How to add an integer to each element in a list?

Question

If I have list=[1,2,3] and I want to add 1 to each element to get the output [2,3,4], how would I do that?

I assume I would use a for loop but not sure exactly how.

Answer 1

A lot of good solutions, here is some benchmarking.

from functools import partial
from operator import add

test_list = list(range(1000)); n = 3;

%timeit _=list(map(lambda x:x+n, test_list))
%timeit _=[x+n for x in test_list]
%timeit _=[x.__add__(n) for x in test_list]
%timeit _=list(map(lambda x:x.__add__(n), test_list))
%timeit _=list(map(partial(add,n), test_list))

Out:
>> 53.6 µs ± 651 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
>> 38.2 µs ± 392 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
>> 80.2 µs ± 851 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
>> 108 µs ± 513 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
>> 53.7 µs ± 332 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

Seems like [x+n for x in test_list] is the fastest solution in pure python.

Answer 2

Just in case anyone was looking for a solution that only uses built-ins and no lambdas:

from functools import partial
from operator import add


my_list = range(1, 4)  # list(my_list) #=> [1, 2, 3]
my_list_plus_one = list(map(partial(add, 1), my_list)  #=> [2, 3, 4]

Answer 3

list = [1,2,3,4,5]

for index in range(len(list)):
      list[index] = list[index] +1

print(list)

Answer 4

import numpy as np

np.add([1, 2, 3], 1).tolist()

which gives

[2, 3, 4]

Answer 5

Many of the answers above are very good. I've also seen some weird answers that will do the job. Also, the last answer seen was through a normal loop. This willingness to give answers leads me to itertools and numpy, which will do the same job in a different way.

Here I present different ways to do the job, not answered above.

import operator
import itertools

x = [3, 5, 6, 7]

integer = 89

"""
Want more vairaint can also use zip_longest from itertools instead just zip
"""
#lazy eval
a = itertools.starmap(operator.add, zip(x, [89] * len(x))) # this is not subscriptable but iterable
print(a)
for i in a:
  print(i, end = ",")


# prepared list
a = list(itertools.starmap(operator.add, zip(x, [89] * len(x)))) # this returns list
print(a)



# With numpy (before this, install numpy if not present with `pip install numpy`)
import numpy

res = numpy.ones(len(x), dtype=int) * integer + x # it returns numpy array
res = numpy.array(x) + integer # you can also use this, infact there are many ways to play around
print(res)
print(res.shape) # prints structure of array, i.e. shape

# if you specifically want a list, then use tolist

res_list = res.tolist()
print(res_list)

Output

>>> <itertools.starmap object at 0x0000028793490AF0> # output by lazy val
>>> 92,94,95,96,                                     # output of iterating above starmap object
>>> [92, 94, 95, 96]                                 # output obtained by casting to list
>>>                   __
>>> # |\ | |  | |\/| |__| \ /
>>> # | \| |__| |  | |     |
>>> [92 94 95 96]                                    # this is numpy.ndarray object
>>> (4,)                                             # shape of array
>>> [92, 94, 95, 96]                                 # this is a list object (doesn't have a shape)

My sole reason to highlight the use of numpy is that one should always do such manipulations with libraries like numpy because it is performance efficient for very large arrays.

Answer 6

Came across a not so efficient, but unique way of doing it. So sharing it across.And yes it requires extra space for another list.

from operator import add
test_list1 = [4, 5, 6, 2, 10]
test_list2 = [1] * len(test_list1)

res_list = list(map(add, test_list1, test_list2))

print(test_list1)
print(test_list2)
print(res_list)

#### Output ####
[4, 5, 6, 2, 10]
[1, 1, 1, 1, 1]
[5, 6, 7, 3, 11]

Answer 7

Edit: this isn't in-place

Firstly don't use the word 'list' for your variable. It shadows the keyword list.

The best way is to do it in place using splicing, note the [:] denotes a splice:

>>> _list=[1,2,3]
>>> _list[:]=[i+1 for i in _list]
>>> _list
[2, 3, 4]

Answer 8

>>> [x.__add__(1) for x in [1, 3, 5]]
3: [2, 4, 6]

My intention here is to expose if the item in the list is an integer it supports various built-in functions.

Answer 9

if you want to use numpy there is another method as follows

import numpy as np
list1 = [1,2,3]
list1 = list(np.asarray(list1) + 1)

Answer 10

Python 2+:

>>> mylist = [1,2,3]
>>> map(lambda x: x + 1, mylist)
[2, 3, 4]

Python 3+:

>>> mylist = [1,2,3]
>>> list(map(lambda x: x + 1, mylist))
[2, 3, 4]

Answer 11

The other answers on list comprehension are probably the best bet for simple addition, but if you have a more complex function that you needed to apply to all the elements then map may be a good fit.

In your example it would be:

>>> map(lambda x:x+1, [1,2,3])
[2,3,4]

Answer 12

>>> mylist = [1,2,3]
>>> [x+1 for x in mylist]
[2, 3, 4]
>>>

list-comprehensions python.

Answer 13

new_list = [x+1 for x in my_list]