CODEWITHSUNDEEP
python
newbie learner
newbie learner

Reputation: 11

A Small programming puzzle

z=[1,2,","]

count=0


for i in z:
    if (",") in z:
        count+=1
        print count

this python code counts all commas in list z.Why? Answer is 3.when it should be one. One can test by changing elements.

If list element comma "," is deleted, program produces blank, not 2 commas as count.

Questions:

  1. can someone answer please why comma gets counted only?
  2. why comma count is 3 when it should be one.

thanks

Upvotes: 1

Views: 102

Answers (4)

Amen
Amen

Reputation: 1633

When you use if (",") in z python checks if the comma is in your list and it will return true each time and when you delete the "," from the list it never produce true and the count would be zero so the correct version would be 

z=[1,2,","]  

count=0

for i in z:
    if i == ',':
        count+=1
print count

and also your print statement should not be in your for body

Upvotes: 0

J Kelly
J Kelly

Reputation: 510

Three times you checked whether ',' is in z. Yes it is. Yes it still is. Yes again. That's 3. 

for i in z:
    if i == ',':
        count+=1
print count

Upvotes: 0

kylieCatt
kylieCatt

Reputation: 11039

The reason you are seeing 3 as your answer is because of this:

if (",") in z:

You are iterating across the list and for each element in the list you are checking if ',' is in the list z, which it is. You count variable is incremented accordingly.

Upvotes: 0

Joran Beasley
Joran Beasley

Reputation: 113988

change it to 

if (",") in i:

really you should use better names (i typically refers to an index into an array or a integer count.. whereas you are iterating over the actual items in the list)

for item in z:
    if (",") in item:
        count+=1
        print count

Upvotes: 2

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