How To Find A Word - First Letter Will Be Capital & Other Will Be Lower

Problem Statement: Filter those words from the complete set of text6, having first letter in upper case and all other letters in lower case. Store the result in variable title_word

Solution 1:

One of the above suggestions did work for me. Sample code below.

title_words = [word for word in text6 if (len(word)==1and word[0].isupper()) or (word[0].isupper() and word[1:].islower()) ]
print(len(title_words))

Solution 2:

I think the problem is with set(text6). I suggest you iterate over text6.tokens.

Update, explanation

The code you've provided is correct.

The issues is that the text can contain same words multiple times. Doing a set(words) will reduce the total available words, so you start with an incomplete data set.

The other responses are not necessary wrong in checking the validity of a word, but they are iterating over the same wrong data set.

Solution 3:

In the question, "Store the result in variable title_words. print the number of words present in title_words."

The result of filtering a list of elements is a list of the same type of elements. In your case, filtering the list text6 (assuming it's a list of strings) would result in a (smaller) list of strings. Your title_words variable should be this filtered list, not the number of strings; the number of strings would just be the length of the list.

It's also ambiguous from the question if capitalized words should be filtered out (ie. removed from the smaller list) or filtered (ie. kept in the list), so try out both to see if you're interpreting it incorrectly.

Solution 4:

Give regular expressions a try:

>>>import re>>>from nltk.book import text6>>>>>>text = ' '.join(set(text6))>>>title_words = re.findall(r'([A-Z]{1}[a-z]+)', text)>>>len(title_words)
461

Solution 5:

There are 50 singleton elements (elements of length one) in text6, however, your code would not pass any as a success, like, 'I' or 'W' etc. Is that correct, or do you require words of minimum length 2?