I am trying to implement the following simple condition with numpy arrays, but the output is wrong.
dt = 1.0
t = np.arange(0.0, 5.0, dt)
x = np.empty_like(t)
if np.where((t >=0) & (t < 3)):
x = 2*t
else:
x=4*t
I get the output below
array([0., 2., 4., 6., 8.])
But I am expecting
array([0., 2., 4., 12., 16.])
Thanks for your help!
Looking in the docs for np.where:
Note: When only condition is provided, this function is a shorthand for
np.asarray(condition).nonzero(). Usingnonzerodirectly should be preferred, as it behaves correctly for subclasses. The rest of this documentation covers only the case where all three arguments are provided.
Since you don't provide the x and y arguments, where acts like nonzero.
nonzero returns a tuple of np.arrays, which is truthy when converted to bool. So your code ends up evaluating as:
if True:
x = 2*t
Instead, you want to use:
x = np.where((t >= 0) & (t < 3), 2*t, 4*t)
The usage of np.where is different
dt = 1.0
t = np.arange(0.0, 5.0, dt)
x = np.empty_like(t)
x = np.where((t >= 0) & (t < 3), 2*t, 4*t)
x
Output
[ 0., 2., 4., 12., 16.]
in your code the if statement is not necessary and causes the problem.
np.where() creates the condition therefore you do not need the if statement.
Here is a working example of your code with the output you want
dt = 1.0
t = np.arange(0.0, 5.0, dt)
x = np.empty_like(t)
np.where((t >=0) & (t < 3),2*t,4*t)
dt = 1.0 t = np.arange(0.0, 4.0, dt) x = np.empty_like(t) if ((t>=0.0).any()) & ((t<1.0).any()): x = 2*t elif ((t>=1.0).any()) & ((t<2.0).any()): x = 3*t elif ((t>=2.0).any()) & ((t<3.0).any()): x = 4*t else: x = 5*t 
dt = 1.0 t = np.arange(0.0, 5.0, dt) x = np.empty_like(t) x = np.where((t >=0) & (t < 1),2*t,np.where((t >=1) & (t < 2),3*t,np.where((t >=2) & (t < 3),4*t,5*t)))