在Python中合并索引数组
假设我有两个numpy数组,格式如下:
x = [[1,2]
[2,4]
[3,6]
[4,NaN]
[5,10]]
y = [[0,-5]
[1,0]
[2,5]
[5,20]
[6,25]]
有没有什么高效的方法可以把它们合并成这样:
xmy = [[0, NaN, -5 ]
[1, 2, 0 ]
[2, 4, 5 ]
[3, 6, NaN]
[4, NaN, NaN]
[5, 10, 20 ]
[6, NaN, 25 ]
我可以用一个简单的函数来搜索找到索引,但这样做不够优雅,而且对于很多数组和大尺寸的情况可能效率不高。任何建议都很感谢。
1 个回答
10
请查看 numpy.lib.recfunctions.join_by
这个功能只适用于结构化数组或递归数组,所以会有一些小问题。
首先,你需要对结构化数组有一点了解。如果你不太清楚,可以看看 这里。
import numpy as np
import numpy.lib.recfunctions
# Define the starting arrays as structured arrays with two fields ('key' and 'field')
dtype = [('key', np.int), ('field', np.float)]
x = np.array([(1, 2),
(2, 4),
(3, 6),
(4, np.NaN),
(5, 10)],
dtype=dtype)
y = np.array([(0, -5),
(1, 0),
(2, 5),
(5, 20),
(6, 25)],
dtype=dtype)
# You want an outer join, rather than the default inner join
# (all values are returned, not just ones with a common key)
join = np.lib.recfunctions.join_by('key', x, y, jointype='outer')
# Now we have a structured array with three fields: 'key', 'field1', and 'field2'
# (since 'field' was in both arrays, it renamed x['field'] to 'field1', and
# y['field'] to 'field2')
# This returns a masked array, if you want it filled with
# NaN's, do the following...
join.fill_value = np.NaN
join = join.filled()
# Just displaying it... Keep in mind that as a structured array,
# it has one dimension, where each row contains the 3 fields
for row in join:
print row
这段代码的输出是:
(0, nan, -5.0)
(1, 2.0, 0.0)
(2, 4.0, 5.0)
(3, 6.0, nan)
(4, nan, nan)
(5, 10.0, 20.0)
(6, nan, 25.0)
希望这对你有帮助!
编辑1:添加了示例
编辑2:实际上不应该用浮点数来连接... 把'key'字段改成了整数。