CODEWITHSUNDEEP
python
500500
500500

Reputation: 1

TypeError: 'int' object is not iterable map(), list() , set() do not solve my problem

I'm writing a program to efficiantly divide by three, but I can extract the individual numbers from the int.

I have already tried map() and list() and set() and they have not worked

num = 1
while True:
    num = num + 1
    threetest = 0
    digits = list(num)                 #(This is line 17)
    for a in range ( 0, len(str(num))):
        threetest = threetest + digits[a]
    if (treetest % 3) != 0:
        ...

Traceback (most recent call last): File "C:\test.py", line 17, in digits = list(num) TypeError: 'int' object is not iterable

I would expect to be able to add up the induvidual digits of a number like 936 and then divide the sum by 3 to efficiantly find if it is a multipule of 3

Upvotes: 0

Views: 555

Answers (3)

shaik moeed
shaik moeed

Reputation: 5785

Check this,

>>> digits = [ast.literal_eval(i) for i in str(a)]
>>> digits
[3, 4, 6]

You can replace,

digits = list(num)

with,

digits = [int(i) for i in str(a)]

Upvotes: 0

GordonAitchJay
GordonAitchJay

Reputation: 4860

To get a list of the digits of type int, try

digits = list(map(int, str(num)))

num is an int, which isn't an iterable. First get a str of num, then iterate over it and make each digit an int, and pack it all in a list.

Also, the code below is more pythonic:

num = 1
while True:
    num += 1
    three_test = 0
    digits = list(map(int, str(num)))
    for digit in range(digits):
        three_test += digit
    if (tree_test % 3) != 0:

Upvotes: 0

Óscar López
Óscar López

Reputation: 236004

You can get the desired effect by replacing this:

threetest = 0
digits = list(num)
for a in range (0, len(str(num))):
    threetest = threetest + digits[a]

With this:

threetest = 0
for digit in str(num):
    threetest += int(digit)

Or even simpler (and way more idiomatic):

threetest = sum(int(digit) for digit in str(num))

Upvotes: 1

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