Python append to list or ignore

Question

I want make an array as follow :

for i in range(0, len(final), 3):
    ar.append(
        (
            final[i] + 100,
            final[i+1] + 100,
            final[i+2] + 100
        )
    )

but there is a simple problem. length of final is about 2682715 and i have exception for list index when reach 2682714 with step=3

final[2682714] = OK
final[2682715] = OK
final[2682716] = ERROR

How can i handle that? we can use try/except IndexError but whole item will be ignored

What should the last element of `ar` be? `(final[2682714], final[2682715])` — , Jan 01 '18 at 23:05
https://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks — Take_Care_, Jan 01 '18 at 23:06
i want two last items and `except` will be ignored them @SeppeMariën — Arash Hatami, Jan 01 '18 at 23:31

Patrick Artner · Accepted Answer · 2018-01-01T23:34:30.857

3

final = range(31)
ar = []
for i in range(0, len(final), 3):
    ar.append(tuple(final[i:i+3]))

print(ar)

Output:

[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 14), (15, 16, 17), (18,
19, 20), (21, 22, 23), (24, 25, 26), (27, 28, 29),(30,)]

Your list length is divisible by 3 - you last element would be the last triplet , not a tuple with only one element...

Mine is not divisible by 3 without rest, so I have an incomplete tuple at the end.

Using the list comprehension tuple(final[i:i+3]) will ensure you do not overstep the boundaries of final and the remaining 1 or 2 numbers are in the last tuple.

edited Jan 01 '18 at 23:34

answered Jan 01 '18 at 23:07

Patrick Artner

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Note, `list(range(31))` isn't actually necessary, since `range` objects are sequences that support `len` and slicing... – juanpa.arrivillaga Jan 01 '18 at 23:15
OK, it's good ... but what about if i want add some number with them ? for example `final[i]+100, final[i+1]+100, final[i+2]+100` we can't use tuple at this situation... is it possible ? – Arash Hatami Jan 01 '18 at 23:22
1

@ArashHatami not specified in your question so far. Do you want to add the same constant to every one? – Patrick Artner Jan 01 '18 at 23:24
2

@ArashHatami `ar.append( tuple([x+100 for x in final[i:i+3]]))` – Patrick Artner Jan 01 '18 at 23:26

TheQuestioner · Answer 2 · 2018-01-01T23:39:29.863

2

you can use list comprehension:

final = (1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15)
ar = []
ar = [final[i:i + 3] for i in range(0, len(final), 3)]
print(ar)

output:

[(1, 2, 3), (4, 5, 7), (8, 9, 10), (11, 12, 13), (14, 15)]

edited Jan 01 '18 at 23:39

answered Jan 01 '18 at 23:09

TheQuestioner

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i updated my answer, it will give you now the last items – TheQuestioner Jan 02 '18 at 00:04

f5r5e5d · Answer 3 · 2018-01-01T23:20:59.460

2

slicing is nice, for one line add a zip

a = [*range(9)]

[*zip(a[::3], a[1::3], a[2::3])]

Out[321]: [(0, 1, 2), (3, 4, 5), (6, 7, 8)]

zip automatically drops extra elements that don't make up a tripple

edited Jan 01 '18 at 23:20

answered Jan 01 '18 at 23:17

f5r5e5d

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thats `0+every 3`, `skip 1+every 3`, `skip 2+every3` and zipped? Neat – Patrick Artner Jan 01 '18 at 23:19

Python append to list or ignore

3 Answers3