Append integer to beginning of list in Python

Question

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

User · Answer

You can use Unpack list      a   5      li    1 2 3       li    a   li           5  1  2  3

User · Answer

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

User · Answer

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  lt first element gt    append pd Series  lt list of other elements gt     tolist

User · Answer

gt  gt  gt var 7  gt  gt  gt array    1 2 3 4 5 6   gt  gt  gt array insert 0 var   gt  gt  gt 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

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New lists can be made by simply adding lists together   list1     value1   value2   value3   list2     value0   newlist list2 list1 print newlist

User · Answer

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

User · Answer

gt  gt  gt  a   5  gt  gt  gt  li    1  2  3   gt  gt  gt   a    li    Don t use  list  as variable name   5  1  2  3

User · Answer

Another way of doing the same    list 0 0     a

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Alternative    gt  gt  gt  from collections import deque   gt  gt  gt  my list   deque    gt  gt  gt  my list append 1          append right  gt  gt  gt  my list append 2          append right  gt  gt  gt  my list append 3          append right  gt  gt  gt  my list appendleft 100    append left  gt  gt  gt  my list  deque  100  1  2  3     gt  gt  gt  my list 0   100      NOTE    collections deque is faster than Python pure list in a loop Relevant-Post

User · Answer

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

[python] Append integer to beginning of list in Python

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