pythonic way to add elements to nested list

Question

I have a nest list:

listSchedule = [[list1], [list2], etc..]]

I have another list and I want to append to each nested list the element of this list if the first element of each matches a string.

I can do it but I wonder if there is a more 'pythonic' way, using list comprehension?

index = 0;
for row in listSchedule:
   if row[0] == 'Download':
      row[3] = myOtherList[index]
      index +=1

Is there a reason you're not just using a dict (or if Download might appear multiple times, a multi-dict like collections.defaultdict(list)) or something? Needing to loop over every element to find a "header" to append data to sounds an awful lot like the case where you'd want a dict. — ShadowRanger
– ShadowRanger, Commented Sep 20, 2018 at 13:20
The more I look at this, the more I suspect an XY problem. Can you provide more details about the problem so we're not just applying touch-ups to a bad design? Even a real minimal reproducible example would be helpful (as written, this code guarantees an IndexError, since it assigns to row[3], but the listSchedule you provided is a list of len 1 lists, so any index but row[0] would error out. — ShadowRanger
– ShadowRanger, Commented Sep 20, 2018 at 14:20

fafl · Accepted Answer · 2018-09-20 13:20:05Z

4

Your code is very readable, I would not change it.

With a list comprehension you could write something like:

for index, row in enumerate([row for row in listSchedule if row[0] == 'Download']):
    row[3] = myOtherList[index]

answered Sep 20, 2018 at 13:20

fafl

7,5074 gold badges32 silver badges54 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

ShadowRanger Over a year ago

Omit the brackets; no need for a temporary list here (enumerate is perfectly happy to work with generators, it doesn't need full sequences). Without the brackets, enumerate(row for row in listSchedule if row[0] == 'Download') enumerates a generator expression that produces the rows lazily, no potentially large temporary lists needed.

Frayal · Accepted Answer · 2018-09-20 13:36:59Z

1

you could try that but make a copy of the otherlist to not lose the info:

[row+[myotherlist.pop(0)] if row[0]=='Download' else row for row in listScheduel]

for example:

list = [['Download',1,2,3],[0,1,2,3],['Download',1,2,3],['Download',1,2,3]]
otherlist = [0,1,2,3,4]
l = [ row+[otherlist.pop(0)] if row[0]=='Download' else row for row in list]

Output:

[['Download', 1, 2, 3, 0],
[0, 1, 2, 3],
['Download', 1, 2, 3, 1],
['Download', 1, 2, 3, 2]]

edited Sep 20, 2018 at 13:36

answered Sep 20, 2018 at 13:22

Frayal

2,17113 silver badges19 bronze badges

Comments

C.Nivs · Accepted Answer · 2018-09-20 13:23:18Z

0

If you want to truly append, why not use row.append(myOtherList[index])? That way you avoid IndexErrors

i = 0

for row in listSchedule:
    if row[0]=="Download":
        row.append(myOtherList[i])
        i+=1

answered Sep 20, 2018 at 13:23

C.Nivs

13.2k3 gold badges21 silver badges48 bronze badges

Comments

Sasha Tsukanov · Accepted Answer · 2018-09-20 14:20:24Z

0

We can use a queue and pop its values one by one when we meet the condition. To avoid copying of data let's implement the queue as a view over myOtherList using an iterator (thanks to ShadowRanger).

queue = iter(myOtherList)

for row in listSchedule:
    if row[0] == "Download":
        row.append(next(iter))

edited Sep 20, 2018 at 14:20

answered Sep 20, 2018 at 13:43

Sasha Tsukanov

1,1251 gold badge9 silver badges20 bronze badges

1 Comment

ShadowRanger Over a year ago

You could save the temporary deque with just queue = iter(myOtherList), and row.append(next(queue)). Assumes myOtherList isn't being otherwise modified in the loop, but otherwise gets the same behavior, with no overhead.

Collectives™ on Stack Overflow

pythonic way to add elements to nested list

4 Answers 4

1 Comment

Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related