I have an 3rd party python method (ANSA) that returns strings based on user gui input with default values to start with.
The default values may look like this and are stored in a dictionary:
Label Value
Userdata1 123
Userdata2 'abc'
Userdata3 123.456
... ...
Now, these are all returned as a dictionary with string values by the 3rd party "back_box" function.
Is it, based on the class of the default values, possible to change the new string class back to the default classes?
Here is something that may make it a bit clearer:
data = {'Userdata1':123,
'Userdata2':'abc',
'Userdata3':123.456}
new_data = black_box(data) # Outputs dictionary with the values as strings
for i, var_name in enumerate(data):
original_class = type(data[var_name])
new_data[var_name] = new_data[var_name].StrToClass(original_class)
So what i am looking for is a method/function/magic trick called strToClass in the above example.
I know it may be a risky way to do it, but any ideas about pythonic ways to do it?
(python3 btw)
Just keep the types of the original values of the dictionary in a separate new dictionary, then use that dictionary to transform them back --
>>> my_dict = {'Userdata1':123,
'Userdata2':'abc',
'Userdata3':123.456}
>>> type_dict = {k:type(v) for k,v in my_dict.iteritems()}
>>> type_dict
{'Userdata2': <type 'str'>, 'Userdata3': <type 'float'>, 'Userdata1': <type 'int'>}
>>> str_dict = {k:str(v) for k,v in my_dict.iteritems()}
>>> str_dict
{'Userdata2': 'abc', 'Userdata3': '123.456', 'Userdata1': '123'}
>>> new_dict = {k:type_dict[k](v) for k,v in str_dict.iteritems()}
>>> new_dict
{'Userdata2': 'abc', 'Userdata3': 123.456, 'Userdata1': 123}
Or, using your pseudocode
for i, var_name in enumerate(data):
original_class = type(data[var_name])
new_data[var_name] = original_class(new_data[var_name])
I believe the magic function you're looking for is ast.literal_eval.
for i, var_name in enumerate(data):
new_data[var_name] = ast.literal_eval(new_data[var_name])
Never name a variable def
For complicated classes you'll need more work, however this should do nicely for basic types, and types that have a representation identical to other basic types
d = {'Userdata1':123,
'Userdata2':'abc',
'Userdata3':123.456}
# create a list containing each type
var_type = [ type(d[key]) for key in sorted(d.keys())]
d = black_box(d)
import ast
for vt, key in zip(var_type, sorted(d.keys())):
# if the variable was a string you're already done
if not vt == str:
# cast from string to generic type
d[key] = ast.literal_eval(d[key])
# cast from generic to original type
try:
d[key] = vt(d[key])
except:
print "Error converting dictionary value with key: " + key
print "Unable to cast {} as {}".format(d[key], vt)
Related
I have a string '{"News":"news", "News":"politics", "News":"breaking", "News":"stories"}' that I am trying to convert to a dictionary. I have used both json.loads() and ast.literal_eval() to convert them, but it seems like both methods only take the last key value pair.
Is there a one line solution for this problem? Or would I need something more complex?
Assuming that the string is otherwise valid JSON, you could use the object_pairs_hook parameter to the JSON decoder:
import json
def multidict(l):
result = {}
for k, v in l:
result.setdefault(k, []).append(v)
return result
data = '{"News":"news", "News":"politics", ' \
'"News":"breaking", "News":"stories"}'
data = json.loads(data, object_pairs_hook=multidict)
assert data == {'News': ['news', 'politics', 'breaking', 'stories']}
Dictionaries can not have more than one instance for every key. That's why you can not produce a dictionary with 4 identical keys, News for that case.
Consider changing the data format to something that meet your specifications, like a dictionary with a list values:
{"News": ["news", "politics", "breaking", "stories"]}
You could use defaultdict with a list:
>>> from collections import defaultdict
>>> dictionary = defaultdict(list)
>>> values = '{"News":"news", "News":"politics", "News":"breaking", "News":"stories"}'
>>> for pair in values.strip('{}').split(','):
... key, value = pair.strip().split(':')
... key = key.strip('"')
... value = value.strip('"')
... dictionary[key].append(value)
Result:
>>> dictionary
{'News': ['news', 'politics', 'breaking', 'stories']}
I'm using ConfigParser which returns a dictionary of configuration data as such:
{'general': {'UserKey': 'thisisatestkey'}}
If I want to simply print the value of the UserKey key (in this case thisisatestkey), then I generally just do a print "Your key is: {0}".format(mydictvar.get('UserKey')).
If I just print out the raw dict to a string I get the above. If I use the print statement above I get result of None since there is no key in the root of the dict called UserKey. If I .get('general') I just get: {'UserKey': 'thisisatestkey'}
Obviously I could do a fore loop like so:
keydic = cp.get_config_data()
for m, k in keydic.iteritems():
for s, v in k.iteritems():
userkey = v
and then print userkey which works fine. But I want to know how I can just avoid having to do the entire for loop first and just print the darned value right inline? Thanks!
You can use
mydictvar['general']['UserKey']
Or, if keys might be missing
mydictvar.get('general', {}).get('UserKey')
mydictvar['general'] returns a dictionary object; you can then just apply [...] to that value to retrieve the next key.
This works in string formatting too:
>>> mydictvar = {'general': {'UserKey': 'thisisatestkey'}}
>>> print "Your key is: {0[general][UserKey]}".format(mydictvar)
Your key is: thisisatestkey
simply without loop:
>>> my_dict = {'general': {'UserKey': 'thisisatestkey'}}
>>> my_dict['general']['UserKey']
'thisisatestkey'
[Python 3.4.2]
I know this question sounds ridiculous, but I can't figure out where I'm messing up. I'm trying to add keys and values to a dictionary by using strings instead of quoted text. So instead of this,
dict['key'] = value
this:
dict[key] = value
When I run the command above, I get this error:
TypeError: 'str' object does not support item assignment
I think Python is thinking that I'm trying to create a string, not add to a dictionary. I'm guessing I'm using the wrong syntax. This is what I'm trying to do:
dict[string_for_key][string_for_value] = string_for_deeper_value
I want this^ command to do this:
dict = {string_for_key: string_for_value: string_for_deeper_value}
I'm getting this error:
TypeError: 'str' object does not support item assignment
I should probably give some more context. I'm:
creating one dictionary
creating a copy of it (because I need to edit the dictionary while iterating through it)
iterating through the first dictionary while running some queries
trying to assign a query's result as a value for each "key: value" in the dictionary.
Here's a picture to show what I mean:
key: value: query_as_new_value
-----EDIT-----
Sorry, I should have clarified: the dictionary's name is not actually 'dict'; I called it 'dict' in my question to show that it was a dictionary.
-----EDIT-----
I'll just post the whole process I'm writing in my script. The error occurs during the last command of the function. Commented out at the very bottom are some other things I've tried.
from collections import defaultdict
global query_line, pericope_p, pericope_f, pericope_e, pericope_g
def _pre_query(self, typ):
with open(self) as f:
i = 1
for line in f:
if i == query_line:
break
i += 1
target = repr(line.strip())
###skipping some code
pericope_dict_post[self][typ] = line.strip()
#^Outputs error TypeError: 'str' object does not support item assignment
return
pericope_dict_pre = {'pericope-p.txt': 'pericope_p',
'pericope-f.txt': 'pericope_f',
'pericope-e.txt': 'pericope_e',
'pericope-g.txt': 'pericope_g'}
pericope_dict_post = defaultdict(dict)
#pericope_dict_post = defaultdict(list)
#pericope_dict_post = {}
for key, value in pericope_dict_pre.items():
pericope_dict_post[key] = value
#^Works
#pericope_dict_post.update({key: value})
#^Also works
#pericope_dict_post.append(key)
#^AttributeError: 'dict' object has no attribute 'append'
#pericope_dict_post[key].append(value)
#^AttributeError: 'dict' object has no attribute 'append'
_pre_query(key, value)
-----FINAL EDIT-----
Matthias helped me figure it out, although acushner had the solution too. I was trying to make the dictionary three "levels" deep, but Python dictionaries cannot work this way. Instead, I needed to create a nested dictionary. To use an illustration, I was trying to do {key: value: value} when I needed to do {key: {key: value}}.
To apply this to my code, I need to create the [second] dictionary with all three strings at once. So instead of this:
my_dict[key] = value1
my_dict[key][value1] = value2
I need to do this:
my_dict[key][value1] = value2
Thanks a ton for all your help guys!
You could create a dictionary that expands by itself (Python 3 required).
class AutoTree(dict):
"""Dictionary with unlimited levels"""
def __missing__(self, key):
value = self[key] = type(self)()
return value
Use it like this.
data = AutoTree()
data['a']['b'] = 'foo'
print(data)
Result
{'a': {'b': 'foo'}}
Now I'm going to explain your problem with the message TypeError: 'str' object does not support item assignment.
This code will work
from collections import defaultdict
data = defaultdict(dict)
data['a']['b'] = 'c'
data['a'] doesn't exist, so the default value dict is used. Now data['a'] is a dict and this dictionary gets a new value with the key 'b' and the value 'c'.
This code won't work
from collections import defaultdict
data = defaultdict(dict)
data['a'] = 'c'
data['a']['b'] = 'c'
The value of data['a'] is defined as the string 'c'. Now you can only perform string operations with data['a']. You can't use it as a dictionary now and that's why data['a']['b'] = 'c' fails.
first, do not use dict as your variable name as it shadows the built-in of the same name.
second, all you want is a nested dictionary, no?
from collections import defaultdict
d = defaultdict(dict)
d[string_for_key][string_for_value] = 'snth'
another way, as #Matthias suggested, is to create a bottomless dictionary:
dd = lambda: defaultdict(dd)
d = dd()
d[string_for_key][string_for_value] = 'snth'
you can do something like this:
>>> my_dict = {}
>>> key = 'a' # if key is not defined before it will raise NameError
>>> my_dict[key] = [1]
>>> my_dict[key].append(2)
>>> my_dict
{'a': [1, 2]}
Note: dict is inbuilt don't use it as variable name
I was wondering if there was a way to initialize a dictionary in python with keys but no corresponding values until I set them. Such as:
Definition = {'apple': , 'ball': }
and then later i can set them:
Definition[key] = something
I only want to initialize keys but I don't know the corresponding values until I have to set them later. Basically I know what keys I want to add the values as they are found. Thanks.
Use the fromkeys function to initialize a dictionary with any default value. In your case, you will initialize with None since you don't have a default value in mind.
empty_dict = dict.fromkeys(['apple','ball'])
this will initialize empty_dict as:
empty_dict = {'apple': None, 'ball': None}
As an alternative, if you wanted to initialize the dictionary with some default value other than None, you can do:
default_value = 'xyz'
nonempty_dict = dict.fromkeys(['apple','ball'],default_value)
You could initialize them to None.
you could use a defaultdict. It will let you set dictionary values without worrying if the key already exists. If you access a key that has not been initialized yet it will return a value you specify (in the below example it will return None)
from collections import defaultdict
your_dict = defaultdict(lambda : None)
It would be good to know what your purpose is, why you want to initialize the keys in the first place. I am not sure you need to do that at all.
1) If you want to count the number of occurrences of keys, you can just do:
Definition = {}
# ...
Definition[key] = Definition.get(key, 0) + 1
2) If you want to get None (or some other value) later for keys that you did not encounter, again you can just use the get() method:
Definition.get(key) # returns None if key not stored
Definition.get(key, default_other_than_none)
3) For all other purposes, you can just use a list of the expected keys, and check if the keys found later match those.
For example, if you only want to store values for those keys:
expected_keys = ['apple', 'banana']
# ...
if key_found in expected_keys:
Definition[key_found] = value
Or if you want to make sure all expected keys were found:
assert(all(key in Definition for key in expected_keys))
You can initialize the values as empty strings and fill them in later as they are found.
dictionary = {'one':'','two':''}
dictionary['one']=1
dictionary['two']=2
Comprehension could be also convenient in this case:
# from a list
keys = ["k1", "k2"]
d = {k:None for k in keys}
# or from another dict
d1 = {"k1" : 1, "k2" : 2}
d2 = {k:None for k in d1.keys()}
d2
# {'k1': None, 'k2': None}
q = input("Apple")
w = input("Ball")
Definition = {'apple': q, 'ball': w}
Based on the clarifying comment by #user2989027, I think a good solution is the following:
definition = ['apple', 'ball']
data = {'orange':1, 'pear':2, 'apple':3, 'ball':4}
my_data = {}
for k in definition:
try:
my_data[k]=data[k]
except KeyError:
pass
print my_data
I tried not to do anything fancy here. I setup my data and an empty dictionary. I then loop through a list of strings that represent potential keys in my data dictionary. I copy each value from data to my_data, but consider the case where data may not have the key that I want.
Is it possible to assign multiple keys per value in a Python dictionary. One possible solution is to assign value to each key:
dict = {'k1':'v1', 'k2':'v1', 'k3':'v1', 'k4':'v2'}
but this is not memory efficient since my data file is > 2 GB. Otherwise you could make a dictionary of dictionary keys:
key_dic = {'k1':'k1', 'k2':'k1', 'k3':'k1', 'k4':'k4'}
dict = {'k1':'v1', 'k4':'v2'}
main_key = key_dict['k2']
value = dict[main_key]
This is also very time and effort consuming because I have to go through whole dictionary/file twice. Is there any other easy and inbuilt Python solution?
Note: my dictionary values are not simple string (as in the question 'v1', 'v2') rather complex objects (contains different other dictionary/list etc. and not possible to pickle them)
Note: the question seems similar as How can I use both a key and an index for the same dictionary value?
But I am not looking for ordered/indexed dictionary and I am looking for other efficient solutions (if any) other then the two mentioned in this question.
What type are the values?
dict = {'k1':MyClass(1), 'k2':MyClass(1)}
will give duplicate value objects, but
v1 = MyClass(1)
dict = {'k1':v1, 'k2':v1}
results in both keys referring to the same actual object.
In the original question, your values are strings: even though you're declaring the same string twice, I think they'll be interned to the same object in that case
NB. if you're not sure whether you've ended up with duplicates, you can find out like so:
if dict['k1'] is dict['k2']:
print("good: k1 and k2 refer to the same instance")
else:
print("bad: k1 and k2 refer to different instances")
(is check thanks to J.F.Sebastian, replacing id())
Check out this - it's an implementation of exactly what you're asking: multi_key_dict(ionary)
https://pypi.python.org/pypi/multi_key_dict
(sources at https://github.com/formiaczek/python_data_structures/tree/master/multi_key_dict)
(on Unix platforms it possibly comes as a package and you can try to install it with something like:
sudo apt-get install python-multi-key-dict
for Debian, or an equivalent for your distribution)
You can use different types for keys but also keys of the same type. Also you can iterate over items using key types of your choice, e.g.:
m = multi_key_dict()
m['aa', 12] = 12
m['bb', 1] = 'cc and 1'
m['cc', 13] = 'something else'
print m['aa'] # will print '12'
print m[12] # will also print '12'
# but also:
for key, value in m.iteritems(int):
print key, ':', value
# will print:1
# 1 : cc and 1
# 12 : 12
# 13 : something else
# and iterating by string keys:
for key, value in m.iteritems(str):
print key, ':', value
# will print:
# aa : 12
# cc : something else
# bb : cc and 1
m[12] = 20 # now update the value
print m[12] # will print '20' (updated value)
print m['aa'] # will also print '20' (it maps to the same element)
There is no limit to number of keys, so code like:
m['a', 3, 5, 'bb', 33] = 'something'
is valid, and either of keys can be used to refer to so-created value (either to read / write or delete it).
Edit: From version 2.0 it should also work with python3.
Using python 2.7/3 you can combine a tuple, value pair with dictionary comprehension.
keys_values = ( (('k1','k2'), 0), (('k3','k4','k5'), 1) )
d = { key : value for keys, value in keys_values for key in keys }
You can also update the dictionary similarly.
keys_values = ( (('k1',), int), (('k3','k4','k6'), int) )
d.update({ key : value for keys, value in keys_values for key in keys })
I don't think this really gets to the heart of your question but in light of the title, I think this belongs here.
The most straightforward way to do this is to construct your dictionary using the dict.fromkeys() method. It takes a sequence of keys and a value as inputs and then assigns the value to each key.
Your code would be:
dict = dict.fromkeys(['k1', 'k2', 'k3'], 'v1')
dict.update(dict.fromkeys(['k4'], 'v2'))
And the output is:
print(dict)
{'k1': 'v1', 'k2': 'v1', 'k3': 'v1', 'k4': 'v2'}
You can build an auxiliary dictionary of objects that were already created from the parsed data. The key would be the parsed data, the value would be your constructed object -- say the string value should be converted to some specific object. This way you can control when to construct the new object:
existing = {} # auxiliary dictionary for making the duplicates shared
result = {}
for k, v in parsed_data_generator():
obj = existing.setdefault(v, MyClass(v)) # could be made more efficient
result[k] = obj
Then all the result dictionary duplicate value objects will be represented by a single object of the MyClass class. After building the result, the existing auxiliary dictionary can be deleted.
Here the dict.setdefault() may be elegant and brief. But you should test later whether the more talkative solution is not more efficient -- see below. The reason is that MyClass(v) is always created (in the above example) and then thrown away if its duplicate exists:
existing = {} # auxiliary dictionary for making the duplicates shared
result = {}
for k, v in parsed_data_generator():
if v in existing:
obj = existing[v]
else:
obj = MyClass(v)
existing[v] = obj
result[k] = obj
This technique can be used also when v is not converted to anything special. For example, if v is a string, both key and value in the auxiliary dictionary will be of the same value. However, the existence of the dictionary ensures that the object will be shared (which is not always ensured by Python).
I was able to achieve similar functionality using pandas MultiIndex, although in my case the values are scalars:
>>> import numpy
>>> import pandas
>>> keys = [numpy.array(['a', 'b', 'c']), numpy.array([1, 2, 3])]
>>> df = pandas.DataFrame(['val1', 'val2', 'val3'], index=keys)
>>> df.index.names = ['str', 'int']
>>> df.xs('b', axis=0, level='str')
0
int
2 val2
>>> df.xs(3, axis=0, level='int')
0
str
c val3
I'm surprised no one has mentioned using Tuples with dictionaries. This works just fine:
my_dictionary = {}
my_dictionary[('k1', 'k2', 'k3')] = 'v1'
my_dictionary[('k4')] = 'v2'