# All possible replacements of two lists?

## Question:

I have two lists, both of the same length:

```
a = [1,2,3]
b = [4,5,6]
```

How can I get all possible results, from iterating over `a`

and either choosing to replace it with the corresponding element from `b`

, or not doing so?

```
output[0] = [1,2,3] # no replacements
output[1] = [4,2,3] # first item was replaced
output[2] = [1,5,3] # second item was replaced
output[3] = [1,2,6] # third item was replaced
output[4] = [4,5,3] # first and second items were replaced
output[5] = [4,2,6] # first and third items were replaced
output[6] = [1,5,6] # second and third items were replaced
output[7] = [4,5,6] # all items were replaced
```

## Answers:

Each item may independently be replaced or left alone. This can be modeled by a bit being 1 or 0. If you consider each item to be a separate bit, then iterating over all of the possibilities can be mapped to iterating over all of the combinations of *n* bits.

In other words, iterate from 0 to 2^{n}-1 and look at the bit patterns.

```
n = len(a)
for i in range(2**n):
yield [a[j] if i & (1 << j) != 0 else b[j] for j in range(n)]
```

Breaking this down, `i & (1 << j) != 0`

checks if the *j*th bit of *i* is set. If it is, use `a[j]`

, otherwise `b[j]`

.

Result:

```
[1, 2, 3]
[4, 2, 3]
[1, 5, 3]
[4, 5, 3]
[1, 2, 6]
[4, 2, 6]
[1, 5, 6]
[4, 5, 6]
```

Creating 3 lists of two elements would not over-complicate the code at all. `zip`

can “flip the axes” of multiple lists trivially (making X sequences of Y elements into Y sequences of X elements), making it easy to use `itertools.product`

:

```
import itertools
a = [1,2,3]
b = [4,5,6]
# Unpacking result of zip(a, b) means you automatically pass
# (1, 4), (2, 5), (3, 6)
# as the arguments to itertools.product
output = list(itertools.product(*zip(a, b)))
print(*output, sep="n")
```

Which outputs:

```
(1, 2, 3)
(1, 2, 6)
(1, 5, 3)
(1, 5, 6)
(4, 2, 3)
(4, 2, 6)
(4, 5, 3)
(4, 5, 6)
```

Different ordering than your example output, but it’s the same set of possible replacements.

Okay this is similar to the other answers, but taking *a bit from both*. You can model your problem as finding all possible bits of a sequence of given length, and replacing only when there is 1, and otherwise not.

```
from itertools import product
a = [1,2,3]
b = [4,5,6]
## All binary combinations of length of a (or b)
combinations = product([0,1], repeat=len(a))
for combination in combinations:
y = []
for l, i in zip(zip(a,b),combination):
y.append(l[i])
print y
```

All combinations of bits are:

```
(0, 0, 0)
(0, 0, 1)
(0, 1, 0)
(0, 1, 1)
(1, 0, 0)
(1, 0, 1)
(1, 1, 0)
(1, 1, 1)
```

Which results in:

```
[1, 2, 3]
[1, 2, 6]
[1, 5, 3]
[1, 5, 6]
[4, 2, 3]
[4, 2, 6]
[4, 5, 3]
[4, 5, 6]
```