How to convert strings into integers?

PythonStringInteger

Python Problem Overview


I have a tuple of tuples from a MySQL query like this:

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

I'd like 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]]

I tried to achieve it with eval but didn't get any decent result yet.

Python Solutions


Solution 1 - Python

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
2

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]

Solution 2 - Python

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]

Solution 3 - Python

I would rather prefer using only comprehension lists:

[[int(y) for y in x] for x in T1]

Solution 4 - Python

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

Solution 5 - Python

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.

Solution 6 - Python

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

x = "1"

y = "test"

x_test = is_number(x)

print(x_test)

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

y_test = is_number(y)

print(y_test)

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

Solution 7 - Python

T3=[]

for i in range(0,len(T1)):
    T3.append([])
    for j in range(0,len(T1[i])):
        b=int(T1[i][j])
        T3[i].append(b)
    
print T3

Solution 8 - Python

Using list comprehensions:

t2 = [map(int, list(l)) for l in t1]

Solution 9 - Python

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)

Solution 10 - Python

See this function

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

Then

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:'

Solution 11 - Python

In Python 3.5.1 things like these work:

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

and

Enter number:  4.7
4
5

Solution 12 - Python

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.

Usage:

a = '22'
b = int(a)

and

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

Solution 13 - Python

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).

Solution 14 - Python

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

rumpy.random.permutation(x)

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

Solution 15 - Python

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)  
print(new_list)  

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