how can i compare string within a list to an integer number?

Question

my task today is to create a function which takes a list of string and an integer number. If the string within the list is larger then the integer value it is then discarded and deleted from the list. This is what i have so far:

def main(L,n):

    i=0
    while i<(len(L)):
        if L[i]>n:
            L.pop(i)
        else:
            i=i+1
    return L

    #MAIN PROGRAM
    L = ["bob", "dave", "buddy", "tujour"]
    n = int (input("enter an integer value)
    main(L,n)

So really what im trying to do here is to let the user enter a number to then be compared to the list of string values. For example, if the user enters in the number 3 then dave, buddy, and tujour will then be deleted from the list leaving only bob to be printed at the end.

Thanks a million!

All the solutions posted were very helpful. Thanks, I really appreciate it! — CP_nate
– CP_nate, Commented Mar 12, 2015 at 19:25

Malonge · Accepted Answer · 2015-03-12 19:16:41Z

1

Looks like you are doing to much here. Just return a list comprehension that makes use of the appropriate conditional.

def main(L,n):
    return([x for x in L if len(x) <= n])

answered Mar 12, 2015 at 19:16

Malonge

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1 Comment

Padraic Cunningham Over a year ago

This creates a new list which is not the same as the OP's code

Malik Brahimi · Accepted Answer · 2015-03-12 19:20:50Z

0

Just use the built-in filter method, where n is the cut off length:

newList = filter(lambda i:len(i) <= n, oldList)

answered Mar 12, 2015 at 19:20

Malik Brahimi

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Comments

Padraic Cunningham · Accepted Answer · 2015-03-12 19:24:07Z

0

You should not remove elements from a list you are iterating over, you need to copy or use reversed:

L = ["bob", "dave", "buddy", "tujour"]
n = int(input("enter an integer value"))

for name in reversed(L):
    # compare length of name vs n
    if len(name) > n:
        # remove name if length is > n
        L.remove(ele)
print(L)

Making a copy using [:] syntax:

   for name in l[:]:
        # compare length of name vs n
        if len(name) > n:
            # remove name if length is > n
            L.remove(ele)
    print(L)

edited Mar 12, 2015 at 19:24

answered Mar 12, 2015 at 19:18

Padraic Cunningham

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Comments

tanmoy · Accepted Answer · 2015-03-12 20:12:54Z

0

This is a simple solution where you can print L after calling the main function. Hope this helps.

def main(L,n): l=len(L) x=0 while x<l: #if length is greater than n, remove the element and decrease the list size by 1 if len(L[x])>n: L.remove(L[x]) l=l-1 #else move to the next element in the list else: x=x+1

edited Mar 12, 2015 at 20:12

answered Mar 12, 2015 at 19:52

tanmoy

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