Append integer to beginning of list in Python


How do I prepend an integer to the beginning of a list?

[1, 2, 3]  ⟶  [42, 1, 2, 3]
Asked By: gen



>>> x = 42
>>> xs = [1, 2, 3]
>>> [x] + xs
[42, 1, 2, 3]

Note: don’t use list as a variable name.

Answered By: Rohit Jain
>>> x = 42
>>> xs = [1, 2, 3]
>>> xs.insert(0, x)
>>> xs
[42, 1, 2, 3]

How it works:

list.insert(index, value)

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

Answered By: Nullify

Another way of doing the same,

list[0:0] = [a]
Answered By: v2b

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.


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
Answered By: timgeb

New lists can be made by simply adding lists together.

list1 = ['value1','value2','value3']
list2 = ['value0']
Answered By: Erico9001

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()
Answered By: aparna ramesh

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).

Answered By: Dovi Salomon

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

Answered By: sahil panindre

You can use Unpack list:

a = 5

li = [1,2,3]

li = [a, *li]

=> [5, 1, 2, 3]

Answered By: HoangYell


>>> 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]



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

Answered By: Benyamin Jafari
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