2

How can i create a multidimensional list while iterating through a 1d list based upon some condition.

I am iterating over a 1d list and whenever i find a '\n' i should append the so created list with a new list, for example,

 a = [1,2,3,4,5,'\n',6,7,8,9,0,'\n',3,45,6,7,2]

so I want it to be as,

new_list = [[1,2,3,4],[6,7,8,9,0],[3,45,6,7,2]]

how should i do it? Please help

def storeData(k):
    global dataList
    dlist = []
    for y in k:
        if y != '\n':
            dlist.append(y)
        else:
            break
    return dlist

This is what i have tried.

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6
  • Given the input is an empty list. What is the expected result? Commented Jan 26, 2017 at 18:58
  • You should probably change whatever code produced this bizarre input list. Post-processing it is the wrong way to go. Commented Jan 26, 2017 at 18:59
  • @WillemVanOnsem I would assume an empty list, but better if the OP says. Commented Jan 26, 2017 at 18:59
  • i just simply want to convert the 'a' list into the 'new_list' format Commented Jan 26, 2017 at 19:00
  • Your code has a parameter k and a global datalist. Not sure what the global is for - delete it. You want a list of lists but are only using 1 list dlist. Seems like you you need to build a sublist, then add the sublist to dlist whenever you see a '\n'. Commented Jan 26, 2017 at 19:04

4 Answers 4

Reset to default
3

example code:

lst = [[]]
for x in a:
    if x != '\n':
        lst[-1].append(x)
    else:
        lst.append([])
print(lst)

output:

[[1, 2, 3, 4, 5], [6, 7, 8, 9, 0], [3, 45, 6, 7, 2]]
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Comments

1

Using itertools.groupby would do the job (grouping by not being a linefeed):

import itertools

a = [1,2,3,4,5,'\n',6,7,8,9,0,'\n',3,45,6,7,2]

new_list = [list(x) for k,x in itertools.groupby(a,key=lambda x : x!='\n') if k]

print(new_list)

We compare the key truth value to filter out the occurrences of \n

result:

[[1, 2, 3, 4, 5], [6, 7, 8, 9, 0], [3, 45, 6, 7, 2]]
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2 Comments

You could use the currently-ignored key values from groupby to eliminate the second pass.
@user2357112 fixed. It's much better, you're right!
0

This is how I did it but there must be better solutions. 

x = 0
output = [[]]
for item in a:
    if item != "\n":
        output[x].append(item)
    else:
        x += 1
        output.append([])

print(output)
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Comments

0

Here is the basic approach:

EDIT: Good heavens! Silly bug... here's a fix:

>>> sub = []
>>> final = []
>>> for e in a:
...     if e == '\n':
...         final.append(sub)
...         sub = []
...     else:
...         sub.append(e)
... else:
...     final.append(sub)
...
>>> final
[[1, 2, 3, 4, 5], [6, 7, 8, 9, 0], [3, 45, 6, 7, 2]]
>>>

There are other ways, but this is the naive imperative way.

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4 Comments

OPs question and desired result don't quite line up. They want [[1,2,3,4],[6,7,8,9,0],[3,45,6,7,2]] not [[1, 2, 3, 4, 5], [6, 7, 8, 9, 0]]. You need a condition to add sub to the end of final if you run out of items in a without seeing another '\n'.
@Fruitspunchsamurai yes, yes. My mistake.
Why do final.append(sub) in an else? And the ... btw make this really cumbersome to test...
@StefanPochmann My use of else is a personal preference. To my mind, it helps my eyes look at the code as a "logical unit", so that I know that whatever I'm doing in the else is somehow cleaning up whatever is going on in my for-loop. It is obviously not necessary.

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