Here's my array:
import numpy as np
a = np.array([0, 5.0, 0, 5.0, 5.0])
Is it possible to use numpy.where in some way to add a value x to all those entires in a that are less than l?
So something like:
a = a[np.where(a < 5).add(2.5)]
Should return:
array([2.5, 5.0, 2.5, 5.0, 5.0])
a = np.array([0., 5., 0., 5., 5.])
a[np.where(a < 5)] += 2.5
in case you really want to use where or just
a[a < 5] += 2.5
which I usually use for these kind of operations.
You could use np.where to create the array of additions and then simply add to a -
a + np.where(a < l, 2.5,0)
Sample run -
In [16]: a = np.array([1, 5, 4, 5, 5])
In [17]: l = 5
In [18]: a + np.where(a < l, 2.5,0)
Out[18]: array([ 3.5, 5. , 6.5, 5. , 5. ])
int += float) to allow in-place addition.Given that you probably need to change the dtype (from int to float) you need to create a new array. A simple way without explicit .astype or np.where calls is multiplication with a mask:
>>> b = a + (a < 5) * 2.5
>>> b
array([ 2.5, 5. , 2.5, 5. , 5. ])
with np.where this can be changed to a simple expression (using the else-condition, third argument, in where):
>>> a = np.where(a < 5, a + 2.5, a)
>>> a
array([ 2.5, 5. , 2.5, 5. , 5. ])
a += np.where(a < 1, 2.5, 0)
where will return the second argument wherever the condition (first argument) is satisfied and the third argument otherwise.
You can use a "masked array" as an index. Boolean operations, such as a < 1 return such an array.
>>> a<1
array([False, False, False, False, False], dtype=bool)
you can use it as
>>> a[a<1] += 1
The a<1 part selects only the items in a that match the condition. You can operate on this part only then.
If you want to keep a trace of your selection, you can proceed in two steps.
>>> mask = a>1
>>> a[mask] += 1
Also, you can count the items matching the conditions:
>>> print np.sum(mask)
