Assume python dict:
mydict = {'a': 100, 'b': 200, 'c': 300}
I know one of the values:
value = 200
How to remove the 'b': 200 pair from the dict? I need this:
mydict = {'a': 100, 'c': 300}
Use a dictionary comprehension. Note that (as jonrsharpe has stated) this will create a new dictionary which excludes the key:value pair that you want to remove. If you want to delete it from your original dictionary then please see his answer.
>>> d = {'a': 100, 'b': 200, 'c': 300}
>>> val = 200
# Use d.items() for Python 2.x and d.iteritems() for Python 3.x
>>> d2 = {k:v for k,v in d.items() if v != val}
>>> d2
{'a': 100, 'c': 300}
It sounds like you want:
for key, val in list(mydict.items()):
if val == value:
del mydict[key]
break # unless you want to remove multiple occurences
You'll need to loop over every items(), either with dict comprehension:
new_dict = {k:v for k,v in my_dict.items() if predicate(value)}
Or modifying the existing dictionary:
for k,v in my_dict.items():
if not predicate(v):
del my_dict[k]
The simplest i found:
for key in [k for k,v in mydict.items() if v==200]:
del mydict[key]
Related
My dictionary is below, if the values end with csv or json i need to put in another dictionary of same dictionary:
d = {'a': ['1.json', '1.html', '1.csv'], 'B': ['2.json', '2.html', '2.csv']}
Code is below:
d = {}
for k,v in d.items():
for i in v:
if i.split('.')[1] == 'csv' or i.split('.')[1] == 'json':
d[k] = v
My out
{'a': ['1.json', '1.html', '1.csv'], 'B': ['2.json', '2.html', '2.csv']}
Expected out
{'a': ['1.json', '1.csv'], 'B': ['2.json', '2.csv']}
You never change v, so the dictionary is unchanged. You should build a new list only keeping the wanted values:
for k,v in d.items():
v = [i for i in v if i.split('.')[1] in ('csv', 'json')]
d[k] = v
or even better:
d = {k: [i for i in v if i.split('.')[1] in ('csv', 'json')]
for k,v in d.items()}
You can try to use rfind to find the last . in the file name and to add it the list if it's matching CSV or JSON
d = {'a': ['1.json', '1.html', '1.csv'], 'B': ['2.json', '2.html', '2.csv']}
for k, v in d.items():
d[k] = []
for i in v:
if i[i.rfind(".") + 1:] in ('csv', 'json'):
d[k].append(i)
Please test it:
from collections import defaultdict
x = defaultdict(list)
d = {'a': ['1.json', '1.html', '1.csv'], 'B': ['2.json', '2.html', '2.csv']}
for k, v in d.items():
for i in v:
if i.split('.')[-1] in ['csv', 'json']:
x[k].append(i)
print(dict(x))
You can use dict comprehension with list comp on each of the values while filtering with str.endswith.
d = {'a': ['1.json', '1.html', '1.csv'], 'B': ['2.json', '2.html', '2.csv']}
new = {k:[i for i in v if i.endswith(('.csv','.json'))] for k,v in d.items()}
# {'a': ['1.json', '1.csv'], 'B': ['2.json', '2.csv']}
A simple solution would be the following, assuming you want to update the existing dictionary
for k, v in d.items():
for i in v:
if i.rsplit('.')[-1] not in {'csv','json'}:
v.remove(i)
You can use split filename and add condition for checking the file type (using list for comparison could be a better option if you want to add more file types).
Combine this with list Comprehension, which provides a concise way to create lists.
d = {'a': ['1.json', '1.html', '1.csv'], 'B': ['2.json', '2.html', '2.csv']}
filtered_d = {key:[item for item in value if item.split('.')[-1] in ['json','csv']] for key, value in d.items()}
you can check documentation here for more detail on list comprehension.
I would like to know how if there exists any python function to merge two dictionary and combine all values that have a common key.
I have found function to append two dict, to merge two dict but not to combine its values.
Example:
D1 = [{k1: v01}, {k3: v03}, {k4: v04},}],
D2 = [{k1: v11}, {k2: v12}, {k4: v14},}],
this should be the expected result:
D3 = [
{k1: [v01, v11]},
{k2: [ v12]},
{K3: [v03 ]},
{k4: [v04, v14]},
]
There is no built-in function for this but you can use a defaultdict for this:
from collections import defaultdict
d = defaultdict(list)
for other in [d1, d1]:
for k, v in other.items():
d[k].append(v)
A solution, without importing anything:
# First initialize data, done correctly here.
D1 = [{'k1': 'v01'}, {'k3': 'v03'}, {'k4': 'v04'}]
D2 = [{'k1': 'v11'}, {'k2': 'v12'}, {'k4': 'v14'}]
# Get all unique keys
keys = {k for d in [*D1, *D2] for k in d}
# Initialize an empty dict
D3 = {x:[] for x in keys}
# sort to maintain order
D3 = dict(sorted(D3.items()))
#Iterate and extend
for x in [*D1, *D2]:
for k,v in x.items():
D3[k].append(v)
# NOTE: I do not recommend you convert a dictionary into a list of records.
# Nonetheless, here is how it would be done.
# To convert to a list
D3_list = [{k:v} for k,v in D3.items()]
print(D3_list)
# [{'k1': ['v01', 'v11']},
# {'k2': ['v12']},
# {'k3': ['v03']},
# {'k4': ['v04', 'v14']}]
If you meant to use actual dicts, instead of lists of dicts, this is easier.
D1 = dict(k1=1, k3=3, k4=4)
D2 = dict(k1=11, k2=12, k4=14)
There isn't a simple built-in function to do this, but the setdefault method is close.
It tries to get the given key, but creates it if it doesn't exist.
D3 = {}
for k, v in D1.items() | D2.items():
D3.setdefault(k, set()).add(v)
And the result.
{'k4': {4, 14}, 'k1': {1, 11}, 'k3': {3}, 'k2': {12}}
This all assumes the order doesn't matter, just combining sets.
A more generic way to merge dicts together may look like this.
(To answer a similar SO question)
def merge(combiner, dicts):
new_dict = {}
for d in dicts:
for k, v in d.items():
if k in new_dict:
new_dict[k] = combiner(new_dict[k], v)
else:
new_dict[k] = v
return new_dict
x = {'a': 'A', 'b': 'B'}
y = {'b': 'B', 'c': 'C'}
z = {'a': 'A', 'd': 'D'}
merge_dicts(combiner= lambda x, y: f'{x} AND {y}', dicts=(x,y,z))
# {'a': 'A AND A', 'b': 'B AND B', 'c': 'C', 'd': 'D'}
I have two dictionaries:
dict1 = {'a': 1,
'b': 2,
'c': 3,
'd': 4,
'x': 5}
and
dict2 = {'a': 'start',
'b': 'start',
'c': 'end',
'd': 'end'}
I am trying to create a new dictionary that maps the values start and end as keys to a dictionary that would contain the info of dict1, while keeping those that are not present in dict2 as keys, e.g.:
dict3 = {'start': {'a': 1, 'b': 2},
'end': {'c': 3, 'd': 4},
'x': {'x': 5}
}
Use dict.setdefault() to create the nested dictionaries in dict3 if not yet there, and dict.get() to determine the key in the top-level output dictionary:
dict3 = {}
for k, v in dict1.items():
nested = dict3.setdefault(dict2.get(k, k), {})
nested[k] = v
So dict2.get(k, k) will produce the value from dict2 for a given key from dict1, using the key itself as a default. So for the 'x' key, that'll produce 'x' as there is no mapping in dict2 for that key.
Demo:
>>> dict3 = {}
>>> for k, v in dict1.items():
... nested = dict3.setdefault(dict2.get(k, k), {})
... nested[k] = v
...
>>> dict3
{'start': {'a': 1, 'b': 2}, 'end': {'c': 3, 'd': 4}, 'x': {'x': 5}}
I actually figured it out while abstracting the example and typing up my question here (should have maybe done this earlier...). Anyways: Yay!
So here is my solution, in case it may help someone. If someone knows a swifter or more elegant way to do it, I would be glad to learn!
dict3 = dict()
for k, v in dict1.items():
# if the key of dict1 exists also in dict2
if k in dict2.keys():
# get its value (the keys-to-be for the new dict3)
new_key = dict2[k]
# if the new key is already in the new dict
if new_key in dict3.keys():
# appends new dict entry to dict3
dict3[new_key].update({k: v})
# otherwise create a new entry
else:
dict3[new_key] = {k: v}
# if there is no corresponding mapping present
else:
# treat the original key as the new key and add to dict3
no_map = k
dict3[no_map] = {k: v}
I want to delete the keys in my dictionary with the value of 1, but the only way I know of is to manually input keys to delete which is inefficient. Any ideas?
I have a feeling I should use:
for key,value in mydict.items():
For example, if I had the dictionary
mydict = {'a':1, 'b':2, 'c':3, 'd':1, 'e':1}
how would I delete the keys with a value of 1 so that I can print a dictionary purely of keys with values > 1?
You can use a dict comprehension to filter the dictionary:
>>> mydict = {'a':1, 'b':2, 'c':3, 'd':1, 'e':1}
>>> {k:v for k,v in mydict.items() if v != 1}
{'c': 3, 'b': 2}
>>>
Note that you should use mydict.iteritems() if you are on Python 2.x to avoid creating an unnecessary list.
You could go though each of the items (the key value pair) in the dictionary to check the conflict and add value in new Dict.(here.. result) that not contain duplicate.
mydict = {'a':1, 'b':2, 'c':3, 'd':1, 'e':1}
result = {}
for key,value in mydict.items():
if value not in result.values():
result[key] = value
print result
To create a new dictionary with the items which has the value greater than 1 use dict comprehension like below.
>>> mydict = {'a':1, 'b':2, 'c':3, 'd':1, 'e':1}
>>> {key:value for key,value in mydict.items() if value > 1}
{'c': 3, 'b': 2}
how to remove keys in a dict which do not have any value. I have a dict as :
d = {'CB': '', 'CA': [-7.5269999504089355, -2.2330000400543213, 6.748000144958496], 'C': [-8.081000328063965, -3.619999885559082, 6.406000137329102], 'N': [-6.626999855041504, -2.318000078201294, 7.9029998779296875], 'H':''}
I want to remove keys which values are blank. I need output as :
d = {'CA': [-7.5269999504089355, -2.2330000400543213, 6.748000144958496], 'C': [-8.081000328063965, -3.619999885559082, 6.406000137329102], 'N': [-6.626999855041504, -2.318000078201294, 7.9029998779296875]}
how to do this?
Use a dict-comprehension:
>>> {k:v for k, v in d.items() if v != ''}
{'N': [-6.626999855041504, -2.318000078201294, 7.9029998779296875], 'CA': [-7.5269999504089355, -2.2330000400543213, 6.748000144958496], 'C': [-8.081000328063965, -3.619999885559082, 6.406000137329102]}
If by empty you meant any falsy value then use just if v.
If you want to modify the original dict itself(this will affect all the references to the dict object):
for k, v in d.items():
if not v: del d[k]
>> {key:value for key, value in d.items() if value}
If you really need to do this in one-line:
for k in [k for k,v in d.iteritems() if not v]: del d[k]
This is ugly. It makes a list of all of the keys in d that have a "blank" value and then iterates over this list removing the keys from d (to avoid the unsafe removal of items from dictionary during iteration).
I'd suggest that three lines would be more readable:
blanks = [k for k,v in d.iteritems() if not v]
for k in blanks:
del d[k]
If making a fresh dict is acceptable, use a dict comprehension:
d2 = {k:v for k,v in d.iteritems() if not v}