NumPy 在二维数组中逐行搜索一维数组

Question

Say we have x, y, z arrays:

x = np.array([10, 11])

y = np.array([10, 10])

z = np.array([[ 10, 229, 261, 11, 243],
             [  10, 230, 296, 10,  79],
             [  10, 10, 10, 10,  10],
             [  0, 260, 407, 229,  79],
             [  10, 10, 11, 106, 11]])

I need a function that takes x or y array and search for it in z:

myfunc(x, z) # should give following result:
([1, 2, 4], [1, 2, 1])

first list above is index of rows in z where x is found, second list is number of time x occurred in each row.

myfunc(y, z) # should give following result:
([0, 4], [1, 2])

I did search for similar questions and tried to implement those. But couldn't figure out how to count occurrences of 1d array in 2d.

Answer 1

Ok, assuming you don't care about the order of your variables inside x or y you can use Counter to find the number of occurrences and use those to find the minimum of occurrences of you x and y variables. What makes it a bit messy is the fact, that you can have duplicates in your searched for values, so there again use Counter to make it easier. You can't use counter on a 2D-Array however, so you need to iterate over your z.

from collections import Counter

def myfunct(x, z):
    occurences_result = []
    rows_result = []

    for row, i in enumerate(z):
        count = Counter(i)
        count_x = Counter(x)
        occurences = [int(count[x_value]/x_occ) for x_value, x_occ in count_x.items() if x_value in count]
        if len(occurences) == len(count_x):
            occurences_result.append(min(occurences))
            rows_result.append(row)
    return((occurences_result, rows_result))

print(myfunct(y,z))