如何将熊猫数据帧转换为NumPy数组?
DataFrame:
import numpy as np
import pandas as pd
index = [1, 2, 3, 4, 5, 6, 7]
a = [np.nan, np.nan, np.nan, 0.1, 0.1, 0.1, 0.1]
b = [0.2, np.nan, 0.2, 0.2, 0.2, np.nan, np.nan]
c = [np.nan, 0.5, 0.5, np.nan, 0.5, 0.5, np.nan]
df = pd.DataFrame({'A': a, 'B': b, 'C': c}, index=index)
df = df.rename_axis('ID')
给了
label A B C
ID
1 NaN 0.2 NaN
2 NaN NaN 0.5
3 NaN 0.2 0.5
4 0.1 0.2 NaN
5 0.1 0.2 0.5
6 0.1 NaN 0.5
7 0.1 NaN NaN
我想把它转换成一个NumPy数组，像这样:
array([[ nan, 0.2, nan],
[ nan, nan, 0.5],
[ nan, 0.2, 0.5],
[ 0.1, 0.2, nan],
[ 0.1, 0.2, 0.5],
[ 0.1, nan, 0.5],
[ 0.1, nan, nan]])
另外，是否可以像这样保存dtype ?
array([[ 1, nan, 0.2, nan],
[ 2, nan, nan, 0.5],
[ 3, nan, 0.2, 0.5],
[ 4, 0.1, 0.2, nan],
[ 5, 0.1, 0.2, 0.5],
[ 6, 0.1, nan, 0.5],
[ 7, 0.1, nan, nan]],
dtype=[('ID', '<i4'), ('A', '<f8'), ('B', '<f8'), ('B', '<f8')])
DataFrame的一个更简单的例子:
df
gbm nnet reg
0 12.097439 12.047437 12.100953
1 12.109811 12.070209 12.095288
2 11.720734 11.622139 11.740523
3 11.824557 11.926414 11.926527
4 11.800868 11.727730 11.729737
5 12.490984 12.502440 12.530894
USE:
np.array(df.to_records().view(type=np.matrix))
GET:
array([[(0, 12.097439 , 12.047437, 12.10095324),
(1, 12.10981081, 12.070209, 12.09528824),
(2, 11.72073428, 11.622139, 11.74052253),
(3, 11.82455653, 11.926414, 11.92652727),
(4, 11.80086775, 11.72773 , 11.72973699),
(5, 12.49098389, 12.50244 , 12.53089367)]],
dtype=(numpy.record, [('index', '<i8'), ('gbm', '<f8'), ('nnet', '<f4'),
('reg', '<f8')]))
下面是我从pandas DataFrame制作结构数组的方法。
创建数据帧
import pandas as pd
import numpy as np
import six
NaN = float('nan')
ID = [1, 2, 3, 4, 5, 6, 7]
A = [NaN, NaN, NaN, 0.1, 0.1, 0.1, 0.1]
B = [0.2, NaN, 0.2, 0.2, 0.2, NaN, NaN]
C = [NaN, 0.5, 0.5, NaN, 0.5, 0.5, NaN]
columns = {'A':A, 'B':B, 'C':C}
df = pd.DataFrame(columns, index=ID)
df.index.name = 'ID'
print(df)
A B C
ID
1 NaN 0.2 NaN
2 NaN NaN 0.5
3 NaN 0.2 0.5
4 0.1 0.2 NaN
5 0.1 0.2 0.5
6 0.1 NaN 0.5
7 0.1 NaN NaN
定义函数，从pandas数据帧中创建numpy结构数组(而不是记录数组)。
def df_to_sarray(df):
"""
Convert a pandas DataFrame object to a numpy structured array.
This is functionally equivalent to but more efficient than
np.array(df.to_array())
:param df: the data frame to convert
:return: a numpy structured array representation of df
"""
v = df.values
cols = df.columns
if six.PY2: # python 2 needs .encode() but 3 does not
types = [(cols[i].encode(), df[k].dtype.type) for (i, k) in enumerate(cols)]
else:
types = [(cols[i], df[k].dtype.type) for (i, k) in enumerate(cols)]
dtype = np.dtype(types)
z = np.zeros(v.shape[0], dtype)
for (i, k) in enumerate(z.dtype.names):
z[k] = v[:, i]
return z
使用reset_index创建一个新的数据帧，其中包含索引作为其数据的一部分。将该数据帧转换为结构数组。
sa = df_to_sarray(df.reset_index())
sa
array([(1L, nan, 0.2, nan), (2L, nan, nan, 0.5), (3L, nan, 0.2, 0.5),
(4L, 0.1, 0.2, nan), (5L, 0.1, 0.2, 0.5), (6L, 0.1, nan, 0.5),
(7L, 0.1, nan, nan)],
dtype=[('ID', '<i8'), ('A', '<f8'), ('B', '<f8'), ('C', '<f8')])
编辑:更新df_to_sarray以避免在python 3中调用.encode()时出错。感谢Joseph Garvin和halcyon的评论和解决方案。
DataFrame的一个更简单的例子:
df
gbm nnet reg
0 12.097439 12.047437 12.100953
1 12.109811 12.070209 12.095288
2 11.720734 11.622139 11.740523
3 11.824557 11.926414 11.926527
4 11.800868 11.727730 11.729737
5 12.490984 12.502440 12.530894
USE:
np.array(df.to_records().view(type=np.matrix))
GET:
array([[(0, 12.097439 , 12.047437, 12.10095324),
(1, 12.10981081, 12.070209, 12.09528824),
(2, 11.72073428, 11.622139, 11.74052253),
(3, 11.82455653, 11.926414, 11.92652727),
(4, 11.80086775, 11.72773 , 11.72973699),
(5, 12.49098389, 12.50244 , 12.53089367)]],
dtype=(numpy.record, [('index', '<i8'), ('gbm', '<f8'), ('nnet', '<f4'),
('reg', '<f8')]))
