Python - Remove integers in a list comprehension with less than two digits

Question

I am doing one of the Python beginner projects in this subreddit: http://www.reddit.com/r/beginnerprojects and for part of one of the tasks, I need to remove all integers from this list that have less than two digits. I'm not sure where I'm going wrong. I know I could just change the range in the list, but I want to utilise as many skills as possible.

numbers = [x for x in range(1,1001)]

def two_or_more_digits():
    for num in numbers:
        if len(str(num)) < 2:
            numbers.remove(num)
    print (numbers)

I am working in Python 3.

Thank you.

Hi! Are you calling the two_or_more_digits function? It doesn't show in your code — mguijarr, Aug 16 '14 at 14:27
possible duplicate of [Removing Item From List - during iteration - what's wrong with this idiom?](http://stackoverflow.com/questions/2896752/removing-item-from-list-during-iteration-whats-wrong-with-this-idiom) — bereal, Aug 16 '14 at 14:29
Yes, I am. Sorry, forgot to put that in. The print should also be indented 4 spaces. I'll edit that now. — billylevin, Aug 16 '14 at 14:30
For the `range(1, 1001)` you're iterating over - why convert to `str` then take the `len` - a much more simple test is to check that it's `>= 10`... (in fact - `range(10, 1001)` will give you the result you're after from the outset...) — Jon Clements, Aug 16 '14 at 14:31
can numbers contain any combination of different numbers in any order? — Padraic Cunningham, Aug 16 '14 at 14:50
in your own code if you use `for num in numbers[:]` you will also get the desired result, it makes a copy of numbers to iterate over, then you are not iterating over and mutating the same list which is something you should not do — Padraic Cunningham, Aug 16 '14 at 16:55

score 1 · Accepted Answer · answered Aug 16 '14 at 14:39

1

Instead of converting your digit to a string and calculating its length you can also use simple division:

[i for i in range(1,1001) if i // 10 > 0]

this will return a list with ints from 10 to 1000 (inclusive).

answered Aug 16 '14 at 14:39

martin

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WKPlus · Answer 2 · 2014-08-16T14:37:02.327

0

You can use list comprehension:

numbers = [x for x in range(1,1001) if len(str(x)) >= 2]

Or filter:

all_nums = [x for x in range(1,1001)]
needed_nums = filter(lambda i: len(str(i)) >= 2, all_nums]

If all elements all integer, you don't need to convert it to str and filter by length, you can just use x >= 10 instead of len(str(x)) >= 2

edited Aug 16 '14 at 14:37

answered Aug 16 '14 at 14:30

WKPlus

6,955
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why converting to a string to know if an integer is less than 10 ? – mguijarr Aug 16 '14 at 14:31
@mguijarr Yes, we can use `x >= 10` instead of `len(str(x)) >= 2` when x are in [1..1000] range. – WKPlus Aug 16 '14 at 14:35
`list(filter(lambda i: len(str(i)) >= 2, numbers))` to get a list in python 3 – Padraic Cunningham Aug 16 '14 at 14:55

score 0 · Answer 3 · answered Aug 16 '14 at 14:31

You're altering the numbers array while iterating over it... Can that be right?

If you want to grow your skills, probably you want to express the condition as a predicate, and use it either in a filter, or in a list comprehension:

filtered = filter( lambda x: len(str(num)) < 2, numbers )

filtered2 = [x for x in numbers if len(str(x)) < 2]

(or even better: use itertools.ifilter instead)

Python - Remove integers in a list comprehension with less than two digits

3 Answers3