numpy如何迭代数组的列？

Suppose I have and m x n array. I want to pass each column of this array to a function to perform some operation on the entire column. How do I iterate over the columns of the array?

For example, I have a 4 x 3 array like

1  99 2
2  14 5
3  12 7
4  43 1

for column in array:
  some_function(column)

where column would be “1,2,3,4” in the first iteration, “99,14,12,43” in the second, and “2,5,7,1” in the third.

Just iterate over the transposed of your array:

for column in array.T:
   some_function(column)

This should give you a start

>>> for col in range(arr.shape[1]):
    some_function(arr[:,col])


[1 2 3 4]
[99 14 12 43]
[2 5 7 1]

For a three dimensional array you could try:

for c in array.transpose(1, 0, 2):
    do_stuff(c)

See the docs on how array.transpose works. Basically you are specifying which dimension to shift. In this case we are shifting the second dimension (e.g. columns) to the first dimension.

for c in np.hsplit(array, array.shape[1]):
    some_fun(c)

You can also use unzip to iterate through the columns

for col in zip(*array):
   some_function(col)

For example you want to find a mean of each column in matrix. Let’s create the following matrix

mat2 = np.array([1,5,6,7,3,0,3,5,9,10,8,0], dtype=np.float64).reshape(3, 4)

The function for mean is

def my_mean(x):
    return sum(x)/len(x)

To do what is needed and store result in colon vector ‘results’

results = np.zeros(4)
for i in range(0, 4):
    mat2[:, i] = my_mean(mat2[:, i])

results = mat2[1,:]      

The results are: array([4.33333333, 5. , 5.66666667, 4. ])

Alternatively, you can use enumerate. It gives you the column number and the column values as well.

for num, column in enumerate(array.T):
    some_function(column) # column: Gives you the column value as asked in the question
    some_function(num) # num: Gives you the column number