Python append for multiple lists

Question

I've got a problem of appending elements in multiple lists. My program goes like this. where list is just a series of numbers.

for j in range(0, len(list)):
    if (int(list[j][4]) == 0 or int(list[j][:-4]) == 41601000000):
        filelist0.append(list[j])
    if (int(list[j][4]) == 1 or int(list[j][:-4]) == 41602000000):
        filelist1.append(list[j])
    if (int(list[j][4]) == 2 or int(list[j][:-4]) == 41603000000):
        filelist2.append(list[j])
    if (int(list[j][4]) == 3 or int(list[j][:-4]) == 41604000000):
        filelist3.append(list[j])
    if (int(list[j][4]) == 4 or int(list[j][:-4]) == 41605000000):
        filelist4.append(list[j])
    if (int(list[j][4]) == 5 or int(list[j][:-4]) == 41606000000):
        filelist5.append(list[j])
    if (int(list[j][4]) == 6 or int(list[j][:-4]) == 41607000000):
        filelist6.append(list[j])
    if (int(list[j][4]) == 7 or int(list[j][:-4]) == 41608000000):
        filelist7.append(list[j])
    if (int(list[j][4]) == 8 or int(list[j][:-4]) == 41609000000):
        filelist8.append(list[j])

This is ugly. Is there a way to write the above code in a line or two? Obvious I can iterate strings, but not for the name of the lists. (can't iterate filelist[m] for m in range(0, 9)).

Thanks!

Tip: Don't use list as a variable name. Subtle mayhem ensues. — Tim Pietzcker
– Tim Pietzcker, Commented May 1, 2013 at 21:46

Andrew Clark · Accepted Answer · 2013-05-01 21:50:16Z

3

The following should be equivalent:

filelists = [filelist0, filelist1, filelist2, filelist3, filelist4,
             filelist5, filelist6, filelist7, filelist8]
for x in lst:
    for i, filelist in enumerate(filelists):
        if int(x[4]) == i or int(x[:-4]) == 41601000000 + i * 1000000:
            filelist.append(x)

Note that I renamed list to lst, it isn't a good idea to used built-in names for your variables.

edited May 1, 2013 at 21:50

answered May 1, 2013 at 21:35

Andrew Clark

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2 Comments

Fred Foo Over a year ago

+1, but there's no need for the j variable. A loop over x in lst would be even shorter.

John Gordon Over a year ago

Shouldn't that be if int(x[4]) == i or ...

Dan Temkin · Accepted Answer · 2017-10-30 22:50:38Z

1

If you are using Python 2.7 you can actually use the map() function without a variable assigned, e.g.

>>> listdata = [0, 1, 2, 3, 4]
>>> list_o_lists = [ls0, ls1, ls2, ls3, ls4]
>>> map(lambda x, y: x.append(y), listdata, list_o_lists)

Hope this helps, thought it was a bit simpler than the other answers.

edited Oct 30, 2017 at 22:50

answered Jun 19, 2016 at 10:22

Dan Temkin

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Comments

mfitzp · Accepted Answer · 2013-05-01 21:45:03Z

0

Why not put the lists in a dict? You can use defaultdict to automatically create the item if it's not already in there. The increment number on the end is done by %d into a string then using int to convert back to a number. Still ugly, but there you go.

from collections import defaultdict

filelist = defaultdict(list)

for j in range(0, len(list)):
    for fl in range(0, 9) #0-8
        if (int(list[j][4]) == fl or int(list[j][:-4]) == int('4160%d000000' % fl+1)):
            filelist[fl].append(list[j])

edited May 1, 2013 at 21:45

answered May 1, 2013 at 21:36

mfitzp

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Comments

kiriloff · Accepted Answer · 2013-05-01 22:05:03Z

0

This solution with one list of lists

filelists = []

list = [[1,3,2,6,5,7,6],[3,5,7,8,5,5,4]]

for i in range(8):
    filelists.append([])
    for j in range(0, len(list)):
        if (int(list[j][4]) == i or int(list[j][-4]) == 41601000000):
            print 'added'
            filelists[i-1].append(list[j])

Result is

>>> filelists
[[], [], [], [], [[1, 3, 2, 6, 5, 7, 6], [3, 5, 7, 8, 5, 5, 4]], [], [], []]

answered May 1, 2013 at 22:05

kiriloff

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Comments

dugres · Accepted Answer · 2013-05-01 21:47:44Z

-2

filelist = [[] for _ in range(9)]
for j, item in enumerate(item_list):
    for i in range(9):
        if (int(item[4])==i or
            int(item[-4])==int('4160{}000000'/format(i+1))
        ):
            filelist[i].append(j)
            break

answered May 1, 2013 at 21:47

dugres

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