How do I convert all strings in a list of lists to integers?


I have a tuple of tuples containing strings:

T1 = (('13', '17', '18', '21', '32'),
      ('07', '11', '13', '14', '28'),
      ('01', '05', '06', '08', '15', '16'))

I want to convert all the string elements into integers and put them back into a list of lists:

T2 = [[13, 17, 18, 21, 32],
      [7, 11, 13, 14, 28],
      [1, 5, 6, 8, 15, 16]]


Asked By: elfuego1



int() is the Python standard built-in function to convert a string into an integer value. You call it with a string containing a number as the argument, and it returns the number converted to an integer:

>>> int("1") + 1

If you know the structure of your list, T1 (that it simply contains lists, only one level), you could do this in Python 3:

T2 = [list(map(int, x)) for x in T1]

In Python 2:

T2 = [map(int, x) for x in T1]
Answered By: unwind

You can do this with a list comprehension:

T2 = [[int(column) for column in row] for row in T1]

The inner list comprehension ([int(column) for column in row]) builds a list of ints from a sequence of int-able objects, like decimal strings, in row. The outer list comprehension ([... for row in T1])) builds a list of the results of the inner list comprehension applied to each item in T1.

The code snippet will fail if any of the rows contain objects that can’t be converted by int. You’ll need a smarter function if you want to process rows containing non-decimal strings.

If you know the structure of the rows, you can replace the inner list comprehension with a call to a function of the row. Eg.

T2 = [parse_a_row_of_T1(row) for row in T1]
Answered By: Will Harris

If it’s only a tuple of tuples, something like rows=[map(int, row) for row in rows] will do the trick. (There’s a list comprehension and a call to map(f, lst), which is equal to [f(a) for a in lst], in there.)

Eval is not what you want to do, in case there’s something like __import__("os").unlink("importantsystemfile") in your database for some reason.
Always validate your input (if with nothing else, the exception int() will raise if you have bad input).

Answered By: AKX

Using list comprehensions:

t2 = [map(int, list(l)) for l in t1]
Answered By: Jon

I would agree with everyone’s answers so far but the problem is that if you do not have all integers, they will crash.

If you wanted to exclude non-integers then

T1 = (('13', '17', '18', '21', '32'),
      ('07', '11', '13', '14', '28'),
      ('01', '05', '06', '08', '15', '16'))
new_list = list(list(int(a) for a in b) for b in T1 if a.isdigit())

This yields only actual digits. The reason I don’t use direct list comprehensions is because list comprehension leaks their internal variables.

Answered By: Christian Witts

Using only list comprehensions:

[[int(y) for y in x] for x in T1]
Answered By: odwl

for i in range(0,len(T1)):
    for j in range(0,len(T1[i])):

print T3
Answered By: P.Hunter

Instead of putting int( ), put float( ) which will let you use decimals along with integers.

Answered By: weir99

Try this.

x = "1"

x is a string because it has quotes around it, but it has a number in it.

x = int(x)

Since x has the number 1 in it, I can turn it in to a integer.

To see if a string is a number, you can do this.

def is_number(var):
        if var == int(var):
            return True
    except Exception:
        return False

x = "1"

y = "test"

x_test = is_number(x)


It should print to IDLE True because x is a number.

y_test = is_number(y)


It should print to IDLE False because y in not a number.

Answered By: TheRedstoneLemon

In Python 3.5.1 things like these work:

c = input('Enter number:')
print (int(float(c)))
print (round(float(c)))


Enter number:  4.7
Answered By: George Moraitis

Yet another functional solution for Python 2:

from functools import partial

map(partial(map, int), T1)

Python 3 will be a little bit messy though:

list(map(list, map(partial(map, int), T1)))

we can fix this with a wrapper

def oldmap(f, iterable):
    return list(map(f, iterable))

oldmap(partial(oldmap, int), T1)
Answered By: Eli Korvigo

See this function

def parse_int(s):
        res = int(eval(str(s)))
        if type(res) == int:
            return res


val = parse_int('10')  # Return 10
val = parse_int('0')  # Return 0
val = parse_int('10.5')  # Return 10
val = parse_int('0.0')  # Return 0
val = parse_int('Ten')  # Return None

You can also check

if val == None:  # True if input value can not be converted
    pass  # Note: Don't use 'if not val:'
Answered By: Shameem

You can do something like this:

T1 = (('13', '17', '18', '21', '32'),  
     ('07', '11', '13', '14', '28'),  
     ('01', '05', '06', '08', '15', '16'))  
new_list = list(list(int(a) for a in b if a.isdigit()) for b in T1)  
Answered By: Amit Rawat

I want to share an available option that doesn’t seem to be mentioned here yet:


Will generate a random permutation of array x. Not exactly what you asked for, but it is a potential solution to similar questions.

Answered By: samvoit4

Python has built in function int(string) and optional parameter base.

if your string contains an Integer value, it will convert that to the corresponding Integer value.
However if you have decimnal number as string you’ll need float() to convert it.


a = '22'
b = int(a)


if a = '22.22'
b = int(a) '''will give error, invalid literal for int().'''
b = float(a) '''will convert the string.'''
Answered By: ToM
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