Python sort array by another positions array

Question

Assume I have two arrays, the first one containing int data, the second one containing positions

a = [11, 22, 44, 55]

b = [0, 1, 10, 11]

i.e. I want a[i] to be be moved to position b[i] for all i. If I haven't specified a position, then insert a -1

i.e

sorted_a = [11, 22,-1,-1,-1,-1,-1,-1,-1,-1, 44, 55]
            ^   ^                            ^   ^
            0   1                            10  11

Another example:

a = [int1, int2, int3]

b = [5, 3, 1]

sorted_a = [-1, int3, -1, int2, -1, int1]

Here's what I've tried:

def sort_array_by_second(a, b):

   sorted = []

   for e1 in a:
      sorted.appendAt(b[e1])

  return sorted

Which I've obviously messed up.

Yes, they are always the same length. I'm using 0-based indexing, also — Mark Kennedy
– Mark Kennedy, Commented Dec 12, 2013 at 0:12

Paulo Bu · Accepted Answer · 2013-12-12 00:32:20Z

7

Something like this:

res = [-1]*(max(b)+1)   # create a list of required size with only -1's

for i, v in zip(b, a):
    res[i] = v

The idea behind the algorithm:

Create the resulting list with a size capable of holding up to the largest index in b
Populate this list with -1
Iterate through b elements
Set elements in res[b[i]] with its proper value a[i]

This will leave the resulting list with -1 in every position other than the indexes contained in b, which will have their corresponding value of a.

edited Dec 12, 2013 at 0:32

answered Dec 11, 2013 at 23:12

Paulo Bu

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14 Comments

John La Rooy Over a year ago

I think you are just mean to derive the ordering from b, the values don't matter

John La Rooy Over a year ago

On second thoughts - maybe " I want a[i] to be be moved to position b[i] for all i " means you are right

Paulo Bu Over a year ago

I'm sorry I don't understand you quite well. The OP isn't very clear about the algorithm, I just tried to illustrate him one way to do it.

Mark Kennedy Over a year ago

Hi, thanks for the quick response. I get a IndexError: tuple index out of range

jfs Over a year ago

you could use for index, value in zip(b, a)

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qwwqwwq · Accepted Answer · 2013-12-11 23:13:22Z

1

I would use a custom key function as an argument to sort. This will sort the values according to the corresponding value in the other list:

to_be_sorted = ['int1', 'int2', 'int3', 'int4', 'int5']
sort_keys = [4, 5, 1, 2, 3]

sort_key_dict = dict(zip(to_be_sorted, sort_keys))

to_be_sorted.sort(key = lambda x: sort_key_dict[x])

This has the benefit of not counting on the values in sort_keys to be valid integer indexes, which is not a very stable thing to bank on.

answered Dec 11, 2013 at 23:13

qwwqwwq

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John La Rooy · Accepted Answer · 2013-12-11 23:19:18Z

1

>>> a = ["int1", "int2", "int3", "int4", "int5"]
>>> b = [4, 5, 1, 2, 3]
>>> sorted(a, key=lambda x, it=iter(sorted(b)): b.index(next(it)))
['int4', 'int5', 'int1', 'int2', 'int3']

answered Dec 11, 2013 at 23:19

John La Rooy

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Comments

GMPrazzoli · Accepted Answer · 2013-12-11 23:19:34Z

1

Paulo Bu answer is the best pythonic way. If you want to stick with a function like yours:

def sort_array_by_second(a, b):
   sorted = []
   for n in b:
      sorted.append(a[n-1]) 
  return sorted

will do the trick.

answered Dec 11, 2013 at 23:19

GMPrazzoli

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Comments

koffein · Accepted Answer · 2013-12-11 23:25:03Z

1

Sorts A by the values of B:

A = ['int1', 'int2', 'int3', 'int4', 'int5']
B = [4, 5, 1, 2, 3]

from operator import itemgetter
C = [a for a, b in sorted(zip(A, B), key = itemgetter(1))]

print C

Output

['int3', 'int4', 'int5', 'int1', 'int2']

answered Dec 11, 2013 at 23:25

koffein

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jfs · Accepted Answer · 2013-12-12 00:27:30Z

1

a = [11, 22, 44, 55] # values
b = [0, 1, 10, 11]  # indexes to sort by

sorted_a = [-1] * (max(b) + 1)
for index, value in zip(b, a):
    sorted_a[index] = value

print(sorted_a)
# -> [11, 22, -1, -1, -1, -1, -1, -1, -1, -1, 44, 55]

answered Dec 12, 2013 at 0:27

jfs

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