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+---- Thread: [Python] python (easy task help) (/showthread.php?tid=5180343)
python (easy task help) - Gameh3x - Feb 22, 2016
I also ran into trouble with another task. I need to write a program that asks the user for 2 words, if they are the same it prints "same", if not it prints the word that has more letter "e" in it.
1)So I start it off with this right:
word1 = raw_input("Enter the first word")
word2 = raw_input("Enter the second word")
2)Now if they are the same it should be
if word1 == word2
print ("Same")
3)How do i make it check which word has more letters "e" in it?
* - JBlock - Feb 22, 2016
by same you mean the same amount of letters or the exact same word?
* - Gameh3x - Feb 22, 2016
(Feb 22, 2016 01:53 PM)JBlock Wrote: by same you mean the same amount of letters or the exact same word?
Exact same word. But lower/caps doesn't matter. So DOG is still equal to dog :)
* - JBlock - Feb 22, 2016
was coded in python 3.5, so its input instead of raw_input
🤝
Code
word1 = input("first word")
word2= input("second word")
def checkyourselfbeforeyouwreckyourself(word1,word2):
count1=0
count2=0
if(word1 == word2):
return("same")
else:
for x in word1:
if(x =="e"):
count1+=1
for x in word2:
if(x =="e"):
count2+=1
if(count1>count2):
return(word1)
elif(count2>count1):
return(word2)
else:
return("both words had the same amount of the letter \"e\"")
var=checkyourselfbeforeyouwreckyourself(word1,word2)
print(var)* - Gameh3x - Feb 22, 2016
(Feb 22, 2016 02:08 PM)JBlock Wrote: was coded in python 3.5, so its input instead of raw_input
🤝
Codeword1 = input("first word")
word2= input("second word")
def checkyourselfbeforeyouwreckyourself(word1,word2):
count1=0
count2=0
if(word1 == word2):
return("same")
else:
for x in word1:
if(x =="e"):
count1+=1
for x in word2:
if(x =="e"):
count2+=1
if(count1>count2):
return(word1)
elif(count2>count1):
return(word2)
else:
return("both words had the same amount of the letter \"e\"")
var=checkyourselfbeforeyouwreckyourself(word1,word2)
print(var)
Thanks!
for x in word1:
if(x =="e"):
count1+=1
for x in word2:
if(x =="e"):
count2+=1
So whenever there's an "e" in the word it adds +1 to the count1 variable right? :)
It's actually pretty simple now when I see the solution :)
Also what's the difference between python 3.5 and 2.7? I'm pretty sure we use 2.7 in our class.
* - khalouni - Feb 22, 2016
whenever there is an "e" in the word1 it adds +1 to count1
whenever there is an "e" in the word2 it adds +1 to count2