Counting: How do I add a zero if a word does not occur in a list?

Counting: How do I add a zero if a word does not occur in a list?

Problem Description:

I would like to find keywords from a list, but return a zero if the word does not exist (in this case: part). In this example, collabor occurs 4 times and part 0 times.

My current output is

[[‘collabor’, 4]]

But what I would like to have is

[[‘collabor’, 4], [‘part’, 0]]

str1 = ["collabor", "part"]
x10 = []
for y in wordlist:
    for string in str1:
        if y.find(string) != -1:
            x10.append(y)

from collections import Counter
x11 = Counter(x10) 
your_list = [list(i) for i in x11.items()]
rowssorted = sorted(your_list, key=lambda x: x[0])
print(rowssorted)

Solution – 1

Although you have not clearly written your problem and requirements,I think I understood the task.

I assume that you have a set of words that may or may not occur in a given list and you want to print the count of those words based on the occurrence in the given list.

Code:

constants=["part","collabor"]
wordlist = ["collabor", "collabor"]
d={}
for const in constants:
    d[const]=0
for word in wordlist:
    if word in d:
        d[word]+=1
    else:
        d[word]=0

from collections import Counter
x11 = Counter(d) 
your_list = [list(i) for i in x11.items()]
rowssorted = sorted(your_list, key=lambda x: x[0])
print(rowssorted)

output:

[['collabor', 2], ['part', 0]]

This approach gives the required output.

In python, to get the count of occurrence dictionary is popular.

Hope it helps!