Integers, should be entered one per line, how to make 'hit return when done'?

Question

nums = []

num = 0
valid_list = []
while num !=  '':
    num = input('Enter numbers :')
    nums.append(num)
nums2 = [(nums[i]) for i in range(0,len(nums))]
for i in range(len(nums2)):
    if int(nums2[i]) < 101 and int(nums2[i]) > 0:
        valid_list.append(nums2[i])

print(valid_list)

Sorry dont really know how to ask question corectly, but i hope those who get my idea will help, Thanks. So i am tryed to make program wich fills list with integers entered by user, than chech if they fit 1-101 and if they fit put those number in valid_list, The problem is while num != '': (BTW inters must be entered one by one, and it must stop when return is hited)

Rather than just put random words in the tags, it would help immensely if you would add a tag for the language you're using. — Ken White, Nov 17 '14 at 20:48
Take a look here: http://stackoverflow.com/questions/7255463/exit-while-loop-by-user-hitting-enter-key — Ayb4btu, Nov 17 '14 at 20:54

score 0 · Answer 1 · edited May 23 '17 at 12:05

Without having actually tested it, I think this is what you want:

nums = []

num = 0
valid_list = []

while true:
    num = input('Enter numbers :')

    if not num:
        break

    nums.append(num)

nums2 = [(nums[i]) for i in range(0,len(nums))]

for i in range(len(nums2)):
    if int(nums2[i]) < 101 and int(nums2[i]) > 0:
        valid_list.append(nums2[i])

print(valid_list)

From Python 'If not' syntax the if not num syntax is explained as executing if num is any kind of zero or empty container, or False.

Integers, should be entered one per line, how to make 'hit return when done'?

1 Answers1