How to convert a list of multiple integers into a single integer?

Question

How do I convert a list in Python 3.5 such as:

x=[1, 3, 5]

to an int of 135 (a whole int)?

Answer 1

If you don't like map you can always use a list comprehension:

s = [str(i) for i in x]
r = int("".join(s))

Answer 2

Here is a more mathematical way that does not have to convert back and forth to string. Note that it will only work if 0 <= i <= 9.

>>> x = [1, 3, 5]
>>> sum(d * 10**i for i, d in enumerate(x[::-1]))
135

The idea is to multiply each element in the list by its corresponding power of 10 and then to sum the result.

Answer 3

If you have a list of ints and you want to join them together, you can use map with str to convert them to strings, join them on the empty string and then cast back to ints with int.

In code, this looks like this:

r = int("".join(map(str, x)))

and r now has the wanted value of 135.

This, of course, is a limited approach that comes with some conditions. It requires the list in question to contain nothing else but positive ints (as your sample) or strings representing ints, else the steps of conversion to string might fail or the joining of (negative) numbers will be clunky.

Answer 4

Using only math (no conversions to or from strings), you can use the reduce function (functools.reduce in Python 3)

b = reduce(lambda total, d: 10*total + d, x, 0)

This makes use of Horner's rule, which factors the polynomial representing the number to reduce the number of multiplications. For example,

1357 = 1*10*10*10 + 3*10*10 + 5*10 + 7     # 6 multiplications
     = ((1*10 + 3)*10 + 5)*10 + 7          # 3 multiplications

As a result, this is faster than computing powers of 10 or creating a string and converting the result to an integer.

>>> timeit.timeit('reduce(lambda t,d: 10*t+d, x, 0)', 'from functools import reduce; x=[1,3,5,7]')
0.7217515400843695
>>> timeit.timeit('int("".join(map(str, [1,3,5,7])))')
1.425914661027491
>>> timeit.timeit('sum(d * 10**i for i, d in enumerate(x[::-1]))', 'x=[1,3,5,7]')
1.897974518011324

---

In fairness, string conversion is faster once the number of digits gets larger.

>>> import timeit

# 30 digits
>>> setup='from functools import reduce; x=[5, 2, 6, 8, 4, 6, 6, 4, 8, 0, 3, 1, 7, 6, 8, 2, 9, 9, 9, 5, 4, 5, 5, 4, 3, 6, 9, 2, 2, 1]' 
>>> print(timeit.timeit('reduce(lambda t,d: 10*t+d, x, 0)', setup))
6.520374411018565
>>> print(timeit.timeit('int("".join(map(str, x)))', setup))
6.797425839002244
>>> print(timeit.timeit('sum(d * 10**i for i, d in enumerate(x[::-1]))', setup))
19.430233853985555

# 60 digits
>>> setup='from functools import reduce; x=2*[5, 2, 6, 8, 4, 6, 6, 4, 8, 0, 3, 1, 7, 6, 8, 2, 9, 9, 9, 5, 4, 5, 5, 4, 3, 6, 9, 2, 2, 1]' 
>>> print(timeit.timeit('reduce(lambda t,d: 10*t+d, x, 0)', setup))
13.648188541992567
>>> print(timeit.timeit('int("".join(map(str, x)))', setup))
12.864593736943789
>>> print(timeit.timeit('sum(d * 10**i for i, d in enumerate(x[::-1]))', setup))
44.141602706047706

# 120 digits!
>>> setup='from functools import reduce; x=4*[5, 2, 6, 8, 4, 6, 6, 4, 8, 0, 3, 1, 7, 6, 8, 2, 9, 9, 9, 5, 4, 5, 5, 4, 3, 6, 9, 2, 2, 1]' 
>>> print(timeit.timeit('reduce(lambda t,d: 10*t+d, x, 0)', setup))
28.364255172084086
>>> print(timeit.timeit('int("".join(map(str, x)))', setup))
25.184791765059344
>>> print(timeit.timeit('sum(d * 10**i for i, d in enumerate(x[::-1]))', setup))
99.88558598596137