cᴏʟᴅsᴘᴇᴇᴅ - 30 days ago 9

Python Question

As a challenge, I've given myself this problem:

Given 2 lists, A, and B, where B is a shuffled version of A, the idea is to figure out the shuffled indices.

For example:

`A = [10, 40, 30, 2]`

B = [30, 2, 10, 40]

result = [2, 3, 0, 1]

A[2] A[3] A[0] A[1]

|| || || ||

30 2 10 40

Note that ties for identical elements can be resolved arbitrarily.

I've come up with a solution that involves the use of a dictionary to store indices. What other possible solutions does this problem have? A solution using a library also works. Numpy, pandas, anything is fine.

Answer Source

As an improvement over your current solution, you could use `collections.defaultdict`

and avoid `dict.setdefault`

:

```
from collections import defaultdict
A = [10, 40, 30, 2]
B = [30, 2, 10, 40]
idx = defaultdict(list)
for i, l in enumerate(A):
idx[l].append(i)
res = [idx[l].pop() for l in B]
print(res)
```

Here are the timings for the two methods using the sample input given:

**Script used for testing**

```
from timeit import timeit
setup = """
from collections import defaultdict;
idx1 = defaultdict(list); idx2 = {}
A = [10, 40, 30, 2]
B = [30, 2, 10, 40]
"""
me = """
for i, l in enumerate(A):
idx1[l].append(i)
res = [idx1[l].pop() for l in B]
"""
coldspeed = """
for i, l in enumerate(A):
idx2.setdefault(l, []).append(i)
res = [idx2[l].pop() for l in B]
"""
print(timeit(setup=setup, stmt=me))
print(timeit(setup=setup, stmt=coldspeed))
```

**Results**

```
original: 2.601998388010543
modified: 2.0607256239745766
```

So it appears that using `defaultdict`

actually yields a slight speed increase. This actually makes since though since `defaultdict`

is implemented in C rather than Python. Not to mention that the attribute lookup of the original solution - `idx.setdefault1`

- is costly.