numpy.std(a, axis=None, dtype=None, out=None, ddof=0, keepdims=<no value>)
[source]
Compute the standard deviation along the specified axis.
Returns the standard deviation, a measure of the spread of a distribution, of the array elements. The standard deviation is computed for the flattened array by default, otherwise over the specified axis.
Parameters: 


Returns: 

The standard deviation is the square root of the average of the squared deviations from the mean, i.e., std = sqrt(mean(abs(x  x.mean())**2))
.
The average squared deviation is normally calculated as x.sum() / N
, where N = len(x)
. If, however, ddof
is specified, the divisor N  ddof
is used instead. In standard statistical practice, ddof=1
provides an unbiased estimator of the variance of the infinite population. ddof=0
provides a maximum likelihood estimate of the variance for normally distributed variables. The standard deviation computed in this function is the square root of the estimated variance, so even with ddof=1
, it will not be an unbiased estimate of the standard deviation per se.
Note that, for complex numbers, std
takes the absolute value before squaring, so that the result is always real and nonnegative.
For floatingpoint input, the std is computed using the same precision the input has. Depending on the input data, this can cause the results to be inaccurate, especially for float32 (see example below). Specifying a higheraccuracy accumulator using the dtype
keyword can alleviate this issue.
>>> a = np.array([[1, 2], [3, 4]]) >>> np.std(a) 1.1180339887498949 # may vary >>> np.std(a, axis=0) array([1., 1.]) >>> np.std(a, axis=1) array([0.5, 0.5])
In single precision, std() can be inaccurate:
>>> a = np.zeros((2, 512*512), dtype=np.float32) >>> a[0, :] = 1.0 >>> a[1, :] = 0.1 >>> np.std(a) 0.45000005
Computing the standard deviation in float64 is more accurate:
>>> np.std(a, dtype=np.float64) 0.44999999925494177 # may vary
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https://docs.scipy.org/doc/numpy1.17.0/reference/generated/numpy.std.html