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Creating numpy array with different steps

I would like to create an array like the following:

# 0 | 4 | 8 | 16 | 32

In which each element, except the first, is a double previous one. I can create this less through iteration, etc.

However, since Python provides many one-line functions, I was wondering if I have a way to do this.

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There may be one line, but this is more explicit:

x = np.multiply.accumulate( np.ones( 10 )*2)
x[0] = 0


OR

x  = 2**np.arange(1,10)
x[0] = 0
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import numpy as np

In this case, you can use a list comprehension to evaluate your strength function ( 2^n

), then generate numpy.array

.



>>> np.array([0] + [2**i for i in range(2, 10)])
array([  0,   4,   8,  16,  32,  64, 128, 256, 512])
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You can use numpy.logspace

to get ranges at log spacing. Use a keyword argument base=N

to set the exponent base:

In [27]: np.logspace(0, 10, 11, base=2).astype(int)
Out[27]: array([   1,    2,    4,    8,   16,   32,   64,  128,  256,  512, 1024])

I like this method because the name of the function "logspace" makes it clear that I am going to use a range with a log interval (as opposed to linear).

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