计算列表字典中的频率

Question

I have a dictionary with a bunch of lists, ie:

{"0": [0, 0, 0, 0, 1, 1], "1": [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1], etc...}

How can I count the frequencies of the numbers in each list? and output a dictionary in the format:

{"0": {"0": 4, "1": 2}, "1": {"0": 12, "1": 4}, etc..}

Answer 1

use collections.Counter :

from collections import Counter

spam = {"0": [0, 0, 0, 0, 1, 1], "1": [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1]}
eggs = {key:Counter(value) for key, value in spam.items()}
print(eggs)

output:

{'0': Counter({0: 4, 1: 2}), '1': Counter({0: 12, 1: 4})}

of course you can convert Counter to dict if you prefer

Answer 2

You can always make a counter yourself, that just counts for every element in the list how many appearances it has (notice that you can use an integer as a dictionary key and not just a string). Like this:

a = {"0": [0, 0, 0, 0, 1, 1], "1": [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1]}
different_a = dict()
for key in a.keys():
  counting_dict = dict()
  for e in a[key]:
    if str(e) in counting_dict:
      counting_dict[str(e)] += 1
    else:
      counting_dict[str(e)] = 1
  different_a[key] = counting_dict

But let's look at an nicer way than just counting the number of elements.

My favorite is to use collections.Counter:

from collections import Counter
new_a = {key: Counter(value) for key, value in a.items()}

It turns the list into a set (a sort of list where each element can only appear once and the order of elements has no meaning), with the added feature that it counts the number of appearances. Read more about this data structure here .

Answer 3

Try this (d is your dict):

d={"0": [0, 0, 0, 0, 1, 1], "1": [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1]}

res={i:{'0':k.count(0), '1':k.count(1)} for i,k in d.items()}

>>>print(res)

{'0': {'0': 4, '1': 2}, '1': {'0': 12, '1': 4}}

Answer 4

You can do this

d = {"0": [0, 0, 0, 0, 1, 1], "1": [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1],}

a = dict()

for k,v in d.items():
    count = dict()
    for x in set(v):
        count[str(x)] = v.count(x)
    a[k] = count
    
print(a)

Output:

{'0': {'0': 4, '1': 2}, '1': {'0': 12, '1': 4}}

Or You can do it using Dictionary Comprehension:

d = {"0": [0, 0, 0, 0, 1, 1], "1": [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1],}

a = {k:{str(x):v.count(x) for x in set(v)} for k,v in d.items()}

print(a)

Output:

{'0': {'0': 4, '1': 2}, '1': {'0': 12, '1': 4}}