Append integer to beginning of list in Python

The Question :

557 people think this question is useful

I have an integer and a list. I would like to make a new list of them beginning with the variable and ending with the list. Writing a + list I get errors. The compiler handles a as integer, thus I cannot use append, or extend either. How would you do this?

The Question Comments :

The Answer 1

875 people think this answer is useful
>>>var=7
>>>array = [1,2,3,4,5,6]
>>>array.insert(0,var)
>>>array
[7, 1, 2, 3, 4, 5, 6]

How it works:

array.insert(index, value)

Insert an item at a given position. The first argument is the index of the element before which to insert, so array.insert(0, x) inserts at the front of the list, and array.insert(len(array), x) is equivalent to array.append(x).Negative values are treated as being relative to the end of the array.

The Answer 2

542 people think this answer is useful
>>> a = 5
>>> li = [1, 2, 3]
>>> [a] + li  # Don't use 'list' as variable name.
[5, 1, 2, 3]

The Answer 3

106 people think this answer is useful

Note that if you are trying to do that operation often, especially in loops, a list is the wrong data structure.

Lists are not optimized for modifications at the front, and somelist.insert(0, something) is an O(n) operation.

somelist.pop(0) and del somelist[0] are also O(n) operations.

The correct data structure to use is a deque from the collections module. deques expose an interface that is similar to those of lists, but are optimized for modifications from both endpoints. They have an appendleft method for insertions at the front.

Demo:

In [1]: lst = [0]*1000
In [2]: timeit -n1000 lst.insert(0, 1)
1000 loops, best of 3: 794 ns per loop
In [3]: from collections import deque
In [4]: deq = deque([0]*1000)
In [5]: timeit -n1000 deq.appendleft(1)
1000 loops, best of 3: 73 ns per loop

The Answer 4

44 people think this answer is useful

Another way of doing the same,

list[0:0] = [a]

The Answer 5

15 people think this answer is useful

You can use Unpack list:

a = 5

li = [1,2,3]

li = [a, *li]

=> [5, 1, 2, 3]

The Answer 6

5 people think this answer is useful

Based on some (minimal) benchmarks using the timeit module it seems that the following has similar if not better performance than the accepted answer

new_lst = [a, *lst]

As with [a] + list this will create a new list and not mutate lst.

If your intention is to mutate the list then use lst.insert(0, a).

The Answer 7

3 people think this answer is useful

Alternative:

>>> from collections import deque

>>> my_list = deque()
>>> my_list.append(1)       # append right
>>> my_list.append(2)       # append right
>>> my_list.append(3)       # append right
>>> my_list.appendleft(100) # append left
>>> my_list

deque([100, 1, 2, 3])

>>> my_list[0]

100


[NOTE]:

collections.deque is faster than Python pure list in a loop Relevant-Post.

The Answer 8

2 people think this answer is useful

New lists can be made by simply adding lists together.

list1 = ['value1','value2','value3']
list2 = ['value0']
newlist=list2+list1
print(newlist)

The Answer 9

2 people think this answer is useful
list_1.insert(0,ur_data)

make sure that ur_data is of string type so if u have data= int(5) convert it to ur_data = str(data)

The Answer 10

1 people think this answer is useful

None of these worked for me. I converted the first element to be part of a series (a single element series), and converted the second element also to be a series, and used append function.

l = ((pd.Series(<first element>)).append(pd.Series(<list of other elements>))).tolist()

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