CODEWITHSUNDEEP
pythonalgorithmlistsorting
nr.o
nr.o

Reputation: 165

Sort list in place using another list of index

given 

a = [1,4,5,3,2,6,0]
b = ['b','e','f','d','c','g','a']

order b in place, the expected order of b is available in the corresponding positional element of a.

output will be 

['a','b','c','d','e','f','g']

try for other similar input sets.

a = [4,0,1,3,2]
b = ['E','A','B','D','C']

I can get it done using a third list, even sorted() creates a third list, but the key is to sort b in place

print sorted(b,key=lambda bi : a[b.index(bi)])

core of the problem is how to prevent iterating over items in b that were already iterated.

Upvotes: 0

Views: 128

Answers (4)

M4rtini
M4rtini

Reputation: 13539

def sorter(a,b):
    for i in range(len(a)):
        while i != a[i]:
            ai = a[i]
            b[i], b[ai], a[i], a[ai] = b[ai], b[i], a[ai], a[i]
    return b

Upvotes: 1

John La Rooy
John La Rooy

Reputation: 304137

The key is to realise that the items in b aren't much use to the key function. You are interested in their counterparts in a. To do this inplace, means you can't just use zip to pair the items up. Here I use the default argument trick to get an iterator over a into the lambda function.

>>> a = [1,4,5,3,2,6,0]
>>> b = ['b','e','f','d','c','g','a']
>>> b.sort(key=lambda x, it=iter(a): next(it))
>>> b
['a', 'b', 'c', 'd', 'e', 'f', 'g']

Upvotes: 5

Alfe
Alfe

Reputation: 59416

Try this:

zip(*sorted(zip(a, b)))[1]

Should give:

('a', 'b', 'c', 'd', 'e', 'f', 'g')

Since during sorting the b itself appears to be empty (see my question about that), you can use that piece of code to do it in-place:

b.sort(key=lambda x, b=b[:]: a[b.index(x)])

This uses a copy of the b to search in during sorting. This is certainly not very good for performance, so don't blame me ;-)

Upvotes: 6

Abhishek Bansal
Abhishek Bansal

Reputation: 12715

Simple bubble sort:

for i in range( len(a) ):
    for j in range(len(a)-1-i):
         if (a[j] > a[j+1]):
             #swap a[j] & a[j+1]
             #swap b[j] & b[j+1]

Upvotes: 0

Related Questions