# What does numpy.random.seed(0) do?

`np.random.seed(0)`

makes the random numbers predictable

`>>> numpy.random.seed(0) ; numpy.random.rand(4)array([ 0.55, 0.72, 0.6 , 0.54])>>> numpy.random.seed(0) ; numpy.random.rand(4)array([ 0.55, 0.72, 0.6 , 0.54])`

With the seed reset (every time), the *same* set of numbers will appear every time.

If the random seed is not reset, *different* numbers appear with every invocation:

`>>> numpy.random.rand(4)array([ 0.42, 0.65, 0.44, 0.89])>>> numpy.random.rand(4)array([ 0.96, 0.38, 0.79, 0.53])`

(pseudo-)random numbers work by starting with a number (the seed), multiplying it by a large number, adding an offset, then taking modulo of that sum. The resulting number is then used as the seed to generate the next "random" number. When you set the seed (every time), it does the same thing every time, giving you the same numbers.

If you want seemingly random numbers, do not set the seed. If you have code that uses random numbers that you want to debug, however, it can be very helpful to set the seed before each run so that the code does the same thing every time you run it.

To get the most random numbers for each run, call `numpy.random.seed()`

. This will cause numpy to set the seed to a random number obtained from `/dev/urandom`

or its Windows analog or, if neither of those is available, it will use the clock.

For more information on using seeds to generate pseudo-random numbers, see wikipedia.

If you set the `np.random.seed(a_fixed_number)`

every time you call the numpy's other random function, the result will be the same:

`>>> import numpy as np>>> np.random.seed(0) >>> perm = np.random.permutation(10) >>> print perm [2 8 4 9 1 6 7 3 0 5]>>> np.random.seed(0) >>> print np.random.permutation(10) [2 8 4 9 1 6 7 3 0 5]>>> np.random.seed(0) >>> print np.random.permutation(10) [2 8 4 9 1 6 7 3 0 5]>>> np.random.seed(0) >>> print np.random.permutation(10) [2 8 4 9 1 6 7 3 0 5]>>> np.random.seed(0) >>> print np.random.rand(4) [0.5488135 0.71518937 0.60276338 0.54488318]>>> np.random.seed(0) >>> print np.random.rand(4) [0.5488135 0.71518937 0.60276338 0.54488318]`

However, if you just call it once and use various random functions, the results will still be different:

`>>> import numpy as np>>> np.random.seed(0) >>> perm = np.random.permutation(10)>>> print perm [2 8 4 9 1 6 7 3 0 5]>>> np.random.seed(0) >>> print np.random.permutation(10)[2 8 4 9 1 6 7 3 0 5]>>> print np.random.permutation(10) [3 5 1 2 9 8 0 6 7 4]>>> print np.random.permutation(10) [2 3 8 4 5 1 0 6 9 7]>>> print np.random.rand(4) [0.64817187 0.36824154 0.95715516 0.14035078]>>> print np.random.rand(4) [0.87008726 0.47360805 0.80091075 0.52047748]`

As noted, numpy.random.seed(0) sets the random seed to 0, so the pseudo random numbers you get from random will start from the same point. This can be good for debuging in some cases. HOWEVER, after some reading, this seems to be the wrong way to go at it, if you have threads because it is not thread safe.

from differences-between-numpy-random-and-random-random-in-python:

For numpy.random.seed(), the main difficulty is that it is notthread-safe - that is, it's not safe to use if you have many differentthreads of execution, because it's not guaranteed to work if twodifferent threads are executing the function at the same time. Ifyou're not using threads, and if you can reasonably expect that youwon't need to rewrite your program this way in the future,numpy.random.seed() should be fine for testing purposes. If there'sany reason to suspect that you may need threads in the future, it'smuch safer in the long run to do as suggested, and to make a localinstance of the numpy.random.Random class. As far as I can tell,random.random.seed() is thread-safe (or at least, I haven't found anyevidence to the contrary).

example of how to go about this:

`from numpy.random import RandomStateprng = RandomState()print prng.permutation(10)prng = RandomState()print prng.permutation(10)prng = RandomState(42)print prng.permutation(10)prng = RandomState(42)print prng.permutation(10)`

may give:

[3 0 4 6 8 2 1 9 7 5]

[1 6 9 0 2 7 8 3 5 4]

[8 1 5 0 7 2 9 4 3 6]

[8 1 5 0 7 2 9 4 3 6]

Lastly, note that there might be cases where initializing to 0 (as opposed to a seed that has not all bits 0) may result to non-uniform distributions for some few first iterations because of the way xor works, but this depends on the algorithm, and is beyond my current worries and the scope of this question.