我认为这是您想要的：
data = np.array([[ 4057, 8, 1374],
[ 4057, 9, 759],
[ 4057, 11, 96],
[89205, 16, 146],
[89205, 17, 154],
[89205, 18, 244]])
rows, row_pos = np.unique(data[:, 0], return_inverse=True)
cols, col_pos = np.unique(data[:, 1], return_inverse=True)
pivot_table = np.zeros((len(rows), len(cols)), dtype=data.dtype)
pivot_table[row_pos, col_pos] = data[:, 2]
>>> pivot_table
array([[1374, 759, 96, 0, 0, 0],
[ 0, 0, 0, 146, 154, 244]])
>>> rows
array([ 4057, 89205])
>>> cols
array([ 8, 9, 11, 16, 17, 18])
这种方法有一些局限性，主要是，如果您对相同的行/列组合重复输入，则不会将它们加在一起，而只会保留一个（可能是最后一个）。如果您想将它们全部加在一起，尽管有些麻烦，但是您可能会滥用scipy的稀疏模块：
data = np.array([[ 4057, 8, 1374],
[ 4057, 9, 759],
[ 4057, 11, 96],
[89205, 16, 146],
[89205, 17, 154],
[89205, 18, 244],
[ 4057, 11, 4]])
rows, row_pos = np.unique(data[:, 0], return_inverse=True)
cols, col_pos = np.unique(data[:, 1], return_inverse=True)
pivot_table = np.zeros((len(rows), len(cols)), dtype=data.dtype)
pivot_table[row_pos, col_pos] = data[:, 2]
>>> pivot_table # the element at [0, 2] should be 100!!!
array([[1374, 759, 4, 0, 0, 0],
[ 0, 0, 0, 146, 154, 244]])
import scipy.sparse as sps
pivot_table = sps.coo_matrix((data[:, 2], (row_pos, col_pos)),
shape=(len(rows), len(cols))).A
>>> pivot_table # Now repeated elements are added together
array([[1374, 759, 100, 0, 0, 0],
[ 0, 0, 0, 146, 154, 244]])
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