Question

我有以下数组：

arr1 = np.array(['x','y','z'])

arr2 = np.array(['y','a','b','z','c','x'])

arr3 = np.array([['y','d1','d2','d3'],
                 ['a','d4','d5','d6'],
                 ['b','d7','d8','d9'],
                 ['z','d10','d11','d12'],
                 ['c','d13','d14','d15'],
                 ['x','d16','d17','d18']])

我希望得到以下内容：

answer = np.array([[x,d16,d17,d18],
                   [y,d1,d2,d3]
                   [z,d10,d11,d12]])

我该怎么做？

Answer 1

我认为这应该有用，或者至少让你朝着正确的方向前进。

idxs = np.where( arr3[:,0] == arr1 )
answer = arr3[idxs, :]

了解np.where：http://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html

Answer 2

inds = [ where( arr3[:,0] == val)[0] for val in arr1 ]
inds = [ i[0] for i in inds if i.size ] 
answer = arr3[inds]

Answer 3

取第一行（next()），其中第一行等于e1的元素arr1。

array([
    next(row3 for row3 in arr3 if row3[0]==e1)
      for e1 in arr1])

输出：

array([['x', 'd16', 'd17', 'd18'],
       ['y', 'd1', 'd2', 'd3'],
       ['z', 'd10', 'd11', 'd12']], 
  dtype='|S3')

Answer 4

numpy_indexed包（免责声明：我是其作者）具有有效解决此问题的功能。您可以将其视为list.index的向量化等价物。如果你问我，应该是numpy本身;也许在将来的版本中！

import numpy_indexed as npi
answer = arr3[npi.indices(arr3[:,0], arr1)]

Numpy在数组中查找与另一个数组python中的值对应的行

4 个答案: