I want to create a numpy array.
T = 200
I want to create an array from 0 to 199, in which each value will be divided by 200.
l = [0, 1/200, 2/200, ...]
Numpy have any such method for calculation?
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4 Answers
Alternatively one can use linspace:
>>> np.linspace(0, 1., 200, endpoint=False)
array([ 0. , 0.005, 0.01 , 0.015, 0.02 , 0.025, 0.03 , 0.035,
0.04 , 0.045, 0.05 , 0.055, 0.06 , 0.065, 0.07 , 0.075,
...
0.92 , 0.925, 0.93 , 0.935, 0.94 , 0.945, 0.95 , 0.955,
0.96 , 0.965, 0.97 , 0.975, 0.98 , 0.985, 0.99 , 0.995])
Comments
Use np.arange:
>>> import numpy as np
>>> np.arange(200, dtype=np.float)/200
array([ 0. , 0.005, 0.01 , 0.015, 0.02 , 0.025, 0.03 , 0.035,
0.04 , 0.045, 0.05 , 0.055, 0.06 , 0.065, 0.07 , 0.075,
0.08 , 0.085, 0.09 , 0.095, 0.1 , 0.105, 0.11 , 0.115,
...
0.88 , 0.885, 0.89 , 0.895, 0.9 , 0.905, 0.91 , 0.915,
0.92 , 0.925, 0.93 , 0.935, 0.94 , 0.945, 0.95 , 0.955,
0.96 , 0.965, 0.97 , 0.975, 0.98 , 0.985, 0.99 , 0.995])
answered Jan 13, 2014 at 10:27
Ashwini Chaudhary
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Comments
T = 200.0
l = [x / float(T) for x in range(200)]
Comments
import numpy as np
T = 200
np.linspace(0.0, 1.0 - 1.0 / float(T), T)
Personally I prefer linspace for creating evenly spaced arrays in general. It is more complex in this case as the endpoint depends on the number of points T.
1 Comment
Warren Weckesser
If you have a recent enough numpy, you can use the
endpoint argument of linspace. See @alko's answer.