我正在从matplotlib.mlab.csv2rec函数中重新获得。我的期望是它有2个维度,如'x',但它有1个维度,如'y'。有没有办法从y获得x?
>>> import numpy as np
>>> from datetime import date
>>> x=np.array([(date(2000,1,1),0,1),
... (date(2000,1,1),1,1),
... (date(2000,1,1),1,0),
... (date(2000,1,1),0,0),
... ])
>>> x
array([[2000-01-01, 0, 1],
[2000-01-01, 1, 1],
[2000-01-01, 1, 0],
[2000-01-01, 0, 0]], dtype=object)
>>> y = np.rec.fromrecords( x )
>>> y
rec.array([(datetime.date(2000, 1, 1), 0, 1),
(datetime.date(2000, 1, 1), 1, 1),
(datetime.date(2000, 1, 1), 1, 0), (datetime.date(2000, 1, 1), 0, 0)],
dtype=[('f0', '|O4'), ('f1', '<i4'), ('f2', '<i4')])
>>> x.ndim
2
>>> y.ndim
1
>>> x.shape
(4, 3)
>>> y.ndim
1
>>> y.shape
(4,)
>>>
答案 0 :(得分:2)
你可以通过熊猫来实现:
import pandas as pd
pd.DataFrame(y).values
array([[2000-01-01, 0, 1],
[2000-01-01, 1, 1],
[2000-01-01, 1, 0],
[2000-01-01, 0, 0]], dtype=object)
但如果我是你,我会考虑在熊猫中做我的项目。对于大熊猫,对命名列的支持比普通numpy更深入。
>>> z = pd.DataFrame.from_records(y, index="f0")
>>> z
f1 f2
f0
2000-01-01 0 1
2000-01-01 1 1
2000-01-01 1 0
2000-01-01 0 0
>>> z["f1"]
f0
2000-01-01 0
2000-01-01 1
2000-01-01 1
2000-01-01 0
Name: f1
答案 1 :(得分:0)
听起来很奇怪,但是......我可以使用matplotlib.mlab.rec2csv保存到csv,然后使用numpy.loadtxt读取到ndarray。我的情况比较简单,因为我已经有了csv文件。这是一个如何工作的例子。
>>> a = np.loadtxt( 'name.csv', skiprows=1, delimiter=',', converters = {0: lambda x: 0} )
>>> a
array([[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0.29, 0.29, 0.43, 0.29, 0. ],
[ 0. , 0.71, 0.29, 0.57, 0. , 0. ],
[ 0. , 1. , 0.57, 0.71, 0. , 0. ],
[ 0. , 0.43, 0.29, 0.14, 0.14, 0. ],
[ 0. , 1. , 0.43, 0.71, 0. , 0. ],
[ 0. , 0.57, 0.57, 0.29, 0.14, 0. ],
[ 0. , 1.43, 0.43, 0.86, 0.43, 0. ],
[ 0. , 1. , 0.71, 0.57, 0. , 0. ],
[ 0. , 1.14, 0.57, 0.29, 0. , 0. ],
[ 0. , 1.43, 0.29, 0.71, 0.29, 0.29],
[ 0. , 1.14, 0.43, 1. , 0.29, 0.29],
[ 0. , 0.43, 1.14, 0.86, 0.43, 0.14],
[ 0. , 1.14, 0.86, 0.86, 0.29, 0.29]])
>>> t = a.any( axis = 1 )
>>> t
array([False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, True, True,
True, True, True, True, True, True, True, True, True,
True, True], dtype=bool)
>>> a.ndim
2
同样在我的情况下,我不需要第一列来做出决定。