Doing columns with a for loop on a nested list

Question

I'm trying to get:

 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
31   32   33   34   35   36   37   38   39   40
41   42   43   44   45   46   47   48   49   50
51   52   53   54   55   56   57   58   59   60
61   62   63   64   65   66   67   68   69   70
71   72   73   74   75   76   77   78   79   80
81   82   83   84   85   86   87   88   89   90
91   92   93   94   95   96   97   98   99  100

from this:

# nums is a range object
nums = list(range(1, 101))
chunks = []
print(nums)
print()

for i in range(0, len(nums), 10):
    chunks.append(nums[i:i+10])

in Python.... I can't for the life of me solve it.

You want to print the numbers that way or have a list of lists containing 10 numbers each? — gaganso
– gaganso, Commented Jul 2, 2016 at 6:44

Natecat · Accepted Answer · 2016-07-02 07:14:23Z

1

You can just use nested loops to print chunks out like that. Here's an example:

for i in chunks:
    for j in i:
        print(j, end=" ")
    print()

edited Jul 2, 2016 at 7:14

answered Jul 2, 2016 at 6:48

Natecat

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Comments

niemmi · Accepted Answer · 2016-07-02 07:30:59Z

1

This should produce exactly the same output as in the question:

for chunk in chunks:
    print(''.join(str(x).rjust(5 if i else 2) for i, x in enumerate(chunk)))

Update: For every chunk it will first convert numbers to strings with str and then right justify them with rjust that uses space as default fill char. Since the first number on the row has width of 2 and rest of them have width of 5 enumerate is used to track the index so that correct argument can be passed to rjust. Enumerate returns tuples in form (index, item) and the index is then used in 5 if i else 2 to determine if number is first or not so width of 2 or 5 can be used respectively. Finally all the substrings are joined together for a row which is printed to screen.

edited Jul 2, 2016 at 7:30

answered Jul 2, 2016 at 6:57

niemmi

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5 Comments

rdmtinez Over a year ago

this works perfectly, except that I have no idea why

Sebastiaan Mannem Over a year ago

Well, enumerate(chunk) creates a new list containing an index and the element for every element in the enumerated list... for i,x in loops through the enumerated list str(x) creates a string from a integer rjust adds spaces so that the string is (5 or 2) large, where (5 or 2) depends on the index in the enumeration (element 0 adds 5, all the other add 2). dont know if its clearer, but understand the steps, add this up, and you get how it works in total...

niemmi Over a year ago

@RicardoMartinez Added a short explanation.

niemmi Over a year ago

@SebastiaanMannem: Almost correct but you got the param passed to rjust wrong, 5 if i else 2 results to 2 in case i==0 and 5 in all other cases.

Sebastiaan Mannem Over a year ago

@ricardoMartinez: Would it help you if one was to change the one-liner into a block of code (maybe even with addition of some lines of comment)?

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