How to change list of string to list of elements (numbers)?

Question

When I use:

with open("test.txt") as file
    read = csv.reader(file)
    for i in read:
        print(i)

and I have got something like that:

['[0', ' 0', ' 1', ' 0]']

and I need:

[0, 0, 1, 0]

Any advises please?

Answer 1

ad = ['[0', ' 0', ' 1', ' 0]']
for i in range(len(ad)):
    ad[i] = int(ad[i].replace("[", "").replace("]", ""))
print ad
[0, 0, 1, 0]

Answer 2

Normally, the easiest solution would be a list comprehension where new = [int(a) for a in old]. However, in your case, the first and last elements of your list actually have brackets inside of them too.

Instead, you need to do something like:

new = [int("".join(filter(str.isdigit, a))) for a in old]

This is a pretty big one liner so lets break it down.

1. The list comprehension iterates through each element in your list (I called it old) and names it a
2. A is passed into the filter command with the function str.isdigit. This basically removes any character that isn't a digit. The issue with this, is that it returns an iterator and not a simple value.
3. To fix the iterator problem, I wrapped the filter command with a "".join() command to convert it to a simple string value. This string will only have the number.
4. Finally, we can wrap that entire thing with the int command which will transform your value into an int.

If you don't like the one-liner, it can also be done this way:

new = []
for a in old:
    filtered = filter(str.isdigit, a)
    num_str = "".join(filtered)
    num.append(int(num_str))

It's the same thing but a bit more verbose.

Answer 3

def parseToInt(s): 
    ...:     return int((s.replace('[', '')).replace(']', ''))

list(map(lambda x: parseToINt(x), a))