python字典中每个键的多个变量计数频率

Question

I managed to merge two lists (one representing the key, the other one the values) in a dictionary, in order to further count the frequency of the values. That gives me aa dictionary with multiples values in a list for each key :

I now want to count, for each key, the total number of DIFFERENT VALUES, for eg on my printscreen, for the key 1068, there are 2 different keys (70803, 70805). I tried this:

However, it only gives me the TOTAL times the values appears,which is not what I want. As I am a very beginer in that, I have no idea how to proceed (especially with that dictionary with lists inside)

Thanks to all!

Loïc

Answer 1

If I understand your question correctly, the following should do the trick:

for k, v in d.items():
    print(k, len(set(v)))

For each list of values you create a set of the elements and get its length.

Answer 2

here is my code: first, the dictionnary I obtain from the two lists:

from collections import defaultdict
keys = ['31', '53', '57', '57', '57', '82', '82', '82', '82', '82', '82',         '82', '82', '82','82']
values = ['70803', '70901', '70801', '70801', '70801', '70801', '70804',  '70812', '70812', '70812', '70812', '70813', '70813', '70813', '70813']

d = {}
for k,v in zip(keys, values):
 d.setdefault(k, []).append(v)
d

which return me:

 {'31': ['70803'],
'53': ['70901'],
'57': ['70801', '70801', '70801'],
'82': ['70801',
'70804',
'70812',
'70812',
'70812',
'70812',
'70813',
'70813',
'70813',
'70813']}

I then want to count for each key, the number of different values it has. When I try to proceed:

from collections import Counter
from collections import defaultdict
d = {'31': ['70803'],
'53': ['70901'],
'57': ['70801', '70801', '70801'],
'82': ['70801',
'70804',
'70812',
'70812',
'70812',
'70812',
'70813',
'70813',
'70813',
'70813']}
for k, v in d.items():
   print(k, len(set(v))

It returns me:

File "<ipython-input-5-f243768fffa2>", line 248
print(k, len(set(v))
                    ^

SyntaxError: unexpected EOF while parsing

So here I am :) I don't know how to proceed further