Re: In Python 3, how to append a nested dictionary to a shelve file with the added difficulty of using a for loop?
Aaron Christensen wrote:
> Thanks for the response! Several things you stated definitely got me
> thinking. I really appreciate the response. I used what you said and I
> am able to accomplish what I needed.
Perhaps it becomes clearer when you write two helper functions
def read_record(key):
return db[key]
def write_record(key, value):
db[key] = value
Changing a person's data then becomes
person_data = read_record("person_1")
person_data["age"] = 123
write_record("person_1", person_data)
Also, I was mostly paraphrasing
>>> help(shelve)
so you might read that, too.
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In Python 3, how to append a nested dictionary to a shelve file with the added difficulty of using a for loop?
Hello,
I am trying to figure out how to populate a shelve file with a nested
dictionary.
These are my requirements:
-Create shelve file called people.db
-Append the shelve file with new people (person_1, person_2, etc.).
-Use a for loop to iterate through 'attributes' so that I do not need to write
out the lengthy code line by line to populate to the shelve file.
-Need to reference shelve file data for future use
Here is the key/value format that I would like to append to the shelve file.
person_1 = { 'name': 'Bob', 'type': 'employee', 'attributes':
[{'game': 'basketball', 'high score': '100', 'time': '3.34'},
{'game': 'bridge', 'high score': '10', 'time': '30.34'},
{'game': 'foosball', 'high score': '2', 'time': '24'}]
'''
50+ other attributes
'''
}
# Example: s['person_1]['attributes'][2]['time'] would call out '24'.
# 's' is from 's = shelve.open('people')'
I have a dictionary dictPeople.py file (created using pprint() that contains
the information of person_1, etc. And I am extracting only a small percentage
of the data that is needed.
I have tried the following, but I get an invalid key error.
import shelve, dictPeople
s = shelve.open('people')
person = 'person_1'
s[person]['name'] = dictPeople.person_1['name']
s[person]['type'] = dictPeople.person_1['type']
# I need to use this for loop because there are 50+ attributes.
for attribute in range(0, len(dictPeople['attributes'][attribute])):
s[person]['attributes'][attribute]['game'] = \
dictPeople['attributes'][attribute]['game']
s[person]['attributes'][attribute]['high score'] = \
dictPeople['attributes'][attribute]['high score']
s[person]['attributes'][attribute]['time'] = \
dictPeople['attributes'][attribute]['time']
It turns out, I get the key error because I am not allowed to reference a
key/value pair the same way that I can with a dictionary. However, the crazy
thing is that I can call values in the db file using the same exact format when
trying to write.
For example: I can read data from the .db file using:
x = s['person_1']['name']
print(x)
BUT! I cannot write to that .db file using that exact format or I get an
invalid key error. Makes no sense!
s['person_1']['name'] = 'Bob'
# Returns invalid key entry. Makes no sense.
Therefore, I tried to extract data and populate the db file using the following:
s[person] = { 'name': dictPeople.person_1['name'],
'type': dictPeople.person_1['type'],
for attribute in range(0, len(dictPeople['attributes'][attribute])):
['game': dictPeople['attributes'][attribute]['game'],
'high score': dictPeople['attributes'][attribute]['high score'],
'time': dictPeople['attributes'][attribute]['time']]
}
But, this obvously doesn't work because of the for loop. How can I do this?
-I am trying to figure out how to extract data from the dictionary
dictPeople.py file and store it in the people.db file.
-I am trying to add new people to the people.db file as more people become
available.
-I need to use the for loop because of the 50+ attributes I need to add.
-My future steps would be to modify some of the data values that I extract and
used to populate the people.db file, but my first step is to, at least,
extract, and then populate the people.db file.
Any help or guidance is greatly appreciated. Thank you for your time and
reading my question.
Thanks!
Aaron
--
https://mail.python.org/mailman/listinfo/python-list
In Python 3, how to append a nested dictionary to a shelve file with the added difficulty of using a for loop?
Hello,
I am trying to figure out how to populate a shelve file with a nested
dictionary.
These are my requirements:
-Create shelve file called people.db
-Append the shelve file with new people (person_1, person_2, etc.).
-Use a for loop to iterate through 'attributes' so that I do not need to
write out the lengthy code line by line to populate to the shelve file.
-Need to reference shelve file data for future use
Here is the key/value format that I would like to append to the shelve
file.
person_1 = { 'name': 'Bob', 'type': 'employee', 'attributes':
[{'game': 'basketball', 'high score': '100', 'time': '3.34'},
{'game': 'bridge', 'high score': '10', 'time': '30.34'},
{'game': 'foosball', 'high score': '2', 'time': '24'}]
'''
50+ other attributes
'''
}
# Example: s['person_1]['attributes'][2]['time'] would call out '24'.
# 's' is from 's = shelve.open('people')'
I have a dictionary dictPeople.py file (created using pprint() that
contains the information of person_1, etc. And I am extracting only a small
percentage of the data that is needed.
I have tried the following, but I get an invalid key error.
import shelve, dictPeople
s = shelve.open('people')
person = 'person_1'
s[person]['name'] = dictPeople.person_1['name']
s[person]['type'] = dictPeople.person_1['type']
# I need to use this for loop because there are 50+ attributes.
for attribute in range(0, len(dictPeople['attributes'][attribute])):
s[person]['attributes'][attribute]['game'] = \
dictPeople['attributes'][attribute]['game']
s[person]['attributes'][attribute]['high score'] = \
dictPeople['attributes'][attribute]['high score']
s[person]['attributes'][attribute]['time'] = \
dictPeople['attributes'][attribute]['time']
It turns out, I get the key error because I am not allowed to reference a
key/value pair the same way that I can with a dictionary. However, the
crazy thing is that I can call values in the db file using the same exact
format when trying to write.
For example: I can read data from the .db file using:
x = s['person_1']['name']
print(x)
BUT! I cannot write to that .db file using that exact format or I get an
invalid key error. Makes no sense!
s['person_1']['name'] = 'Bob'
# Returns invalid key entry. Makes no sense.
Therefore, I tried to extract data and populate the db file using the
following:
s[person] = { 'name': dictPeople.person_1['name'],
'type': dictPeople.person_1['type'],
for attribute in range(0, len(dictPeople['attributes'][attribute])):
['game': dictPeople['attributes'][attribute]['game'],
'high score': dictPeople['attributes'][attribute]['high score'],
'time': dictPeople['attributes'][attribute]['time']]
}
But, this obvously doesn't work because of the for loop. How can I do this?
-I am trying to figure out how to extract data from the dictionary
dictPeople.py file and store it in the people.db file.
-I am trying to add new people to the people.db file as more people become
available.
-I need to use the for loop because of the 50+ attributes I need to add.
-My future steps would be to modify some of the data values that I extract
and used to populate the people.db file, but my first step is to, at least,
extract, and then populate the people.db file.
Any help or guidance is greatly appreciated. Thank you for your time and
reading my question.
Thanks!
Aaron
--
https://mail.python.org/mailman/listinfo/python-list
Re: In Python 3, how to append a nested dictionary to a shelve file with the added difficulty of using a for loop?
Aaron Christensen wrote:
> Hello,
>
> I am trying to figure out how to populate a shelve file with a nested
> dictionary.
>
> These are my requirements:
>
> -Create shelve file called people.db
> -Append the shelve file with new people (person_1, person_2, etc.).
> -Use a for loop to iterate through 'attributes' so that I do not need to
> write out the lengthy code line by line to populate to the shelve file.
> -Need to reference shelve file data for future use
>
> Here is the key/value format that I would like to append to the shelve
> file.
>
> person_1 = { 'name': 'Bob', 'type': 'employee', 'attributes':
> [{'game': 'basketball', 'high score': '100', 'time': '3.34'},
> {'game': 'bridge', 'high score': '10', 'time': '30.34'},
> {'game': 'foosball', 'high score': '2', 'time': '24'}]
> '''
> 50+ other attributes
> '''
> }
>
> # Example: s['person_1]['attributes'][2]['time'] would call out '24'.
> # 's' is from 's = shelve.open('people')'
> I have a dictionary dictPeople.py file (created using pprint() that
> contains the information of person_1, etc. And I am extracting only a
> small percentage of the data that is needed.
>
> I have tried the following, but I get an invalid key error.
>
> import shelve, dictPeople
> s = shelve.open('people')
> person = 'person_1'
> s[person]['name'] = dictPeople.person_1['name']
> s[person]['type'] = dictPeople.person_1['type']
> # I need to use this for loop because there are 50+ attributes.
> for attribute in range(0, len(dictPeople['attributes'][attribute])):
> s[person]['attributes'][attribute]['game'] = \
> dictPeople['attributes'][attribute]['game']
> s[person]['attributes'][attribute]['high score'] = \
> dictPeople['attributes'][attribute]['high score']
> s[person]['attributes'][attribute]['time'] = \
> dictPeople['attributes'][attribute]['time']
> It turns out, I get the key error because I am not allowed to reference a
> key/value pair the same way that I can with a dictionary. However, the
> crazy thing is that I can call values in the db file using the same exact
> format when trying to write.
>
> For example: I can read data from the .db file using:
>
> x = s['person_1']['name']
> print(x)
> BUT! I cannot write to that .db file using that exact format or I get an
> invalid key error. Makes no sense!
>
> s['person_1']['name'] = 'Bob'
> # Returns invalid key entry. Makes no sense.
> Therefore, I tried to extract data and populate the db file using the
> following:
>
> s[person] = { 'name': dictPeople.person_1['name'],
> 'type': dictPeople.person_1['type'],
> for attribute in range(0, len(dictPeople['attributes'][attribute])):
> ['game': dictPeople['attributes'][attribute]['game'],
> 'high score': dictPeople['attributes'][attribute]['high score'],
> 'time': dictPeople['attributes'][attribute]['time']]
>
> }
> But, this obvously doesn't work because of the for loop. How can I do
> this?
>
> -I am trying to figure out how to extract data from the dictionary
> dictPeople.py file and store it in the people.db file.
> -I am trying to add new people to the people.db file as more people become
> available.
> -I need to use the for loop because of the 50+ attributes I need to add.
> -My future steps would be to modify some of the data values that I extract
> and used to populate the people.db file, but my first step is to, at
> least, extract, and then populate the people.db file.
>
> Any help or guidance is greatly appreciated. Thank you for your time and
> reading my question.
You don't need these loops. The problem is that when you read the dict from
the db
db = shelve.open(...)
person = db["person_1"]
person is a Python object completely independent of the shelve and the
shelve has no way to trace changes to that object:
$ python3
Python 3.4.3 (default, Oct 14 2015, 20:28:29)
[GCC 4.8.4] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import shelve
>>> db = shelve.open("tmp.shelve")
>>> person_1 = { 'name': 'Bob', 'type': 'employee', 'attributes':
... [{'game': 'basketball', 'high score': '100', 'time': '3.34'},
... {'game': 'bridge', 'high score': '10', 'time': '30.34'},
... {'game': 'foosball', 'high score': '2', 'time': '24'}]
... }
>>> db["person_1"] = person_1
>>> db["person_1"]["age"] = 42
>>> db.close()
>>> db = shelve.open("tmp.shelve")
>>> db["person_1"]
{'name': 'Bob', 'type': 'employee', 'attributes': [{'time': '3.34', 'high
score': '100', 'game': 'basketball'}, {'time': '30.34', 'high score': '10',
'game': 'bridge'}, {'time': '24', 'high score': '2', 'game': 'foosball'}]}
When you look at the data you see that there is no "age" key. The straight-
forward fix is to always use an assignment
db[key] = new_value
when you want a change:
>>> person = db["person_1"]
>>> person["age"] = 42
>>> db["person_1"] = person
>>> db.close()
That way the change will survive:
>>> db =
Re: In Python 3, how to append a nested dictionary to a shelve file with the added difficulty of using a for loop?
Hi Peter,
Thanks for the response! Several things you stated definitely got me
thinking. I really appreciate the response. I used what you said and I am
able to accomplish what I needed.
Thanks!
Aaron
On Mon, Dec 21, 2015 at 7:23 PM, Peter Otten <__pete...@web.de> wrote:
> Aaron Christensen wrote:
>
> > Hello,
> >
> > I am trying to figure out how to populate a shelve file with a nested
> > dictionary.
> >
> > These are my requirements:
> >
> > -Create shelve file called people.db
> > -Append the shelve file with new people (person_1, person_2, etc.).
> > -Use a for loop to iterate through 'attributes' so that I do not need to
> > write out the lengthy code line by line to populate to the shelve file.
> > -Need to reference shelve file data for future use
> >
> > Here is the key/value format that I would like to append to the shelve
> > file.
> >
> > person_1 = { 'name': 'Bob', 'type': 'employee', 'attributes':
> > [{'game': 'basketball', 'high score': '100', 'time': '3.34'},
> > {'game': 'bridge', 'high score': '10', 'time': '30.34'},
> > {'game': 'foosball', 'high score': '2', 'time': '24'}]
> > '''
> > 50+ other attributes
> > '''
> > }
> >
> > # Example: s['person_1]['attributes'][2]['time'] would call out '24'.
> > # 's' is from 's = shelve.open('people')'
> > I have a dictionary dictPeople.py file (created using pprint() that
> > contains the information of person_1, etc. And I am extracting only a
> > small percentage of the data that is needed.
> >
> > I have tried the following, but I get an invalid key error.
> >
> > import shelve, dictPeople
> > s = shelve.open('people')
> > person = 'person_1'
> > s[person]['name'] = dictPeople.person_1['name']
> > s[person]['type'] = dictPeople.person_1['type']
> > # I need to use this for loop because there are 50+ attributes.
> > for attribute in range(0, len(dictPeople['attributes'][attribute])):
> > s[person]['attributes'][attribute]['game'] = \
> > dictPeople['attributes'][attribute]['game']
> > s[person]['attributes'][attribute]['high score'] = \
> > dictPeople['attributes'][attribute]['high score']
> > s[person]['attributes'][attribute]['time'] = \
> > dictPeople['attributes'][attribute]['time']
> > It turns out, I get the key error because I am not allowed to reference a
> > key/value pair the same way that I can with a dictionary. However, the
> > crazy thing is that I can call values in the db file using the same exact
> > format when trying to write.
> >
> > For example: I can read data from the .db file using:
> >
> > x = s['person_1']['name']
> > print(x)
> > BUT! I cannot write to that .db file using that exact format or I get an
> > invalid key error. Makes no sense!
> >
> > s['person_1']['name'] = 'Bob'
> > # Returns invalid key entry. Makes no sense.
> > Therefore, I tried to extract data and populate the db file using the
> > following:
> >
> > s[person] = { 'name': dictPeople.person_1['name'],
> > 'type': dictPeople.person_1['type'],
> > for attribute in range(0, len(dictPeople['attributes'][attribute])):
> > ['game': dictPeople['attributes'][attribute]['game'],
> > 'high score': dictPeople['attributes'][attribute]['high score'],
> > 'time': dictPeople['attributes'][attribute]['time']]
> >
> > }
> > But, this obvously doesn't work because of the for loop. How can I do
> > this?
> >
> > -I am trying to figure out how to extract data from the dictionary
> > dictPeople.py file and store it in the people.db file.
> > -I am trying to add new people to the people.db file as more people
> become
> > available.
> > -I need to use the for loop because of the 50+ attributes I need to add.
> > -My future steps would be to modify some of the data values that I
> extract
> > and used to populate the people.db file, but my first step is to, at
> > least, extract, and then populate the people.db file.
> >
> > Any help or guidance is greatly appreciated. Thank you for your time and
> > reading my question.
>
> You don't need these loops. The problem is that when you read the dict from
> the db
>
> db = shelve.open(...)
> person = db["person_1"]
>
> person is a Python object completely independent of the shelve and the
> shelve has no way to trace changes to that object:
>
> $ python3
> Python 3.4.3 (default, Oct 14 2015, 20:28:29)
> [GCC 4.8.4] on linux
> Type "help", "copyright", "credits" or "license" for more information.
> >>> import shelve
> >>> db = shelve.open("tmp.shelve")
> >>> person_1 = { 'name': 'Bob', 'type': 'employee', 'attributes':
> ... [{'game': 'basketball', 'high score': '100', 'time': '3.34'},
> ... {'game': 'bridge', 'high score': '10', 'time': '30.34'},
> ... {'game': 'foosball', 'high score': '2', 'time': '24'}]
> ... }
> >>> db["person_1"] = person_1
> >>> db["person_1"]["age"] = 42
> >>> db.close()
> >>> db = shelve.open("tmp.shelve")
> >>> db["person_1"]
> {'name': 'Bob', 'type': 'employee',
How to append to a dictionary
I have some code here:
groups = {'IRISH' : 'green', 'AMERICAN' : 'blue'}
I want to add another key: 'ITALIAN' : 'orange'
How do I append this to 'groups'?
Thanks,
Harlin Seritt
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Re: How to append to a dictionary
groups = {'IRISH' : 'green', 'AMERICAN' : 'blue'}
I want to add another key: 'ITALIAN' : 'orange'
How do I append this to 'groups'?
groups['ITALIAN'] = 'orange'
as described at
http://docs.python.org/tut/node7.html#SECTION00750
-tkc
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Re: How to append to a dictionary
Harlin Seritt enlightened us with:
I have some code here:
groups = {'IRISH' : 'green', 'AMERICAN' : 'blue'}
I want to add another key: 'ITALIAN' : 'orange'
How do I append this to 'groups'?
groups['ITALIAN'] = 'orange'
Sybren
--
The problem with the world is stupidity. Not saying there should be a
capital punishment for stupidity, but why don't we just take the
safety labels off of everything and let the problem solve itself?
Frank Zappa
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Re: How to append to a dictionary
Harlin Seritt [EMAIL PROTECTED] writes:
groups = {'IRISH' : 'green', 'AMERICAN' : 'blue'}
I want to add another key: 'ITALIAN' : 'orange'
How do I append this to 'groups'?
Dictionary items have no implicit sequence, so you don't append to
one.
Assigning any value to a key in the dictionary will create that key if
it doesn't exist.
groups = {'IRISH': 'green', 'AMERICAN': 'blue'}
groups['ITALIAN'] = 'orange'
print groups
{'IRISH': 'green', 'AMERICAN': 'blue', 'ITALIAN': 'orange'}
Please enjoy your trip through the Python tutorial, working through
the examples and understanding each one.
URL:http://docs.python.org/tut/
--
\It is the responsibility of intellectuals to tell the truth |
`\and expose lies. -- Noam Chomsky |
_o__) |
Ben Finney
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