Check if float is next to any float stored in the array

Question

I need to check if a given float is close, within a given tolerance, to any float in an array of floats.

import numpy as np # My float a = 0.27 # The tolerance t = 0.01 # Array of floats arr_f = np.arange(0.05, 0.75, 0.008) 

Is there an easy way to do this? Something like if a in arr_f: but given some tolerance in the difference?

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Add

By "allow tolerance" I mean this in the following sense:

 for i in arr_f: if abs(a - i) <= t: print 'float a is in arr_f within tolerance t' break 

Answer 1

How about using np.isclose ?

 >>> np.isclose(arr_f, a, atol=0.01).any() True 

np.isclose compares two objects by element to see if the values ​​are within a given tolerance (the atol keyword argument is given here). The function returns a boolean array.

Answer 2

If you are only interested in the True / False result, then this should work:

 In [1]: (abs(arr_f - a) < t).any() Out[1]: True 

Explanation: abs(arr_f - a) < t returns a boolean array for which any() is called to find out if its value is True .

EDIT . Comparing this approach and the one suggested in another answer shows that this bit is a bit faster:

 In [37]: arr_f = np.arange(0.05, 0.75, 0.008) In [38]: timeit (abs(arr_f - a) < t).any() 100000 loops, best of 3: 11.5 µs per loop In [39]: timeit np.isclose(arr_f, a, atol=t).any() 10000 loops, best of 3: 44.7 µs per loop In [40]: arr_f = np.arange(0.05, 1000000, 0.008) In [41]: timeit (abs(arr_f - a) < t).any() 1 loops, best of 3: 646 ms per loop In [42]: timeit np.isclose(arr_f, a, atol=t).any() 1 loops, best of 3: 802 ms per loop 

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An alternative solution that also returns the corresponding indexes is as follows:

 In [5]: np.where(abs(arr_f - a) < t)[0] Out[5]: array([27, 28]) 

This means that the values ​​located at indices 27 and 28 arr_f are within the required range, and indeed:

 In [9]: arr_f[27] Out[9]: 0.26600000000000001 In [10]: arr_f[28] Out[10]: 0.27400000000000002 

Using this approach, you can also get the result True / False :

 In [11]: np.where(abs(arr_f - a) < t)[0].any() Out[11]: True