Multi tool use

Clash Royale CLAN TAG#URR8PPP

Remove words in relation to size and not in other list

I am trying to delete words that have three letters but that do not belong to a list. I tried to do this:

text = 'je ne suis pas une table'
not_shortword = 'n', 'pas', 'ne'
remove_shortword = ''.join(shortword for shortword in text if len(shortword) > 4 and shortword not in not_shortword)


My output:

'je e suis pas ue table'

Good output:

ne suis pas table

"je"

length == 3

not in not_shortword

3 Answers
3

In addition to missing split() on text, you've got your conditions a little mixed up.

Try:

split()

text

' '.join(
shortword for shortword in text.split(" ")
if len(shortword) >= 3 or shortword in not_shortword
)
# 'ne suis pas table'


The split() in your case breaks in an array of element by the space (one element of array = one word). Looping through it it the solution

text = 'je ne suis pas une table'
not_shortword = 'n', 'pas', 'ne'

rep = ' '.join(shortword for shortword in text.split() if len(shortword ) >= 4 or shortword in not_shortword)

print(rep)
#returns : "ne suis pas table"


Make sure you are splitting the text into a list of words with .split():

.split()

remove_shortword = ' '.join(word for word in text.split() if len(word) >= 3 and word not in not_shortword)


Then join with a space instead of an empty string.

'je une'

je une

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