如何计算项目在另一个列表的列表中出现的次数

Question

suppose I have two lists as follows:

list1 = ["a","b","a","a","b","a","b","a","b","b","b"]
list2 = ["pos","neg","pos","neu","neg","pos","pos","pos","neg","neu","pos"]

I want to count the number of times "pos" , "neg" and "neu" has occurred for each item in list1 .

So the number of times "pos","neg" and "neu" occurs with "a" , and for "b" . For example, the first element in list1 , "a" has a "pos" value because list2[0] is for "pos" .

What is the best approach for this? I feel there is a much better solution out there in comparison to what I have done at the moment. I can see that if more unique items exist in list1 my approach will not be feasible.

list1 = ["a","b","a","a","b","a","b","a","b","b","b"]
list2 = ["pos","neg","pos","neu","neg","pos","pos","pos","neg","neu","pos"]

a_pos = 0
a_neg = 0
a_neu = 0
b_pos = 0
b_neg = 0
b_neu = 0

for i in range(len(list1)):
    if list1[i] == "a":
        if list2[i] == "pos":
            a_pos +=1
        elif list2[i] == "neg":
            a_neg +=1
        else:
            a_neu +=1
    if list1[i] == "b":
        if list2[i] == "pos":
            b_pos +=1
        elif list2[i] == "neg":
            b_neg +=1
        else:
            b_neu +=1       

print(a_pos,a_neg,a_neu)
print(b_pos,b_neg,b_neu)

Answer 1

You could use Counter with zip :

from collections import Counter
Counter(zip(list1, list2))

Counter({('a', 'pos'): 4,
         ('b', 'neg'): 3,
         ('a', 'neu'): 1,
         ('b', 'pos'): 2,
         ('b', 'neu'): 1})

---

Where zip is creating an iterable with the elements from both lists interleaved:

[('a', 'pos'), ('b', 'neg'), ('a', 'pos'),...

So the above works because zip is returning tuples, which are hashable , a necessary condition for Counter to work as its elements are stored as a dictionary

Answer 2

You can zip the two lists together, and then use collections.Counter to count your co-occurences

from collections import Counter
list1 = ["a","b","a","a","b","a","b","a","b","b","b"]
list2 = ["pos","neg","pos","neu","neg","pos","pos","pos","neg","neu","pos"]

print(Counter(zip(list1, list2)))

The output will be

{('a', 'pos'): 4, ('b', 'neg'): 3, ('a', 'neu'): 1, ('b', 'pos'): 2, ('b', 'neu'): 1}

To break it down, zip takes both of your lists, and create an iterator with elements each from each of the lists interleaved

In [1]: from collections import Counter 
   ...: list1 = ["a","b","a","a","b","a","b","a","b","b","b"] 
   ...: list2 = ["pos","neg","pos","neu","neg","pos","pos","pos","neg","neu","pos"]                                                                                                                     

In [2]: list(zip(list1,list2))                                                                                                                                                                          
Out[2]: 
[('a', 'pos'),
 ('b', 'neg'),
 ('a', 'pos'),
 ('a', 'neu'),
 ('b', 'neg'),
 ('a', 'pos'),
 ('b', 'pos'),
 ('a', 'pos'),
 ('b', 'neg'),
 ('b', 'neu'),
 ('b', 'pos')]

We then take this output, and put in into a Counter , which calculate the frequency of each item in the iterator and provides us a dictionary, this is possible because the key of the dictionary is a tuple which is a hashable type.

In [3]: Counter(list(zip(list1,list2)))                                                                                                                                                                 
Out[3]: 
Counter({('a', 'pos'): 4,
         ('b', 'neg'): 3,
         ('a', 'neu'): 1,
         ('b', 'pos'): 2,
         ('b', 'neu'): 1})