So I have this block of code
dictionary = {
'key1': {'a': 1, 'b': 2, 'c': 10},
'key2': {'d': 1, 'e': 1, 'c': 11},
'key3': {'d': 2, 'b': 1, 'g': 12}}
and
list1 = (a,b,c)
What I want to do is run a loop that finds the maximums of all the items in the list and returns the key. So for example, the maximum of 'c' would return 'key2', the maximum of 'b' would return 'key1', etc.
So far I have
for value in list1:
m = max(dictionary, key=lambda v: dictionary[v][value])
print(m + "\n")
But this only works if the same subkey exists in all keys in the dictionary. Any ideas on what to do?
Use float('-inf') when the key is missing:
m = max(dictionary, key=lambda v: dictionary[v].get(value, float('-inf')))
Negative infinity is guaranteed to be smaller than any existing value in the dictionaries, ensuring that nested dictionaries with the specific key missing are ignored.
Demo:
>>> dictionary = {
... 'key1': {'a': 1, 'b': 2, 'c': 10},
... 'key2': {'d': 1, 'e': 1, 'c': 11},
... 'key3': {'d': 2, 'b': 1, 'g': 12}}
>>> list1 = ('a', 'b', 'c')
>>> for value in list1:
... print(value, max(dictionary, key=lambda v: dictionary[v].get(value, float('-inf'))))
...
a key1
b key1
c key2
However, it'll be more efficient if you looped over all your dictionary values just once instead:
maximi = dict.fromkeys(list1, (None, float('-inf')))
for key, nested in dictionary.items():
for k in nested.keys() & maximi: # intersection of keys
if maximi[k][0] is None or dictionary[maximi[k][0]][k] < nested[k]:
maximi[k] = (key, nested[k])
for value in list1:
print(value, maximi[value][0])
That's presuming you are using Python 3; in Python 2, replace .items() with .iteritems() and .keys() with .viewkeys().
Demo:
>>> maximi = dict.fromkeys(list1, (None, float('-inf')))
>>> for key, nested in dictionary.items():
... for k in nested.keys() & maximi: # intersection of keys
... if maximi[k][0] is None or dictionary[maximi[k][0]][k] < nested[k]:
... maximi[k] = (key, nested[k])
...
>>> maximi
{'a': ('key1', 1), 'b': ('key1', 2), 'c': ('key2', 11)}
>>> for value in list1:
... print(value, maximi[value][0])
...
a key1
b key1
c key2
Related
I want a solution to make all the keys of a dictionary have a unique value, and to do that delete the values as minimum as possible to have each value unique. For example:
my_dict = {'c': 3, 'e': 3, 'a': 2, 'f': 2, 'd': 2}
for the above dictionary I need to sub 2 from 'f' and 3 of times from 'e' and 1 time from 'd'. and result would be 6 which means {'c':3, 'a':2, 'd':1}. Removing keys is not a problem.
note we could remove 'c' rather than 'e' or 'a' rather than f'' it's not important which key should be decreed or be removed , what matters is having unique values
This is what I have tried:
for k, v in my_dict.items():
c = 0
while len(my_dict.values()) > len(set(my_dict.values())):
my_dict[k] = my_dict[k] -1
c += 1
It is not the result you were expecting, but it meets the requirements.
my_dict = {'c': 3, 'e': 3, 'a': 2, 'f': 2, 'd': 2}
to_remove = []
result = {}
for key, value in my_dict.items():
while value > 0:
if value not in to_remove:
to_remove.append(value)
result[key] = value
break
else:
value -= 1
result
Simple approach:
my_dict = {'c': 3, 'e': 3, 'a': 2, 'f': 2, 'd': 2}
rd = {v: k for k, v in my_dict.items()}
my_dict = {v: k for k, v in rd.items()}
print(my_dict)
I have a dictionary of values that gives the number of occurrences of a value in a list. How can I return a new dictionary that divides the former dictionary into separate dictionaries based on the value?
In other words, I want to sort this dictionary:
>>> a = {'A':2, 'B':3, 'C':4, 'D':2, 'E':3}
to this one.
b = {2: {'A', 'D'}, 3: {'B', 'E'}, 4: {'C'}}
How do I approach the problem?
from collections import defaultdict
a = {'A': 2, 'B': 3, 'C': 4, 'D': 2, 'E': 3}
b = defaultdict(set)
for k, v in a.items():
b[v].add(k)
This is what you'll get:
defaultdict(<class 'set'>, {2: {'D', 'A'}, 3: {'B', 'E'}, 4: {'C'}})
You can convert b to a normal dict afterwards with b = dict(b).
if you are a python beginner like me, you probably wanna try this
a = {'A': 2 , 'B': 3 , 'C' : 4 , 'D' : 2, 'E' : 3}
b = {}
for key in a:
lst = []
new_key = a[key]
if new_key not in b:
lst.append(key)
b[new_key] = lst
else:
b[new_key].append(key)
print(b)
It uses the mutable property of python dictionary to achieve the result you want.
I have a large list containing many dictionaries. Within each dictionary I want to iterate over 3 particular keys and then dump into a new list. The keys are the same for each dict.
For example, I'd like to grab keys c, d, e from all the dicts in List below, output to List2.
List = [{'a':1, 'b':2, 'c':3, 'd':4, 'e':5, 'f':6...},
{'a':10, 'b':20, 'c':30, 'd':40, 'e':50, 'f':60...},
{'a':100, 'b':200, 'c':300, 'd':400, 'e':500, 'f':600...},]
List2 = [{'c':3, 'd':4, 'e':5},
{'c':30, 'd':40, 'e':50},
{'c':300, 'd':400, 'e':500}]
You can use a nested dict comprehension:
keys = ('c', 'd', 'e')
[{k: d[k] for k in keys} for d in List]
If those keys may be missing, you can use a dictionary view object (dict.viewkeys() in Python 2, dict.keys() in Python 3) to find an intersection to only include keys that are actually present:
keys = {'c', 'd', 'e'}
[{k: d[k] for k in d.viewkeys() & keys} for d in List]
Demo:
>>> List = [{'a':1, 'b':2, 'c':3, 'd':4, 'e':5, 'f':6},
... {'a':10, 'b':20, 'c':30, 'd':40, 'e':50, 'f':60},
... {'a':100, 'b':200, 'c':300, 'd':400, 'e':500, 'f':600}]
>>> keys = ('c', 'd', 'e')
>>> [{k: d[k] for k in keys} for d in List]
[{'c': 3, 'e': 5, 'd': 4}, {'c': 30, 'e': 50, 'd': 40}, {'c': 300, 'e': 500, 'd': 400}]
>>> List = [{'a':1, 'b':2, 'd':4, 'e':5, 'f':6},
... {'a':10, 'b':20, 'c':30, 'd':40, 'f':60},
... {'a':100, 'b':200, 'e':500, 'f':600}]
>>> keys = {'c', 'd', 'e'}
>>> [{k: d[k] for k in d.viewkeys() & keys} for d in List]
[{'e': 5, 'd': 4}, {'c': 30, 'd': 40}, {'e': 500}]
Try this code:
keys = ['c', 'd']
for dictionary in List1:
d = {}
for key in dictionary:
if key in keys:
d[key] = dictionary[key]
List2.append(d)
So I have this block of code
dictionary = {
'key1': {'a': 1, 'b': 2, 'c': 10},
'key2': {'d': 1, 'e': 1, 'c': 11},
'key3': {'d': 2, 'b': 1, 'g': 12}}
and
list1 = (a,b,c)
What I want to do is run a loop that finds the maximums of all the items in the list and returns the key. So for example, the maximum of 'c' would return 'key2', the maximum of 'b' would return 'key1', etc.
So far I have
for value in list1:
m = max(dictionary, key=lambda v: dictionary[v][value])
print(m + "\n")
But this only works if the same subkey exists in all keys in the dictionary. Any ideas on what to do?
Use float('-inf') when the key is missing:
m = max(dictionary, key=lambda v: dictionary[v].get(value, float('-inf')))
Negative infinity is guaranteed to be smaller than any existing value in the dictionaries, ensuring that nested dictionaries with the specific key missing are ignored.
Demo:
>>> dictionary = {
... 'key1': {'a': 1, 'b': 2, 'c': 10},
... 'key2': {'d': 1, 'e': 1, 'c': 11},
... 'key3': {'d': 2, 'b': 1, 'g': 12}}
>>> list1 = ('a', 'b', 'c')
>>> for value in list1:
... print(value, max(dictionary, key=lambda v: dictionary[v].get(value, float('-inf'))))
...
a key1
b key1
c key2
However, it'll be more efficient if you looped over all your dictionary values just once instead:
maximi = dict.fromkeys(list1, (None, float('-inf')))
for key, nested in dictionary.items():
for k in nested.keys() & maximi: # intersection of keys
if maximi[k][0] is None or dictionary[maximi[k][0]][k] < nested[k]:
maximi[k] = (key, nested[k])
for value in list1:
print(value, maximi[value][0])
That's presuming you are using Python 3; in Python 2, replace .items() with .iteritems() and .keys() with .viewkeys().
Demo:
>>> maximi = dict.fromkeys(list1, (None, float('-inf')))
>>> for key, nested in dictionary.items():
... for k in nested.keys() & maximi: # intersection of keys
... if maximi[k][0] is None or dictionary[maximi[k][0]][k] < nested[k]:
... maximi[k] = (key, nested[k])
...
>>> maximi
{'a': ('key1', 1), 'b': ('key1', 2), 'c': ('key2', 11)}
>>> for value in list1:
... print(value, maximi[value][0])
...
a key1
b key1
c key2
What is the best way to split a dictionary in half?
d = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
I'm looking to do this:
d1 = {'key1': 1, 'key2': 2, 'key3': 3}
d2 = {'key4': 4, 'key5': 5}
It does not matter which keys/values go into each dictionary. I am simply looking for the simplest way to divide a dictionary into two.
This would work, although I didn't test edge-cases:
>>> d = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
>>> d1 = dict(d.items()[len(d)/2:])
>>> d2 = dict(d.items()[:len(d)/2])
>>> print d1
{'key1': 1, 'key5': 5, 'key4': 4}
>>> print d2
{'key3': 3, 'key2': 2}
In python3:
d = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
d1 = dict(list(d.items())[len(d)//2:])
d2 = dict(list(d.items())[:len(d)//2])
Also note that order of items is not guaranteed
Here's a way to do it using an iterator over the items in the dictionary and itertools.islice:
import itertools
def splitDict(d):
n = len(d) // 2 # length of smaller half
i = iter(d.items()) # alternatively, i = d.iteritems() works in Python 2
d1 = dict(itertools.islice(i, n)) # grab first n items
d2 = dict(i) # grab the rest
return d1, d2
d1 = {key: value for i, (key, value) in enumerate(d.viewitems()) if i % 2 == 0}
d2 = {key: value for i, (key, value) in enumerate(d.viewitems()) if i % 2 == 1}
If you use python +3.3, and want your splitted dictionaries to be the same across different python invocations, do not use .items, since the hash-values of the keys, which determines the order of .items() will change between python invocations.
See Hash randomization
The answer by jone did not work for me. I had to cast to a list before I could index the result of the .items() call. (I am running Python 3.6 in the example)
d = {'one':1, 'two':2, 'three':3, 'four':4, 'five':5}
split_idx = 3
d1 = dict(list(d.items())[:split_idx])
d2 = dict(list(d.items())[split_idx:])
"""
output:
d1
{'one': 1, 'three': 3, 'two': 2}
d2
{'five': 5, 'four': 4}
"""
Note the dicts are not necessarily stored in the order of creation so the indexes may be mixed up.
Here is the function which can be used to split a dictionary to any divisions.
import math
def linch_dict_divider(raw_dict, num):
list_result = []
len_raw_dict = len(raw_dict)
if len_raw_dict > num:
base_num = len_raw_dict / num
addr_num = len_raw_dict % num
for i in range(num):
this_dict = dict()
keys = list()
if addr_num > 0:
keys = raw_dict.keys()[:base_num + 1]
addr_num -= 1
else:
keys = raw_dict.keys()[:base_num]
for key in keys:
this_dict[key] = raw_dict[key]
del raw_dict[key]
list_result.append(this_dict)
else:
for d in raw_dict:
this_dict = dict()
this_dict[d] = raw_dict[d]
list_result.append(this_dict)
return list_result
myDict = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
print myDict
myList = linch_dict_divider(myDict, 2)
print myList
We can do this efficiently with itertools.zip_longest() (note this is itertools.izip_longest() in 2.x):
from itertools import zip_longest
d = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
items1, items2 = zip(*zip_longest(*[iter(d.items())]*2))
d1 = dict(item for item in items1 if item is not None)
d2 = dict(item for item in items2 if item is not None)
Which gives us:
>>> d1
{'key3': 3, 'key1': 1, 'key4': 4}
>>> d2
{'key2': 2, 'key5': 5}
Here's a function that I use in Python 3.8 that can split a dict into a list containing the desired number of parts. If you specify more parts than elements, you'll get some empty dicts in the resulting list.
def split_dict(input_dict: dict, num_parts: int) -> list:
list_len: int = len(input_dict)
return [dict(list(input_dict.items())[i * list_len // num_parts:(i + 1) * list_len // num_parts])
for i in range(num_parts)]
Output:
>>> d = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
>>> split_dict(d, 2)
[{'a': 1, 'b': 2}, {'c': 3, 'd': 4, 'e': 5}]
>>> split_dict(d, 3)
[{'a': 1}, {'b': 2, 'c': 3}, {'d': 4, 'e': 5}]
>>> split_dict(d, 7)
[{}, {'a': 1}, {'b': 2}, {}, {'c': 3}, {'d': 4}, {'e': 5}]
If you used numpy, then you could do this :
def divide_dict(dictionary, chunk_size):
'''
Divide one dictionary into several dictionaries
Return a list, each item is a dictionary
'''
import numpy, collections
count_ar = numpy.linspace(0, len(dictionary), chunk_size+1, dtype= int)
group_lst = []
temp_dict = collections.defaultdict(lambda : None)
i = 1
for key, value in dictionary.items():
temp_dict[key] = value
if i in count_ar:
group_lst.append(temp_dict)
temp_dict = collections.defaultdict(lambda : None)
i += 1
return group_lst