How to find a word in a sentence that starts and ends with the same letter?












1















Is there a way to write "for" loop and "If Else" in python that searches through a sentence and find the number of words that start and end with the same letter?
I've tried writing something like:



sentence = "Mom knock the door"



list = sentence.split()

for word in sentence:

    if ...





python







share|improve this question





























    1















    Is there a way to write "for" loop and "If Else" in python that searches through a sentence and find the number of words that start and end with the same letter?
    I've tried writing something like:



    sentence = "Mom knock the door"
    

    
list = sentence.split()
    
for word in sentence:
    
    if ...





    python







    share|improve this question



























      1












      1








      1








      Is there a way to write "for" loop and "If Else" in python that searches through a sentence and find the number of words that start and end with the same letter?
      I've tried writing something like:



      sentence = "Mom knock the door"
      

      
list = sentence.split()
      
for word in sentence:
      
    if ...





      python







      share|improve this question
















      Is there a way to write "for" loop and "If Else" in python that searches through a sentence and find the number of words that start and end with the same letter?
      I've tried writing something like:



      sentence = "Mom knock the door"
      

      
list = sentence.split()
      
for word in sentence:
      
    if ...





      python




      python






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Dec 29 '18 at 9:23









      Ha Bom

      8872418




      8872418










      asked Dec 29 '18 at 8:44









      FarzanFarzan

      83




      83
























          6 Answers
          6






          active

          oldest

          votes


















          1














          simply compare the character at the beginning of the string with the character at the end of the string like so:



          if word[0] == word[-1]:


          If it shouldn't be case sensitive, lower the word first by calling:



          word = word.lower()





          share|improve this answer
























          • It's cleaner to .lower() the original sentence if case-sensitivity isn't a requirement.

            – Jason Baumgartner
            Dec 29 '18 at 8:53






          • 1





            @Jason That is true. However, I just wanted to be sure that the for-loop works for him, as he already split the string at the beginning.

            – DWuest
            Dec 29 '18 at 8:55



















          1














          words_list = sentence.split()
          
new_words_list = 
          
for word in words_list:
          
    if word[0] == word[-1]:
          
        new_words_list.append(word)
          

          
print('Number of words that starts and ends with same letter - {}'.format(len(new_words_list)))


          Also you can do it with list comprehension:



          new_words_list = [word for word in words_list if word[0] == word[-1]]


          If you want not to have it case sensitive use word[0].lower() and word[-1].lower() instead of word[0] and word[-1]






          share|improve this answer


























          • This solution is case-sensitive. The author didn't specify the same case in the question -- only that it is the same letter. I would use .lower() on the original sentence if case sensitivity isn't a requirement.

            – Jason Baumgartner
            Dec 29 '18 at 8:49











          • The author didn't add if it's required to be case sensitive or not, so I added that option in my answer.

            – Sergey Pugach
            Dec 29 '18 at 8:54



















          1














          The answers above are all smart, I prefer to deal with it in functional programming way, like this:



          sentence = "Mom knock the door"
          

          

          
def is_same_letter_at_begin_end(word):
          
    return word and word[0].lower() == word[-1].lower()
          

          

          
target_words = list(filter(is_same_letter_at_begin_end, sentence.split()))
          
print(target_words)
          
print(len(target_words))





          share|improve this answer


























          • You can just do return word and word[0].lower() == word[-1].lower(), no need for the conditional. Nice use of filter.

            – snakecharmerb
            Dec 30 '18 at 8:56











          • @snakecharmerb thx, I change it.

            – Menglong Li
            Dec 30 '18 at 10:37



















          0














          list = sentence.split(" ")
          
count = 0
          
for word in list:
          
    lc_word = word.lower()
          
    if lc_word[0] == lc_word[-1]:
          
        count +=1





          share|improve this answer

































            0














            lst = sentence.split()
            
num_words = 0
            
for i in lst:
            
    low = i.lower()
            
    if low[0] == low[len(low)-1]:
            
        num_words += 1
            
return num_words





            share|improve this answer



















            • 1





              While this code may answer the question, it is better to explain how to solve the problem and provide the code as an example or reference. Code-only answers can be confusing and lack context.

              – Robert Columbia
              Dec 29 '18 at 11:12



















            0














            list or set comprehension case insensitive:



            sentence = "Mom knock the door, mom"
            
all_words_count = len([ word for word in sentence.split() if word[0].lower() == word[-1].lower() ])
            
uniq_words_count = len({word.lower() for word in sentence.split() if word[0].lower() == word[-1].lower()})
            

            
print(all_words_count) #=> 3
            
print(uniq_words_count) #=> 2





            share|improve this answer























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              6 Answers
              6






              active

              oldest

              votes








              6 Answers
              6






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1














              simply compare the character at the beginning of the string with the character at the end of the string like so:



              if word[0] == word[-1]:


              If it shouldn't be case sensitive, lower the word first by calling:



              word = word.lower()





              share|improve this answer
























              • It's cleaner to .lower() the original sentence if case-sensitivity isn't a requirement.

                – Jason Baumgartner
                Dec 29 '18 at 8:53






              • 1





                @Jason That is true. However, I just wanted to be sure that the for-loop works for him, as he already split the string at the beginning.

                – DWuest
                Dec 29 '18 at 8:55
















              1














              simply compare the character at the beginning of the string with the character at the end of the string like so:



              if word[0] == word[-1]:


              If it shouldn't be case sensitive, lower the word first by calling:



              word = word.lower()





              share|improve this answer
























              • It's cleaner to .lower() the original sentence if case-sensitivity isn't a requirement.

                – Jason Baumgartner
                Dec 29 '18 at 8:53






              • 1





                @Jason That is true. However, I just wanted to be sure that the for-loop works for him, as he already split the string at the beginning.

                – DWuest
                Dec 29 '18 at 8:55














              1












              1








              1







              simply compare the character at the beginning of the string with the character at the end of the string like so:



              if word[0] == word[-1]:


              If it shouldn't be case sensitive, lower the word first by calling:



              word = word.lower()





              share|improve this answer













              simply compare the character at the beginning of the string with the character at the end of the string like so:



              if word[0] == word[-1]:


              If it shouldn't be case sensitive, lower the word first by calling:



              word = word.lower()






              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Dec 29 '18 at 8:48









              DWuestDWuest

              45712




              45712













              • It's cleaner to .lower() the original sentence if case-sensitivity isn't a requirement.

                – Jason Baumgartner
                Dec 29 '18 at 8:53






              • 1





                @Jason That is true. However, I just wanted to be sure that the for-loop works for him, as he already split the string at the beginning.

                – DWuest
                Dec 29 '18 at 8:55



















              • It's cleaner to .lower() the original sentence if case-sensitivity isn't a requirement.

                – Jason Baumgartner
                Dec 29 '18 at 8:53






              • 1





                @Jason That is true. However, I just wanted to be sure that the for-loop works for him, as he already split the string at the beginning.

                – DWuest
                Dec 29 '18 at 8:55

















              It's cleaner to .lower() the original sentence if case-sensitivity isn't a requirement.

              – Jason Baumgartner
              Dec 29 '18 at 8:53





              It's cleaner to .lower() the original sentence if case-sensitivity isn't a requirement.

              – Jason Baumgartner
              Dec 29 '18 at 8:53




              1




              1





              @Jason That is true. However, I just wanted to be sure that the for-loop works for him, as he already split the string at the beginning.

              – DWuest
              Dec 29 '18 at 8:55





              @Jason That is true. However, I just wanted to be sure that the for-loop works for him, as he already split the string at the beginning.

              – DWuest
              Dec 29 '18 at 8:55













              1














              words_list = sentence.split()
              
new_words_list = 
              
for word in words_list:
              
    if word[0] == word[-1]:
              
        new_words_list.append(word)
              

              
print('Number of words that starts and ends with same letter - {}'.format(len(new_words_list)))


              Also you can do it with list comprehension:



              new_words_list = [word for word in words_list if word[0] == word[-1]]


              If you want not to have it case sensitive use word[0].lower() and word[-1].lower() instead of word[0] and word[-1]






              share|improve this answer


























              • This solution is case-sensitive. The author didn't specify the same case in the question -- only that it is the same letter. I would use .lower() on the original sentence if case sensitivity isn't a requirement.

                – Jason Baumgartner
                Dec 29 '18 at 8:49











              • The author didn't add if it's required to be case sensitive or not, so I added that option in my answer.

                – Sergey Pugach
                Dec 29 '18 at 8:54
















              1














              words_list = sentence.split()
              
new_words_list = 
              
for word in words_list:
              
    if word[0] == word[-1]:
              
        new_words_list.append(word)
              

              
print('Number of words that starts and ends with same letter - {}'.format(len(new_words_list)))


              Also you can do it with list comprehension:



              new_words_list = [word for word in words_list if word[0] == word[-1]]


              If you want not to have it case sensitive use word[0].lower() and word[-1].lower() instead of word[0] and word[-1]






              share|improve this answer


























              • This solution is case-sensitive. The author didn't specify the same case in the question -- only that it is the same letter. I would use .lower() on the original sentence if case sensitivity isn't a requirement.

                – Jason Baumgartner
                Dec 29 '18 at 8:49











              • The author didn't add if it's required to be case sensitive or not, so I added that option in my answer.

                – Sergey Pugach
                Dec 29 '18 at 8:54














              1












              1








              1







              words_list = sentence.split()
              
new_words_list = 
              
for word in words_list:
              
    if word[0] == word[-1]:
              
        new_words_list.append(word)
              

              
print('Number of words that starts and ends with same letter - {}'.format(len(new_words_list)))


              Also you can do it with list comprehension:



              new_words_list = [word for word in words_list if word[0] == word[-1]]


              If you want not to have it case sensitive use word[0].lower() and word[-1].lower() instead of word[0] and word[-1]






              share|improve this answer















              words_list = sentence.split()
              
new_words_list = 
              
for word in words_list:
              
    if word[0] == word[-1]:
              
        new_words_list.append(word)
              

              
print('Number of words that starts and ends with same letter - {}'.format(len(new_words_list)))


              Also you can do it with list comprehension:



              new_words_list = [word for word in words_list if word[0] == word[-1]]


              If you want not to have it case sensitive use word[0].lower() and word[-1].lower() instead of word[0] and word[-1]







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Dec 29 '18 at 8:52

























              answered Dec 29 '18 at 8:48









              Sergey PugachSergey Pugach

              1,8021412




              1,8021412













              • This solution is case-sensitive. The author didn't specify the same case in the question -- only that it is the same letter. I would use .lower() on the original sentence if case sensitivity isn't a requirement.

                – Jason Baumgartner
                Dec 29 '18 at 8:49











              • The author didn't add if it's required to be case sensitive or not, so I added that option in my answer.

                – Sergey Pugach
                Dec 29 '18 at 8:54



















              • This solution is case-sensitive. The author didn't specify the same case in the question -- only that it is the same letter. I would use .lower() on the original sentence if case sensitivity isn't a requirement.

                – Jason Baumgartner
                Dec 29 '18 at 8:49











              • The author didn't add if it's required to be case sensitive or not, so I added that option in my answer.

                – Sergey Pugach
                Dec 29 '18 at 8:54

















              This solution is case-sensitive. The author didn't specify the same case in the question -- only that it is the same letter. I would use .lower() on the original sentence if case sensitivity isn't a requirement.

              – Jason Baumgartner
              Dec 29 '18 at 8:49





              This solution is case-sensitive. The author didn't specify the same case in the question -- only that it is the same letter. I would use .lower() on the original sentence if case sensitivity isn't a requirement.

              – Jason Baumgartner
              Dec 29 '18 at 8:49













              The author didn't add if it's required to be case sensitive or not, so I added that option in my answer.

              – Sergey Pugach
              Dec 29 '18 at 8:54





              The author didn't add if it's required to be case sensitive or not, so I added that option in my answer.

              – Sergey Pugach
              Dec 29 '18 at 8:54











              1














              The answers above are all smart, I prefer to deal with it in functional programming way, like this:



              sentence = "Mom knock the door"
              

              

              
def is_same_letter_at_begin_end(word):
              
    return word and word[0].lower() == word[-1].lower()
              

              

              
target_words = list(filter(is_same_letter_at_begin_end, sentence.split()))
              
print(target_words)
              
print(len(target_words))





              share|improve this answer


























              • You can just do return word and word[0].lower() == word[-1].lower(), no need for the conditional. Nice use of filter.

                – snakecharmerb
                Dec 30 '18 at 8:56











              • @snakecharmerb thx, I change it.

                – Menglong Li
                Dec 30 '18 at 10:37
















              1














              The answers above are all smart, I prefer to deal with it in functional programming way, like this:



              sentence = "Mom knock the door"
              

              

              
def is_same_letter_at_begin_end(word):
              
    return word and word[0].lower() == word[-1].lower()
              

              

              
target_words = list(filter(is_same_letter_at_begin_end, sentence.split()))
              
print(target_words)
              
print(len(target_words))





              share|improve this answer


























              • You can just do return word and word[0].lower() == word[-1].lower(), no need for the conditional. Nice use of filter.

                – snakecharmerb
                Dec 30 '18 at 8:56











              • @snakecharmerb thx, I change it.

                – Menglong Li
                Dec 30 '18 at 10:37














              1












              1








              1







              The answers above are all smart, I prefer to deal with it in functional programming way, like this:



              sentence = "Mom knock the door"
              

              

              
def is_same_letter_at_begin_end(word):
              
    return word and word[0].lower() == word[-1].lower()
              

              

              
target_words = list(filter(is_same_letter_at_begin_end, sentence.split()))
              
print(target_words)
              
print(len(target_words))





              share|improve this answer















              The answers above are all smart, I prefer to deal with it in functional programming way, like this:



              sentence = "Mom knock the door"
              

              

              
def is_same_letter_at_begin_end(word):
              
    return word and word[0].lower() == word[-1].lower()
              

              

              
target_words = list(filter(is_same_letter_at_begin_end, sentence.split()))
              
print(target_words)
              
print(len(target_words))






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Dec 30 '18 at 10:37

























              answered Dec 29 '18 at 9:06









              Menglong LiMenglong Li

              1,250414




              1,250414













              • You can just do return word and word[0].lower() == word[-1].lower(), no need for the conditional. Nice use of filter.

                – snakecharmerb
                Dec 30 '18 at 8:56











              • @snakecharmerb thx, I change it.

                – Menglong Li
                Dec 30 '18 at 10:37



















              • You can just do return word and word[0].lower() == word[-1].lower(), no need for the conditional. Nice use of filter.

                – snakecharmerb
                Dec 30 '18 at 8:56











              • @snakecharmerb thx, I change it.

                – Menglong Li
                Dec 30 '18 at 10:37

















              You can just do return word and word[0].lower() == word[-1].lower(), no need for the conditional. Nice use of filter.

              – snakecharmerb
              Dec 30 '18 at 8:56





              You can just do return word and word[0].lower() == word[-1].lower(), no need for the conditional. Nice use of filter.

              – snakecharmerb
              Dec 30 '18 at 8:56













              @snakecharmerb thx, I change it.

              – Menglong Li
              Dec 30 '18 at 10:37





              @snakecharmerb thx, I change it.

              – Menglong Li
              Dec 30 '18 at 10:37











              0














              list = sentence.split(" ")
              
count = 0
              
for word in list:
              
    lc_word = word.lower()
              
    if lc_word[0] == lc_word[-1]:
              
        count +=1





              share|improve this answer






























                0














                list = sentence.split(" ")
                
count = 0
                
for word in list:
                
    lc_word = word.lower()
                
    if lc_word[0] == lc_word[-1]:
                
        count +=1





                share|improve this answer




























                  0












                  0








                  0







                  list = sentence.split(" ")
                  
count = 0
                  
for word in list:
                  
    lc_word = word.lower()
                  
    if lc_word[0] == lc_word[-1]:
                  
        count +=1





                  share|improve this answer















                  list = sentence.split(" ")
                  
count = 0
                  
for word in list:
                  
    lc_word = word.lower()
                  
    if lc_word[0] == lc_word[-1]:
                  
        count +=1






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Dec 29 '18 at 8:57

























                  answered Dec 29 '18 at 8:48









                  Will WardWill Ward

                  5714




                  5714























                      0














                      lst = sentence.split()
                      
num_words = 0
                      
for i in lst:
                      
    low = i.lower()
                      
    if low[0] == low[len(low)-1]:
                      
        num_words += 1
                      
return num_words





                      share|improve this answer



















                      • 1





                        While this code may answer the question, it is better to explain how to solve the problem and provide the code as an example or reference. Code-only answers can be confusing and lack context.

                        – Robert Columbia
                        Dec 29 '18 at 11:12
















                      0














                      lst = sentence.split()
                      
num_words = 0
                      
for i in lst:
                      
    low = i.lower()
                      
    if low[0] == low[len(low)-1]:
                      
        num_words += 1
                      
return num_words





                      share|improve this answer



















                      • 1





                        While this code may answer the question, it is better to explain how to solve the problem and provide the code as an example or reference. Code-only answers can be confusing and lack context.

                        – Robert Columbia
                        Dec 29 '18 at 11:12














                      0












                      0








                      0







                      lst = sentence.split()
                      
num_words = 0
                      
for i in lst:
                      
    low = i.lower()
                      
    if low[0] == low[len(low)-1]:
                      
        num_words += 1
                      
return num_words





                      share|improve this answer













                      lst = sentence.split()
                      
num_words = 0
                      
for i in lst:
                      
    low = i.lower()
                      
    if low[0] == low[len(low)-1]:
                      
        num_words += 1
                      
return num_words






                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Dec 29 '18 at 9:00









                      Aditya GoyalAditya Goyal

                      11




                      11








                      • 1





                        While this code may answer the question, it is better to explain how to solve the problem and provide the code as an example or reference. Code-only answers can be confusing and lack context.

                        – Robert Columbia
                        Dec 29 '18 at 11:12














                      • 1





                        While this code may answer the question, it is better to explain how to solve the problem and provide the code as an example or reference. Code-only answers can be confusing and lack context.

                        – Robert Columbia
                        Dec 29 '18 at 11:12








                      1




                      1





                      While this code may answer the question, it is better to explain how to solve the problem and provide the code as an example or reference. Code-only answers can be confusing and lack context.

                      – Robert Columbia
                      Dec 29 '18 at 11:12





                      While this code may answer the question, it is better to explain how to solve the problem and provide the code as an example or reference. Code-only answers can be confusing and lack context.

                      – Robert Columbia
                      Dec 29 '18 at 11:12











                      0














                      list or set comprehension case insensitive:



                      sentence = "Mom knock the door, mom"
                      
all_words_count = len([ word for word in sentence.split() if word[0].lower() == word[-1].lower() ])
                      
uniq_words_count = len({word.lower() for word in sentence.split() if word[0].lower() == word[-1].lower()})
                      

                      
print(all_words_count) #=> 3
                      
print(uniq_words_count) #=> 2





                      share|improve this answer




























                        0














                        list or set comprehension case insensitive:



                        sentence = "Mom knock the door, mom"
                        
all_words_count = len([ word for word in sentence.split() if word[0].lower() == word[-1].lower() ])
                        
uniq_words_count = len({word.lower() for word in sentence.split() if word[0].lower() == word[-1].lower()})
                        

                        
print(all_words_count) #=> 3
                        
print(uniq_words_count) #=> 2





                        share|improve this answer


























                          0












                          0








                          0







                          list or set comprehension case insensitive:



                          sentence = "Mom knock the door, mom"
                          
all_words_count = len([ word for word in sentence.split() if word[0].lower() == word[-1].lower() ])
                          
uniq_words_count = len({word.lower() for word in sentence.split() if word[0].lower() == word[-1].lower()})
                          

                          
print(all_words_count) #=> 3
                          
print(uniq_words_count) #=> 2





                          share|improve this answer













                          list or set comprehension case insensitive:



                          sentence = "Mom knock the door, mom"
                          
all_words_count = len([ word for word in sentence.split() if word[0].lower() == word[-1].lower() ])
                          
uniq_words_count = len({word.lower() for word in sentence.split() if word[0].lower() == word[-1].lower()})
                          

                          
print(all_words_count) #=> 3
                          
print(uniq_words_count) #=> 2






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Dec 29 '18 at 9:22









                          iGianiGian

                          3,7502622




                          3,7502622






























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