我有一个xy坐标的numpy查找表，其中列0 = xa，1 = ya，2 = xb，3 = yb。我尝试使用xa和ya（cols 0＆amp; 1）作为一对能够查找xb和yb（cols 2＆amp; 3）的元素，这是我想要的实际xy坐标使用。
lookup=
[[0, 0, 0, 0]
[2, 0, 1.98, -0.01]
[4, 0, 3.99, -0.01]
[6, 0, 6.03, -0.01]
[8, 0, 8.02, -0.03]
[10, 0, 9.98, -0.01]
[12, 0, 11.99, 0]
[14, 0, 13.99, 0]
[0, 1, -0.03, 0.88]
[2, 1, 1.95, 0.86]
[4, 1, 3.97, 0.85]
[6, 1, 5.97, 0.87]
[8, 1, 7.96, 0.86]
[10, 1, 9.95, 0.92]
[12, 1, 11.95, 0.92]
[14, 1, 13.97, 0.87]]
我有一个表格，其数据的x和y位置格式为xa ya，我希望使用查找表将其更改为xb yb：
gridloc=
[[6, 0]
[8, 0]
[8, 0]
[10, 0]
[8, 1]
[10, 1]
[12, 1]
[14, 1]
所以我希望结果如下：
newloc=
[[6.03, -0.01]
[8.02, -0.03]
[8.02, -0.03]
[9.98, -0.01]
[7.96, 0.86]
[9.95, 0.92]
[11.95, 0.92]
[13.97, 0.87]]
我已尝试使用此功能尝试创建字典，但收到错误：
mapping = dict(zip(lookup[:,0:2], range(len(lookup))))
Traceback (most recent call last):
File "<ipython-input-12-528fb6616ce0>", line 1, in <module>
mapping = dict(zip(lookup[:,0:2], range(len(lookup))))
TypeError: unhashable type: 'numpy.ndarray'
有人有什么建议吗？我的桌子首先应该是numpy吗？ dict是解决问题的方法吗？
答案 0 :(得分：3)
这是一种Numpythonic方法：
In [89]: mask = np.logical_and(gridloc[:,0] == lookup[:,None,0], gridloc[:,1] == lookup[:,None, 1])
In [90]: ind = np.where(mask)[0]
In [91]: lookup[ind, 2:]
Out[91]:
array([[ 6.030e+00, -1.000e-02],
[ 8.020e+00, -3.000e-02],
[ 8.020e+00, -3.000e-02],
[ 9.980e+00, -1.000e-02],
[ 7.960e+00, 8.600e-01],
[ 9.950e+00, 9.200e-01],
[ 1.195e+01, 9.200e-01],
[ 1.397e+01, 8.700e-01]])
答案 1 :(得分：2)
一种选择是使用Pandas索引功能来实现：
import numpy as np
import pandas as pd
lookup = np.array(
[[0, 0, 0, 0],
[2, 0, 1.98, -0.01],
[4, 0, 3.99, -0.01],
[6, 0, 6.03, -0.01],
[8, 0, 8.02, -0.03],
[10, 0, 9.98, -0.01],
[12, 0, 11.99, 0],
[14, 0, 13.99, 0],
[0, 1, -0.03, 0.88],
[2, 1, 1.95, 0.86],
[4, 1, 3.97, 0.85],
[6, 1, 5.97, 0.87],
[8, 1, 7.96, 0.86],
[10, 1, 9.95, 0.92],
[12, 1, 11.95, 0.92],
[14, 1, 13.97, 0.87]])
gridloc = np.array(
[[6, 0],
[8, 0],
[8, 0],
[10, 0],
[8, 1],
[10, 1],
[12, 1],
[14, 1]])
idx = pd.MultiIndex.from_arrays([lookup[:, 0], lookup[:, 1]], names=('xa', 'ya'))
df = pd.DataFrame(lookup[:, 2:], columns=('xb', 'yb'), index=idx)
# This should work but is not implemented for multidimensional arrays
# newloc = df.loc[gridloc].values
# Converting to list of tuples works
newloc = df.loc[list(map(tuple, gridloc))].values # Add .copy() if need writing
print(newloc)
输出：
[[ 6.03000000e+00 -1.00000000e-02]
[ 8.02000000e+00 -3.00000000e-02]
[ 8.02000000e+00 -3.00000000e-02]
[ 9.98000000e+00 -1.00000000e-02]
[ 7.96000000e+00 8.60000000e-01]
[ 9.95000000e+00 9.20000000e-01]
[ 1.19500000e+01 9.20000000e-01]
[ 1.39700000e+01 8.70000000e-01]]
答案 2 :(得分：1)
首先，列表是可变的，不能用作dict键。这就是您需要将数据转换为元组的原因：
mapping = dict(zip(map(tuple, lookup[:, :2]), map(tuple, lookup[:, 2:])))#
mapping
#{(0.0, 0.0): (0.0, 0.0),
# (0.0, 1.0): (-0.029999999999999999, 0.88),
# (2.0, 0.0): (1.98, -0.01),
# (2.0, 1.0): (1.95, 0.85999999999999999),
# (4.0, 0.0): (3.9900000000000002, -0.01),
# (4.0, 1.0): (3.9700000000000002, 0.84999999999999998),
# (6.0, 0.0): (6.0300000000000002, -0.01),
# (6.0, 1.0): (5.9699999999999998, 0.87),
# (8.0, 0.0): (8.0199999999999996, -0.029999999999999999),
# (8.0, 1.0): (7.96, 0.85999999999999999),
# (10.0, 0.0): (9.9800000000000004, -0.01),
# (10.0, 1.0): (9.9499999999999993, 0.92000000000000004),
# (12.0, 0.0): (11.99, 0.0),
# (12.0, 1.0): (11.949999999999999, 0.92000000000000004),
# (14.0, 0.0): (13.99, 0.0),
# (14.0, 1.0): (13.970000000000001, 0.87)}
现在要实现目标，您需要将gridloc转换为元组列表，然后将mapping映射到它：
gridloc = list(map(mapping.get, map(tuple, gridloc)))
gridloc
#[(6.0300000000000002, -0.01),
# (8.0199999999999996, -0.029999999999999999),
# (8.0199999999999996, -0.029999999999999999),
# (9.9800000000000004, -0.01),
# (7.96, 0.85999999999999999),
# (9.9499999999999993, 0.92000000000000004),
# (11.949999999999999, 0.92000000000000004),
# (13.970000000000001, 0.87)]