so my program will require the user to input a phrase WITHOUT punctuation and the program will return its position in the list.
def Detection():
print(list.index(find))
find_found=list.index(find)
list[find_found]="hide"
list = input("Enter a phrase without punctuation:")
print(list)
list = list.split(" ")
list = [element.lower() for element in list]
find=input("what word must be found?")
find=find.lower()
if find in list:
for find in list:
Detection()
else:
print("its not in the list.")
l = ["enter", 'a', 'phrase']
l.index('a')
which returns index if word is not there in list throws ValueError Exception.
to get indexes for duplicates also
sorted_list = sorted(l)
first_ind = sorted_list.index("enter")
last_ind = len(sorted_list) - sorted_list[::-1].index("enter") - 1
indexes = range(first_ind, last_ind + 1)
i found a way here it is.
sentence =input("phrase to be shortened: ")
print (sentence)
text = input("Choose a word from the sentence above: ")
sentence = sentence.upper()
sentence = sentence.split(" ")
text = text.upper ()# this makes the text in capital letters
def lookfor (text):
indexes = [ idx+1 for word, idx in zip(sentence, range(0,len(sentence))) if text == word ]
print ("Your word has been found in the sentence at these positions", indexes )
if not indexes:
print ("The word that you have typed is not found in the sentence.")
lookfor(text)
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