排序2D numpy数组的部分

Question

So I have a numpy array A of dimensions (8760,12). Basically all the hours of 12 years. I need to sort each month (730 hours) in each year in the array. I haven't found any way to do it inside the array. So my solution was to take out each month, sort it and then create the entire 2d array again. I was thinking of doing something along the lines of what I have below, but it isn't working.

total=np.zeroes([8760,12])
for j in range(1,12):
    for i in range (1,12):
        #here i take out every month of every year
        month=A[730*(i-1):-730*(12-i),(j-1):-(12-j)]
        #here I sort the data
        month_sorted=np.sort(month,axis=0,kind='quicksort')
        #here I try to add the sorted months back into 1 big array
        np.concatenate(total,month_sorted,axis=0)
    np.concatenate(total,month_sorted,axis=1)

Concatenate doesn't work on arrays of different sizes.

And I don't really have a way to place the month of year 2 in row 2 of my array. I guess it should be done with indexing idx or iloc or something like that.

EDIT: My values are integers.

The result should be values ordered from low to high for each 730(hours in a month) values per row. So imagine I would have 3 years instead of 12 and 9 hours instead of 8760 hours which have to be sorted each 3 hours instead of each 730 hours. The array looks like this :

[[30,40,10,20,50,60,80,200,100]
[8,20,5,6,8,1,5,3,2]
[520,840,600,525,430,20,1,506,703]]

And should be converted into :

[[10,30,40,20,50,60,80,100,200]
[5,8,20,1,6,8,2,3,5]
[520,600,840,20,430,525,1,506,703]]

So my current code take out the first part 30,40,10 and sorts it as 10,30,40. But the part that I can't solve is how to create the big array again from all the smaller ones in the 2 loops.

Answer 1

You can use python indexes and assignment instead of concatenate if you create the empty array first.

A = np.random.randint(0,99,(8760,12))
total=np.zeros([8760,12])
for j in range(12):
    for i in range (12):
        total[730*i:730*(i+1),j] = np.sort(A[730*i:730*(i+1),j])

If you want the same thing staring from no array and using concatenate-like function i would do it like this

total2=None
for j in range(12):
    app1 = None
    for i in range (12):
        app = np.sort(A[730*i:730*(i+1),j])
        if app1 is None:
            app1 = app
        else:
            app1 = np.hstack((app1,app))
    if total2 is None:
        total2 = app1
    else:
        total2 = np.vstack((total2,app1))
total2 = np.transpose(total2)

EDIT to answer comment(how to apply same sorting to different array)

bs = 3
B2 = np.empty(B.shape)
for j in range(A.shape[1]):
    for i in range(int(A.shape[0]/bs)):
        A2_order = np.argsort(A[bs * i : bs * (i + 1), j])
        B2[bs * i : bs * (i + 1),j] = B[A2_order+i*bs,j]

Answer 2

You can avoid looping alltogether.

First transpose and reshape the array so that the array indices go from coarse to fine (year->month->hour).

A = np.transpose(A)
A = np.reshape(A, [12, 12, 730])

Now you can select all hours of a month as A[year, month]

Conveniently, the np.sort function by default sorts along the last axis of the array, so you can just call

A = np.sort(A)

and now each list of A[year, month] entries will be sorted.