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

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

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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

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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

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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

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 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

[python] Extract first item of each sublist

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