# 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?

Contents

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!

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