BlueMoon93 - 1 year ago 67

Python Question

Given maximum a list

`l`

`r`

`N`

`binaryL`

`r*N`

`l`

Example, for

`N=2`

`l = [1, 0, 3] --> [01, 00, 11] (in bits)`

becomes

`binaryL = [0, 0, 1, 1, 0, 1]`

where each group of

`r`

`0, 0, 1`

`l`

`1, 0, 1`

`l`

Another option is to just obtain the bits in their order, where

`binaryL = [0, 1, 0, 0, 1, 1]`

In this case, each value is converted to its bits.

For those wondering about performance,

`import random`

from itertools import chain

import time

N=8

l=[random.randrange(1,2**N,1) for _ in range (10000000)]

r=len(l)

a = time.clock()

res1 = []

for i in l:

res1 += [int(b) for b in "{0:b}".format(i).rjust(N, '0')]

b = time.clock()

res2 = list(map(int, chain.from_iterable(bin(i)[2:].zfill(N) for i in l)))

c = time.clock()

res3 = list(map(int, ''.join(bin(i)[2:].zfill(N) for i in l)))

d = time.clock()

res4 = [0] * N * r

for ind, binary in enumerate(map(bin, l)):

for ind_bit, bit in enumerate(binary[2:].zfill(N)):

res4[r * ind_bit + ind] = int(bit)

e = time.clock()

res5 = list(map(int, chain.from_iterable(zip(*[bin(i)[2:].zfill(N) for i in l]))))

f = time.clock()

# res1, res2 and res3 are show bits by value. res4 and res5 shows bits by index

print(res1==res2)

print(res2==res3)

print(res4==res5)

print(b-a)

print(c-b)

print(d-c)

print(e-d)

print(f-e)

prints for 1000 values:

`True`

True

True

0.003963000000000001 # neverwalkaloner

0.0025400000000000006 # Psidom1

0.0023320000000000007 # Psidom2

0.004358000000000001 # Rockybilly

0.0021629999999999983 # Psidom3

and for 10.000.000 values

`True`

True

True

36.333539 # neverwalkaloner

25.674224000000002 # Psidom1

24.49611499999999 # Psidom2

47.370771000000005 # Rockybilly

66.25204 # Psidom3

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Answer Source

Use `bin`

to convert integer to binary representation, and pad the result to specified length with `zfill()`

, flatten the list with `chain`

and convert the string to int with `map`

:

```
from itertools import chain
l = [1, 0, 3]
N = max(l).bit_length() # as commented by @Jon, use this to determine the maximum
# bit length
list(map(int, chain.from_iterable(bin(i)[2:].zfill(N) for i in l)))
# [0, 1, 0, 0, 1, 1]
```

Without using `chain`

, another option could be:

```
list(map(int, ''.join(bin(i)[2:].zfill(N) for i in l)))
# [0, 1, 0, 0, 1, 1]
```

A transposed version:

```
list(map(int, chain.from_iterable(zip(*[bin(i)[2:].zfill(N) for i in l]))))
# [0, 0, 1, 1, 0, 1]
```

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