For loop running infinite python

Question

It's been 5 hours trying to find the issue but unable to identify why for loop for fn in L running infinite.

L=[]
N=int(raw_input())
for i in range(0,N):
    L.append(list(raw_input().split()))
print L

for fn in L:
    if 'insert'==fn[0]:
        L.insert(int(fn[1]),int(fn[2]))
    elif 'append'==fn[0]:
        L.append(int(fn[1]))
    elif 'remove'==fn[0]:
        L.remove(int(fn[1]))
    elif 'pop'==fn[0]:
        L.pop(int(fn[1]))
    elif 'index'==fn[0]:
        L.index(int(fn[1])) 
    elif 'count'==fn[0]:
        L.count(int(fn[1]))
    elif 'sort'==fn[0]:
        L.sort()     
    elif 'reverse'==fn[0]:
        L.reverse() 
    else :
        print  L

Inputs provided to list:

12
insert 0 5
insert 1 10
insert 0 6
print 
remove 6
append 9
append 1
sort 
print
pop
reverse
print

Which for loop inside the for loop? I see two not nested explicit for loops. — Matthias
– Matthias, Commented Jun 9, 2016 at 16:40

Moses Koledoye · Accepted Answer · 2016-06-09 16:40:37Z

7

You're mutating your list in the loop. Outcomes will be very unpredictable. You can instead iterate on a slice of the list:

for fn in L[:]:
    # your code here
    pass

In this way, the loop terminates when the items in the shallow copy (the slice) are exhausted.

answered Jun 9, 2016 at 16:40

Moses Koledoye

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3 Comments

Abhishek Over a year ago

Thanks for the answer. What is slice of list?

Khertan Over a year ago

It s a shortcut, a slice ([:]) without bound, so a copy of the list as slicing a list create a new list. stackoverflow.com/questions/509211/…

Moses Koledoye Over a year ago

@Code-Monk In this context: a temporary shallow copy of the original list. Read more here

Matthias · Accepted Answer · 2016-06-09 16:41:15Z

2

You insert/remove elements while iterating the elements of a list. You also change the list by reversing the list. Both are mutation operations to the list which are not allowed during iteration. The behavior is unspecified in this case.

for fn in L:
    if ...:
        L.insert(...)

answered Jun 9, 2016 at 16:41

Matthias

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Comments

Khertan · Accepted Answer · 2016-06-09 16:46:30Z

1

You mutate the object used for looping ... and better you reverse it :p reverse <-- reverse --> reverse <-- reverse -->

:p

answered Jun 9, 2016 at 16:46

Khertan

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