I have a list, say
a = [3, 4, 5, 6, 7]
And i want to create a numpy array of zeros of that list's length.
If I do
b = np.zeros((len(a), 1))
I get
[[0, 0, 0, 0, 0]]
instead of
[0, 0, 0, 0, 0]
What is the best way to get the latter option?
3 Answers
If you don't want to have to care about shapes, use np.zeros_like:
np.zeros_like(a)
# array([0, 0, 0, 0, 0])
There's also the option of querying np.shape:
np.zeros(np.shape(a))
# array([0., 0., 0., 0., 0.])
Both options should work for ND lists as well.
2 Comments
Gulzar
What do you mean by "don't want to care about shapes"? Why are the options different?
cs95
@Gulzar from your code, it seems like your grief was mostly because you could not initialise
np.zeros with the correct shape. So my first option was to give you something that did not need explicit dealing with the shape at all—let numpy handle that.You passed two-element tuple to zeros, so it produced 2D array, you can simply pass integer to zeros
a = [3, 4, 5, 6, 7]
b = np.zeros(len(a))
print(b) #prints [ 0. 0. 0. 0. 0.]
Comments
You can try this
np.zeros(len(a), dtype=np.int)
It will return
array([0, 0, 0, 0, 0])
b = np.zeros((len(a), 1))[0]