Ggtcf

Trying to figure out how to delete the period from my output. The correct answer to the assignment is: WA. EA. AI. AR. VI
My output is: WA. EA. AI. AR. VI.

I've tried using rstrip to remove the . from the output but that does not work.

Appreciate some insight to this rookie python programmer.

stopwords = ['to', 'a', 'for', 'by', 'an', 'am', 'the', 
'so', 'it', 'and', 'The']
sent = "The water earth and air are vital"
s_sent=sent.split()
acro = ''
for word in s_sent:
 if word not in stopwords:
 acro=acro + str.upper(word[:2])+". " 
 acro=acro.rstrip('.')
print(acro)

You join to save yourself from troubles of handling the left over spaces and
'.'

stopwords = ['to', 'a', 'for', 'by', 'an', 'am', 'the', 
'so', 'it', 'and', 'The']
sent = "The water earth and air are vital"
s_sent=sent.split()
acro = []
for word in s_sent:
 if word not in stopwords:
 acro.append( str.upper(word[:2]))
print('. '.join(acro))

Output

WA. EA. AI. AR. VI

Another simple way is to introduce a flag to check the first value and then smartly append '. ' at the end of last value. In this way you will not have extra trailing characters

smart_flag = True
acro = ''
for word in s_sent:
 if word not in stopwords:
 if smart_flag:
 acro=acro+str.upper(word[:2])
 smart_flag=False
 else:
 acro=acro +". " + str.upper(word[:2]) 

print(acro)

Just for completeness acro.rstrip(' .')) will strip any characters you pass in.By that I mean it will remove '. ' as well as ' .'

This is because your last character in acro is a space character. You have to either rstrip(". ") or rstrip(" .").

This can also be shortened to this:

stopwords = ['to', 'a', 'for', 'by', 'an', 'am', 'the', 'so', 'it', 'and', 'The']
sent = "The water earth and air are vital"

acro = '. '.join(word[:2].upper() for word in sent.split() if word not in stopwords)
print(acro)

Output:

WA. EA. AI. AR. VI

Update:

But, as @mad_ points out, join in this case would like to create a single string, first of exactly the right length, then with the right contents. But to know the right length it would like to find the lengths of all the components it is going to join, so having a list as its parameter would mean it has a size, whereas my original answer with a generator does not have a size until it has been exhausted.

So, better is:

acro = '. '.join([word[:2].upper() for word in sent.split() if word not in stopwords])