You can use `random.uniform`

`import random random.uniform(0, 1) `

ID : 20389

viewed : 28

90

You can use `random.uniform`

`import random random.uniform(0, 1) `

82

`random.random()`

does exactly that

`>>> import random >>> for i in range(10): ... print(random.random()) ... 0.908047338626 0.0199900075962 0.904058545833 0.321508119045 0.657086320195 0.714084413092 0.315924955063 0.696965958019 0.93824013683 0.484207425759 `

If you want *really* random numbers, and to cover the range [0, 1]:

`>>> import os >>> int.from_bytes(os.urandom(8), byteorder="big") / ((1 << 64) - 1) 0.7409674234050893 `

70

I want a random number between 0 and 1, like 0.3452

`random.random()`

is what you are looking for:

random.random() Return the next random floating point number in the range [0.0, 1.0).From python docs:

And, btw, ** Why your try didn't work?**:

Your try was: `random.randrange(0, 1)`

random.randrange() Return a randomly selected element from range(start, stop, step). This is equivalent to choice(range(start, stop, step)), but doesnâ€™t actually build a range object.From python docs:

So, what you are doing here, with `random.randrange(a,b)`

is choosing a random element from `range(a,b)`

; in your case, from `range(0,1)`

, but, guess what!: the only element in `range(0,1)`

, is `0`

, so, the only element you can choose from `range(0,1)`

, is `0`

; that's *why* you were always getting `0`

back.

68

you can use use numpy.random module, you can get array of random number in shape of your choice you want

`>>> import numpy as np >>> np.random.random(1)[0] 0.17425892129128229 >>> np.random.random((3,2)) array([[ 0.7978787 , 0.9784473 ], [ 0.49214277, 0.06749958], [ 0.12944254, 0.80929816]]) >>> np.random.random((3,1)) array([[ 0.86725993], [ 0.36869585], [ 0.2601249 ]]) >>> np.random.random((4,1)) array([[ 0.87161403], [ 0.41976921], [ 0.35714702], [ 0.31166808]]) >>> np.random.random_sample() 0.47108547995356098 `

58

random.randrange(0,2) this works!