Extract first item of each sublist

Question

I am wondering what is the best way to extract the first item of each sublist in a list of lists and append it to a new list  So if I have   lst     a b c    1 2 3    x y z     and I want to pull out a  1 and x and create a separate list from those   I tried   lst2 append x 0  for x in lst

User · Answer

Python includes a function called itemgetter to return the item at a specific index in a list:

from operator import itemgetter

Pass the itemgetter() function the index of the item you want to retrieve. To retrieve the first item, you would use itemgetter(0). The important thing to understand is that itemgetter(0) itself returns a function. If you pass a list to that function, you get the specific item:

itemgetter(0)([10, 20, 30]) # Returns 10

This is useful when you combine it with map(), which takes a function as its first argument, and a list (or any other iterable) as the second argument. It returns the result of calling the function on each object in the iterable:

my_list = [['a', 'b', 'c'], [1, 2, 3], ['x', 'y', 'z']]
list(map(itemgetter(0), my_list)) # Returns ['a', 1, 'x']

Note that map() returns a generator, so the result is passed to list() to get an actual list. In summary, your task could be done like this:

lst2.append(list(map(itemgetter(0), lst)))

This is an alternative method to using a list comprehension, and which method to choose highly depends on context, readability, and preference.

More info: https://docs.python.org/3/library/operator.html#operator.itemgetter

User · Answer

You could use zip    gt  gt  gt  lst   1 2 3   11 12 13   21 22 23    gt  gt  gt  zip  lst  0   1  11  21    Or  Python 3 where zip does not produce a list    gt  gt  gt  list zip  lst   0   1  11  21    Or     gt  gt  gt  next zip  lst    1  11  21    Or   my favorite  use numpy    gt  gt  gt  import numpy as np  gt  gt  gt  a np array   1 2 3   11 12 13   21 22 23     gt  gt  gt  a array    1   2   3           11  12  13           21  22  23     gt  gt  gt  a   0  array   1  11  21

User · Answer

You said that you have an existing list  So I ll go with that    gt  gt  gt  lst1      a   b   c     1 2 3     x   y   z     gt  gt  gt  lst2    1  2  3    Right now you are appending the generator object to your second list    gt  gt  gt  lst2 append item 0  for item in lst   gt  gt  gt  lst2  1  2  3   lt generator object  lt genexpr gt  at 0xb74b3554 gt     But you probably want it to be a list of first items   gt  gt  gt  lst2 append  item 0  for item in lst    gt  gt  gt  lst2  1  2  3    a   1   x      Now we appended the list of first items to the existing list  If you d like to add the items themeselves  not a list of them  to the existing ones  you d use list extend  In that case we don t have to worry about adding a generator  because extend will use that generator to add each item it gets from there  to extend the current list    gt  gt  gt  lst2 extend item 0  for item in lst   gt  gt  gt  lst2  1  2  3   a   1   x     or   gt  gt  gt  lst2    x 0  for x in lst   1  2  3   a   1   x    gt  gt  gt  lst2  1  2  3    https   docs python org 3 4 tutorial datastructures html more-on-lists https   docs python org 3 4 tutorial datastructures html list-comprehensions

User · Answer

Your code is almost correct  The only issue is the usage of list comprehension   If you use like   x 0  for x in lst   it returns a generator object  If you use like   x 0  for x in lst   it return a list   When you append the list comprehension output to a list  the output of list comprehension is the single element of the list   lst      a   b   c     1 2 3     x   y   z    lst2      lst2 append  x 0  for x in lst   print lst2 0    lst2         a   1   x     lst2 0      a   1   x    Please let me know if I am incorrect

User · Answer

lst      a   b   c     1 2 3     x   y   z    outputlist      for values in lst      outputlist append values 0    print outputlist     Output     a   1   x

User · Answer

Had the same issue and got curious about the performance of each solution   Here s is the  timeit   import numpy as np lst      a   b   c     1 2 3     x   y   z      The first numpy-way  transforming the array    timeit list np array lst  T 0   4 9   s    163 ns per loop  mean    std  dev  of 7 runs  100000 loops each    Fully native using list comprehension  as explained by  alecxe     timeit  item 0  for item in lst  379 ns    23 1 ns per loop  mean    std  dev  of 7 runs  1000000 loops each    Another native way using zip  as explained by  dawg     timeit list zip  lst   0  585 ns    7 26 ns per loop  mean    std  dev  of 7 runs  1000000 loops each    Second numpy-way  Also explained by  dawg    timeit list np array lst    0   4 95   s    179 ns per loop  mean    std  dev  of 7 runs  100000 loops each    Surprisingly  well  at least for me  the native way using list comprehension is the fastest and about 10x faster than the numpy-way  Running the two numpy-ways without the final list saves about one   s which is still in the 10x difference   Note that  when I surrounded each code snippet with a call to len  to ensure that Generators run till the end  the timing stayed the same

User · Answer

Using list comprehension    gt  gt  gt  lst      a   b   c     1 2 3     x   y   z     gt  gt  gt  lst2    item 0  for item in lst   gt  gt  gt  lst2   a   1   x

[python] Extract first item of each sublist

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