Is there a less verbose alternative to this:
for x in xrange(array.shape[0]):
for y in xrange(array.shape[1]):
do_stuff(x, y)
I came up with this:
for x, y in itertools.product(map(xrange, array.shape)):
do_stuff(x, y)
Which saves one indentation, but is still pretty ugly.
I'm hoping for something that looks like this pseudocode:
for x, y in array.indices:
do_stuff(x, y)
Does anything like that exist?
I think you're looking for the ndenumerate.
>>> a =numpy.array([[1,2],[3,4],[5,6]])
>>> for (x,y), value in numpy.ndenumerate(a):
... print x,y
...
0 0
0 1
1 0
1 1
2 0
2 1
Regarding the performance. It is a bit slower than a list comprehension.
X = np.zeros((100, 100, 100))
%timeit list([((i,j,k), X[i,j,k]) for i in range(X.shape[0]) for j in range(X.shape[1]) for k in range(X.shape[2])])
1 loop, best of 3: 376 ms per loop
%timeit list(np.ndenumerate(X))
1 loop, best of 3: 570 ms per loop
If you are worried about the performance you could optimise a bit further by looking at the implementation of ndenumerate, which does 2 things, converting to an array and looping. If you know you have an array, you can call the .coords attribute of the flat iterator.
a = X.flat
%timeit list([(a.coords, x) for x in a.flat])
1 loop, best of 3: 305 ms per loop
If you only need the indices, you could try numpy.ndindex:
>>> a = numpy.arange(9).reshape(3, 3)
>>> [(x, y) for x, y in numpy.ndindex(a.shape)]
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
see nditer
import numpy as np
Y = np.array([3,4,5,6])
for y in np.nditer(Y, op_flags=['readwrite']):
y += 3
Y == np.array([6, 7, 8, 9])
y = 3would not work, usey *= 0andy += 3instead.
I see that no good desciption for using numpy.nditer() is here. So, I am gonna go with one. According to NumPy v1.21 dev0 manual, The iterator object nditer, introduced in NumPy 1.6, provides many flexible ways to visit all the elements of one or more arrays in a systematic fashion.
I have to calculate mean_squared_error and I have already calculate y_predicted and I have y_actual from the boston dataset, available with sklearn.
def cal_mse(y_actual, y_predicted):
""" this function will return mean squared error
args:
y_actual (ndarray): np array containing target variable
y_predicted (ndarray): np array containing predictions from DecisionTreeRegressor
returns:
mse (integer)
"""
sq_error = 0
for i in np.nditer(np.arange(y_pred.shape[0])):
sq_error += (y_actual[i] - y_predicted[i])**2
mse = 1/y_actual.shape[0] * sq_error
return mse
Hope this helps :). for further explaination visit
for x, y in itertools.product(*map(xrange, array.shape)):