Numpy 'Logical_Or' For More Than Two Arguments

Fastest way to compare Numpy ndarrays

I think you want this:

np.logical_or.reduce([prices >= x for prices in dictPrices.values()])

This is explained in some detail here: Numpy `logical_or` for more than two arguments

And of course for the second case you can use logical_and instead of logical_or.

Logical OR without using numpy.logical_or

You don't need the subtraction.
The point is that + already behaves like the or operator

>>(a%2==0)+(a<=6)
array([[ True,  True,  True],
       [ True,  True,  True]], dtype=bool)

because "True+True=True".

When you subtract (a<=6)*(a%2==0) you turn all elements which satisfy both conditions into false.

It is easiest when you just do

>>(a<=6)|(a%2==0)
array([[ True,  True,  True],
       [ True,  True,  True]], dtype=bool)

Slice numpy array based on values of other array using more than 2 arguments

You could slice using np.in1d():

In [15]: b[np.in1d(a, my_list)]
Out[15]: array([1, 3, 4, 5])

Multiple conditions using 'or' in numpy array

If numpy overloads & for boolean and you can safely assume that | is boolean or.

area1 = N.where(((A>0) & (A<10)) | ((A>40) & (A<60))),1,0)

Finding Lowest Common Multiple using numpy (for more than two inputs)

You'd use np.lcm.reduce(), and pass it an array of numbers:

>>> np.lcm.reduce([1, 2, 3, 4])
12

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