrong?
>
> The if __name__ == "__main__" block is inside the class declaration
> block, so at the point that it runs the class has not been created
> yet. Try removing the indentation to place it after the class block
> instead.
Thank you that had me bugged I just couldn't see it.
Cheers
Sayth
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Why is my class undefined?
Morning
I haven't ventured into classes much before. When trying to follow some
examples and create my own classes in a jupyter notebook I receive an error
that the class is undefined.
So I created for practise a frog class
class frog(object):
def __init__(self, ftype, word):
self.ftype = ftype
self.word = ftype
def jump(self):
"""
Make frog jump
"""
return "I am jumping"
if __name__ == "__main__":
tree_frog = frog("Tree Frog", "Ribbitt")
print(frog.ftype)
print(frog.word)
I receive this error
NameError Traceback (most recent call last)
in ()
> 1 class frog(object):
2
3 def __init__(self, ftype, word):
4 self.ftype = ftype
5 self.word = ftype
in frog()
12
13 if __name__ == "__main__":
---> 14 tree_frog = frog("Tree Frog", "Ribbitt")
15 print(frog.ftype)
16 print(frog.word)
NameError: name 'frog' is not defined
what exactly am I doing wrong?
Cheers
Sayth
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Re: Test 0 and false since false is 0
> Another option is to test for type(value) == int:
>
> >>> before = ["a",0,0,"b",None,"c","d",0,1,False,0,1,0,3,[],0,1,9,0,0,
> {},0,0,9]
> >>> wanted = ["a","b",None,"c","d",1,False,1,3,[],1,9,
> {},9,0,0,0,0,0,0,0,0,0,0]
> >>> after = sorted(before, key=lambda x: x == 0 and type(x) == int)
> >>> assert str(after) == str(wanted)
> >>> after
> ['a', 'b', None, 'c', 'd', 1, False, 1, 3, [], 1, 9, {}, 9, 0, 0, 0, 0, 0,
> 0, 0, 0, 0, 0]
>
>
> That way float values will be left alone, too:
>
> >>> sorted([0.0, 0, False, [], "x"], key=lambda x: x == 0 and type(x) ==
> int)
> [0.0, False, [], 'x', 0]
I have been reading this solution
> >>> after = sorted(before, key=lambda x: x == 0 and type(x) == int)
it is really good, however I don't understand it enough to reimplement
something like that myself yet.
Though I can that lambda tests for 0 that is equal to an int why does sorted
put them to the end?
Cheers
Sayth
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Re: Test 0 and false since false is 0
On Friday, 7 July 2017 12:46:51 UTC+10, Rick Johnson wrote:
> On Thursday, July 6, 2017 at 9:29:29 PM UTC-5, Sayth Renshaw wrote:
> > I was trying to solve a problem and cannot determine how to filter 0's but
> > not false.
> >
> > Given a list like this
> > ["a",0,0,"b",None,"c","d",0,1,False,0,1,0,3,[],0,1,9,0,0,{},0,0,9]
> >
> > I want to be able to return this list
> > ["a","b",None,"c","d",1,False,1,3,[],1,9,{},9,0,0,0,0,0,0,0,0,0,0]
> >
> > However if I filter like this
> >
> > def move_zeros(array):
> > l1 = [v for v in array if v != 0]
> > l2 = [v for v in array if v == 0]
> > return l1 + l2
> >
> > I get this
> > ['a', 'b', None, 'c', 'd', 1, 1, 3, [], 1, 9, {}, 9, 0, 0, 0, False, 0, 0,
> > 0, 0, 0, 0, 0]
> >
> > I have tried or conditions of v == False etc but then the 0's being false
> > also aren't moved. How can you check this at once?
>
> Yep. This is a common pitfall for noobs, as no logic can
> explain to them why integer 0 should bool False, and integer
> 1 should bool True. But what's really going to cook your
> noodle is when you find out that any integer greater than 1
> bools True. Go figure! They'll say it's for consistency
> sake. But i say it's just a foolish consistency.
>
> You need to learn the subtle difference between `==` and
> `is`.
>
> ## PYTHON 2.x
> >>> 1 == True
> True
> >>> 1 is True
> False
> >>> 0 == False
> True
> >>> 0 is False
> False
Is there an "is not" method that's not != so I can check is not false.
def move_zeros(array):
l1 = [v for v in array if v is False or v != 0]
l2 = [v for v in array if v is not False or v == 0]
return l1 + l2
Cheers
Sayth
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Test 0 and false since false is 0
I was trying to solve a problem and cannot determine how to filter 0's but not
false.
Given a list like this
["a",0,0,"b",None,"c","d",0,1,False,0,1,0,3,[],0,1,9,0,0,{},0,0,9]
I want to be able to return this list
["a","b",None,"c","d",1,False,1,3,[],1,9,{},9,0,0,0,0,0,0,0,0,0,0]
However if I filter like this
def move_zeros(array):
l1 = [v for v in array if v != 0]
l2 = [v for v in array if v == 0]
return l1 + l2
I get this
['a', 'b', None, 'c', 'd', 1, 1, 3, [], 1, 9, {}, 9, 0, 0, 0, False, 0, 0, 0,
0, 0, 0, 0]
I have tried or conditions of v == False etc but then the 0's being false also
aren't moved. How can you check this at once?
Cheers
Sayth
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Re: Generator - Provide parameters to url - requests
Thanks. I left "base" out as i was trying to remove as much uneeded code from example as possible. I had defined it as base = datetime.datetime(2017,1,1) Reading your code this sounds to simple :-). def dates(first, numdays): # generate datetime objects for extra clarity # note there are no implicit arguments like `base` in your code for _ in range(numdays): yield first first += ONE_DAY Thanks Sayth -- https://mail.python.org/mailman/listinfo/python-list
Generator - Provide parameters to url - requests
Hi
I am struggling to figure out how I can create a generator to provide values to
my url. My url needs to insert the year month and day in the url not as params
to the url.
import json
import requests
import datetime
# using this I can create a list of dates for the first 210 days of this year.
base = datetime.datetime(2017,1,1)
def datesRange(max, min):
date_list = (base - datetime.timedelta(days=x) for x in range(max,min))
DAY = date_list.day
MONTH = date_list.month
YEAR = date_list.year
yield DAY, MONTH, YEAR
dateValues = datesRange(-210,0)
def create_url(day, month, year):
https://api.tatts.com/sales/vmax/web/data/racing/{0}/{1}/{2}/sr/full".format(YEAR,MONTH,DAY)
return url
Then I want to insert them in this url one at a time from the generator
try:
r = requests.get(INSERT_URL_HERE)
if r.status_code == requests.codes.ok then:
# open a file for writing using url paramters
with open(SR + DAY + MONTH + YEAR + '.json', 'w') as f:
# Do stuff from here not relevant to question.
I have just gotten lost.
Is there an easier way to go about this?
Cheers
Sayth
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Re: Create a list of dates for same day of week in a year
> > Thoughts or examples? > > > dateutil.rrule is what you may use e.g. > > > In [38]: from dateutil import rrule > > In [39]: from datetime import date > > In [40]: end = date(2017, 12, 31) > > In [41]: rr = rrule.rrule(rrule.WEEKLY, byweekday=[0, 2], until=end) > > In [42]: days = list(rr) > > In [43]: len(days) > Out[43]: 53 > > In [44]: days[:5], days[-5:] > Out[44]: > ([datetime.datetime(2017, 6, 28, 23, 58, 11), > datetime.datetime(2017, 7, 3, 23, 58, 11), > datetime.datetime(2017, 7, 5, 23, 58, 11), > datetime.datetime(2017, 7, 10, 23, 58, 11), > datetime.datetime(2017, 7, 12, 23, 58, 11)], > [datetime.datetime(2017, 12, 13, 23, 58, 11), > datetime.datetime(2017, 12, 18, 23, 58, 11), > datetime.datetime(2017, 12, 20, 23, 58, 11), > datetime.datetime(2017, 12, 25, 23, 58, 11), > datetime.datetime(2017, 12, 27, 23, 58, 11)]) > > In [45]: Thanks. I am just researching now the format that has come out. unclear what 58 represents. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Create a list of dates for same day of week in a year
> > Is there an obvious method I am missing in creating a list of dates? I want > > to get a list of each Saturday and each Wednesday for the year 2017. > > > > It seems and maybe this is where I am wrong but doesn't the datetime > > library already know the dates if yes is there an easy way to query it? > > > > Sorta-kinda. > > >>> import datetime > >>> today = datetime.date.today() > >>> monday = today - datetime.timedelta(days=today.weekday()) > >>> wednesday = monday + datetime.timedelta(days=2) > >>> saturday = monday + datetime.timedelta(days=5) > >>> week = datetime.timedelta(days=7) > >>> next_wed = wednesday + week > > You can mess around with that. If you want to find the beginning of > the year, you could start by using datetime.date(2017, 1, 1) instead > of today(), and then locate the previous Monday the same way. > > So yes, all the information is available, but no, there's no easy way > to say "give me all the Saturdays in 2017". You'd have to iterate. > > ChrisA Thanks ChrisA could using the calendar and a for loop work better? As below test a date to see if the value = 5 so loop months 1-12 and then days 1-31 and just throw an error when the date doesn't exist 30th of Feb etc In [19]: import calendar In [23]: calendar.weekday(2017,6,24) Out[23]: 5 Or I have set the Saturday to be the firstday of the week https://docs.python.org/2/library/calendar.html is there a way to loop for all firstdayofweeks in year and return dates out that way? Not sure how to get all dates in the year as an object for iteration though. The calendar iter items end at months https://docs.python.org/2/library/calendar.html#calendar.Calendar.itermonthdays In [20]: calendar.setfirstweekday(calendar.SATURDAY) In [21]: calendar.firstweekday() Out[21]: 5 Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Create a list of dates for same day of week in a year
Afternoon Is there an obvious method I am missing in creating a list of dates? I want to get a list of each Saturday and each Wednesday for the year 2017. It seems and maybe this is where I am wrong but doesn't the datetime library already know the dates if yes is there an easy way to query it? Also checked out Python Arrow http://arrow.readthedocs.io/en/latest/ as it expands on the python standard library but I couldn't find an obvious example of this. Thoughts or examples? Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Better way to do this dict comprehesion
Peter I really like this The complete code: >>> from collections import Counter >>> def find_it(seq): ... [result] = [k for k, v in Counter(seq).items() if v % 3 == 0] ... return result ... >>> test_seq = [20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5] >>> find_it(test_seq) But what may be the smallest thing in this i had no idea I could do [result] = blah and get a generator on the return variable that seems insane. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Better way to do this dict comprehesion
> > But the given problem states there will always only be one number appearing > > an odd number of times given that is there a neater way to get the answer? > > Take a step back for a moment. Are you trying to find something that > appears an odd number of times, or a number of times that counts by > three? First figure that out, THEN see if there's a better way to do > what you're doing. A number that appears an odd number of times. > > To find an unpaired number in linear time with minimal space, try > stepping through the list and either adding to a set or removing from > it. At the end, your set should contain exactly one element. I'll let > you write the actual code :) > > ChrisA ChrisA the way it sounds through the list is like filter with map and a lambda. http://www.python-course.eu/lambda.php Haven't had a good go yet... will see how it goes. Thanks Sayth -- https://mail.python.org/mailman/listinfo/python-list
Better way to do this dict comprehesion
Hi I have got this dictionary comprehension and it works but how can I do it better? from collections import Counter def find_it(seq): counts = dict(Counter(seq)) a = [(k, v) for k,v in counts.items() if v % 3 == 0] return a[0][0] test_seq = [20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5] so this returns 5 which is great and the point of the problem I was doing. Can also do it like this def find_it(seq): counts = dict(Counter(seq)) a = [(k) for k,v in counts.items() if v % 3 == 0] return a[0] But the given problem states there will always only be one number appearing an odd number of times given that is there a neater way to get the answer? Thanks Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: How to flatten only one sub list of list of lists
> Replace the slice row[index:index+1] with row[index], either by building a > new list or in place: > > >>> def show(data): > ...for item in data: print(item) > ... > >>> def flatten_one(rows, index): > ... return [r[:index] + r[index] + r[index+1:] for r in rows] > ... > >>> def flatten_inplace(rows, index): > ... for row in rows: > ... row[index:index+1] = row[index] > ... > >>> flatten_inplace(data, 5) > >>> show(data) > ['46295', 'Montauk', '3', '60', '85', '19', '5', '1', '0 $277790.00'] > ['46295', 'Dark Eyes', '5', '59', '83', '6', '4', '1', '0 $105625.00'] > ['46295', 'Machinegun Jubs', '6', '53', '77', '6', '2', '1', '1 $71685.00'] > ['46295', 'Zara Bay', '1', '53', '77', '12', '2', '3', '3 $112645.00'] I went for the one I can understand which was inplace def flatten_inplace(rows, index): for row in rows: row[index:index + 1] = row[index] return rows See now if I can make it more adaptable to use it in some other situations, quite useful. Thanks Sayth -- https://mail.python.org/mailman/listinfo/python-list
How to flatten only one sub list of list of lists
How can I flatten just a specific sublist of each list in a list of lists? So if I had this data [ ['46295', 'Montauk', '3', '60', '85', ['19', '5', '1', '0 $277790.00']], ['46295', 'Dark Eyes', '5', '59', '83', ['6', '4', '1', '0 $105625.00']], ['46295', 'Machinegun Jubs', '6', '53', '77', ['6', '2', '1', '1 $71685.00']], ['46295', 'Zara Bay', '1', '53', '77', ['12', '2', '3', '3 $112645.00']]] How can I make it be [ ['46295', 'Montauk', '3', '60', '85', '19', '5', '1', '0 $277790.00'], ['46295', 'Dark Eyes', '5', '59', '83', '6', '4', '1', '0 $105625.00'], ['46295', 'Machinegun Jubs', '6', '53', '77', '6', '2', '1', '1 $71685.00'], ['46295', 'Zara Bay', '1', '53', '77', '12', '2', '3', '3 $112645.00']] Been looking around but most solutions just entirely flatten everything. This was popular on SO but yeah it flattens everything I want to be more selective def flatten(lst): for elem in lst: if type(elem) in (tuple, list): for i in flatten(elem): yield i else: yield elem What I am thinking is that if for each list the sublist should be at index 1, so [0][1] [1][1] [2][1] for item in list: item[1] - somehow flatten. Thoughts? Sayth -- https://mail.python.org/mailman/listinfo/python-list
Is there a good process or library for validating changes to XML format
Afternoon Is there a good library or way I could use to check that the author of the XML doc I am using doesn't make small changes to structure over releases? Not fully over this with XML but thought that XSD may be what I need, if I search "python XSD" I get a main result for PyXB and generateDS (https://pythonhosted.org/generateDS/). Both seem to be libraries for generating bindings to structures for parsing so maybe I am searching the wrong thing. What is the right thing to search? Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Screwing Up looping in Generator
For completeness I was close this is the working code. def get_list_of_names(generator_arg): name_set = set() for name in generator_arg: base = os.path.basename(name.name) filename = os.path.splitext(base)[0] name_set.add(filename) return name_set def data_attr(roots): """Get the root object and iter items.""" for name in names: directory = "output" write_name = name + ".csv" with open(os.path.join(directory, write_name), 'w', newline='') as csvf: race_writer = csv.writer(csvf, delimiter=',' ) thanks for your time and assistance. It's much appreciated Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Screwing Up looping in Generator
Untested as i wrote this in notepad at work but, if i first use the generator to create a set of filenames and then iterate it then call the generator anew to process file may work? Good idea or better available? def get_list_of_names(generator_arg): name_set = set() for name in generator_arg: base = os.path.basename(name.name) filename = os.path.splitext(base)[0] name_set.add(filename) return name_set for file_name in name_set: directory = "output" with open(os.path.join(directory, filename, 'w', newline='') as csvf: for file in rootobs: # create and write csv Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Screwing Up looping in Generator
So can I call the generator twice and receive the same file twice in 2 for
loops?
Once to get the files name and the second to process?
for file in rootobs:
base = os.path.basename(file.name)
write_to = os.path.join("output", os.path.splitext(base)[0] + ".csv")
with open(write_to, 'w', newline='') as csvf:
for file in rootobs:
# create and write csv
Cheers
Sayth
--
https://mail.python.org/mailman/listinfo/python-list
Re: Screwing Up looping in Generator
On Wednesday, 4 January 2017 12:36:10 UTC+11, Sayth Renshaw wrote:
> So can I call the generator twice and receive the same file twice in 2 for
loops?
>
> Once to get the files name and the second to process?
>
> for file in rootobs:
> base = os.path.basename(file.name)
> write_to = os.path.join("output", os.path.splitext(base)[0] + ".csv")
> with open(write_to, 'w', newline='') as csvf:
> for file in rootobs:
> # create and write csv
>
> Cheers
>
> Sayth
I just need it to write after each file however the
with open(#file) as csvf:
Keeps it all open until every file processed in an output file with the name of
the first file in the generator.
Sayth
--
https://mail.python.org/mailman/listinfo/python-list
Re: Is there a good process or library for validating changes to XML format?
It definitely has more features than i knew http://xmlsoft.org/xmllint.html Essentially thigh it appears to be aimed at checking validity and compliance of xml. I why to check the structure of 1 xml file against the previous known structure to ensure there are no changes. Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Is there a good process or library for validating changes to XML format?
Afternoon Is there a good library or way I could use to check that the author of the XML doc I am using doesn't make small changes to structure over releases? Not fully over this with XML but thought that XSD may be what I need, if I search "python XSD" I get a main result for PyXB and generateDS (https://pythonhosted.org/generateDS/). Both seem to be libraries for generating bindings to structures for parsing so maybe I am searching the wrong thing. What is the right thing to search? Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Screwing Up looping in Generator
For completeness I was close this is the working code. def get_list_of_names(generator_arg): name_set = set() for name in generator_arg: base = os.path.basename(name.name) filename = os.path.splitext(base)[0] name_set.add(filename) return name_set def data_attr(roots): """Get the root object and iter items.""" for name in names: directory = "output" write_name = name + ".csv" with open(os.path.join(directory, write_name), 'w', newline='') as csvf: race_writer = csv.writer(csvf, delimiter=',' ) thanks for your time and assistance. It's much appreciated Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Screwing Up looping in Generator
Untested as i wrote this in notepad at work but, if i first use the generator to create a set of filenames and then iterate it then call the generator anew to process file may work? Good idea or better available? def get_list_of_names(generator_arg): name_set = set() for name in generator_arg: base = os.path.basename(name.name) filename = os.path.splitext(base)[0] name_set.add(filename) return name_set for file_name in name_set: directory = "output" with open(os.path.join(directory, filename, 'w', newline='') as csvf: for file in rootobs: # create and write csv Sayth -- https://mail.python.org/mailman/listinfo/python-list
Re: Screwing Up looping in Generator
On Wednesday, 4 January 2017 12:36:10 UTC+11, Sayth Renshaw wrote:
> So can I call the generator twice and receive the same file twice in 2 for
> loops?
>
> Once to get the files name and the second to process?
>
> for file in rootobs:
> base = os.path.basename(file.name)
> write_to = os.path.join("output", os.path.splitext(base)[0] + ".csv")
> with open(write_to, 'w', newline='') as csvf:
> for file in rootobs:
> # create and write csv
>
> Cheers
>
> Sayth
I just need it to write after each file however the
with open(#file) as csvf:
Keeps it all open until every file processed in an output file with the name of
the first file in the generator.
Sayth
--
https://mail.python.org/mailman/listinfo/python-list
Re: Screwing Up looping in Generator
So can I call the generator twice and receive the same file twice in 2 for
loops?
Once to get the files name and the second to process?
for file in rootobs:
base = os.path.basename(file.name)
write_to = os.path.join("output", os.path.splitext(base)[0] + ".csv")
with open(write_to, 'w', newline='') as csvf:
for file in rootobs:
# create and write csv
Cheers
Sayth
--
https://mail.python.org/mailman/listinfo/python-list
