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for loop that replaces element based on its value

I have a list of integers and I am trying to define a function which loops through every element to check if they are less than 5 and returns the list in string according to their contents.

intlist=[12, 10, 11, 23, 25, 2]

def clear(x):
for i in x:
if i < 5:
x[i] = 0
return str(x)
else:
return str(x)

print clear(intlist)


My code is not working as intended, could anyone enlighten me?

If they are, I am to change all elements in the list to '0'. The outcome should look something like this.

intlist=[0, 0, 0, 0, 0, 0]


However if none of the elements are less than 5, the output should remain the same.

intlist=[12, 10, 11, 23, 25, 2]


if-else

4 Answers
4

Welcome to StackOverflow! Your code has some logic problem. Here's my fix:

intlist=[12, 10, 11, 23, 25, 2]

def clear(x):
for i in x:
if i < 5: # If this is satisfied once, return string with n times of '0'
x = [0]*len(x) # This creates a list of zeros with same length as before
return str(x)

print clear(intlist)


Besides, in your example the element 2 is less than 5, so the output should be 000000

2

000000

You can do it one one-line, creating a list of zeros if any one of the elements in list is less than 5 or retaining the list otherwise.

5

intlist = [12, 10, 11, 23, 25, 2]

intlist = [0] * len(intlist) if any(x for x in intlist if x < 5) else intlist

print(intlist)
# [0, 0, 0, 0, 0, 0]


intlist=[12, 10, 11, 23, 25, 2]

def clear(x):
if (any(i<5 for i in intlist)): return [0]*len(intlist)
else: return(x)
print(clear(intlist)) # [0, 0, 0, 0, 0, 0]


check for the any item is <5 then return all zeros or keep the list as it is.

Try this out if you want to use function as well as an if-else block :

intlist=[12, 10, 11, 23, 25, 2]

def clear(x):
for i in range(len(x)):
if x[i]<5 :
x[i] = 0
else:
continue
return x

print(clear(intlist)) # Python 3
print clear(intlist) # Python 2


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