Question

有一个数据框df，其中一些列描述了给定bin中的某些单位及其输出，如下所示：

df = pd.DataFrame({'bin_dir' : pd.cut(np.rad2deg(np.random.vonmises(np.pi,0.03,100)) % 360,np.arange(0,365,5)),
                   'Unit' : np.tile(np.arange(1,11),10),
                   'value' : np.random.randn(100)*1000+3600})

我现在要创建一个列col1，当单位为1,3,5时，其值为1，bin_dir为（350,355），（355,360），（0,5），（5,10）和2时为单位是2,4,9，dir_bin是（350,355），（355,360），（0,5），（5,10）

怎么能这样做？在dplyr中，我可以将mutate与嵌套的ifelse语句一起使用。如果解决方案可以合并到链式命令中会很好：）

由于

Answer 1

您可以使用嵌套的np.where()：

import re
import pandas as pd

In [50]: bins = re.findall(r'\(.*?\]', '(350, 355], (355, 360], (0, 5], (5, 10]')
    ...: bin_mask = df.bin_dir.isin(bins)
    ...: unit_mask1 = df.Unit.isin([1,3,5])
    ...: unit_mask2 = df.Unit.isin([2,4,9])
    ...:

In [51]: df.assign(col1=
    ...:     np.where(bin_mask & unit_mask1,
    ...:              1,
    ...:              np.where(bin_mask & unit_mask2, 2, np.nan)
    ...:     )
    ...: )
    ...:
Out[51]:
    Unit     bin_dir        value  col1
0      1  (195, 200]  1228.056261   NaN
1      2  (125, 130]  3246.052662   NaN
2      3  (150, 155]  3128.356490   NaN
3      4  (215, 220]  2900.812099   NaN
4      5  (110, 115]  4324.152904   NaN
5      6  (150, 155]  4783.110204   NaN
6      7  (240, 245]  4810.120258   NaN
7      8  (210, 215]  4307.576911   NaN
8      9    (15, 20]  3043.099987   NaN
9     10      (0, 5]  4633.435048   NaN
10     1  (145, 150]  3401.690163   NaN
11     2  (320, 325]  4224.314088   NaN
12     3  (350, 355]  4037.081806   1.0
13     4  (295, 300]  3096.652374   NaN
14     5  (235, 240]  4738.227922   NaN
15     6  (235, 240]  1973.561204   NaN
16     7  (270, 275]  3500.619163   NaN
17     8    (45, 50]  4234.621801   NaN
18     9  (255, 260]  4267.575087   NaN
19    10  (320, 325]  3031.733130   NaN
20     1  (235, 240]  3137.832272   NaN
21     2  (330, 335]  4113.654195   NaN
22     3  (265, 270]  3060.886390   NaN
23     4  (290, 295]  2836.105371   NaN
24     5  (255, 260]  2756.894839   NaN
..   ...         ...          ...   ...
75     6  (325, 330]  2471.775169   NaN
76     7    (70, 75]  4463.964881   NaN
77     8  (110, 115]  5681.124294   NaN
78     9  (135, 140]  2500.650717   NaN
79    10  (225, 230]  2936.364153   NaN
80     1  (280, 285]  1138.591459   NaN
81     2  (250, 255]  3121.142300   NaN
82     3  (150, 155]  2991.257906   NaN
83     4  (160, 165]  3078.156743   NaN
84     5  (130, 135]  4335.076559   NaN
85     6    (85, 90]  4970.471290   NaN
86     7  (335, 340]  3207.906304   NaN
87     8  (350, 355]  3605.474926   NaN
88     9  (125, 130]  4922.963220   NaN
89    10    (60, 65]  3121.061944   NaN
90     1  (105, 110]  3092.191627   NaN
91     2      (0, 5]  3693.602055   2.0
92     3  (195, 200]  2291.508096   NaN
93     4    (40, 45]  4628.409801   NaN
94     5  (215, 220]  3327.321452   NaN
95     6  (110, 115]  4347.471046   NaN
96     7  (110, 115]  4494.707840   NaN
97     8  (110, 115]  3545.460851   NaN
98     9    (55, 60]  2831.042251   NaN
99    10    (30, 35]  3705.225870   NaN

[100 rows x 4 columns]

当然，你可以在没有预先计算的面具的情况下做到这一点：

In [52]: df.assign(col1=
    ...:     np.where(df.bin_dir.isin(bins) & df.Unit.isin([1,3,5]),
    ...:              1,
    ...:              np.where(df.bin_dir.isin(bins) & df.Unit.isin([2,4,9]),
    ...:                       2,
    ...:                       np.nan
    ...:              )
    ...:     )
    ...: )
    ...:
Out[52]:
    Unit     bin_dir        value  col1
0      1  (195, 200]  1228.056261   NaN
1      2  (125, 130]  3246.052662   NaN
2      3  (150, 155]  3128.356490   NaN
3      4  (215, 220]  2900.812099   NaN
4      5  (110, 115]  4324.152904   NaN
5      6  (150, 155]  4783.110204   NaN
6      7  (240, 245]  4810.120258   NaN
7      8  (210, 215]  4307.576911   NaN
8      9    (15, 20]  3043.099987   NaN
9     10      (0, 5]  4633.435048   NaN
10     1  (145, 150]  3401.690163   NaN
11     2  (320, 325]  4224.314088   NaN
12     3  (350, 355]  4037.081806   1.0
13     4  (295, 300]  3096.652374   NaN
14     5  (235, 240]  4738.227922   NaN
15     6  (235, 240]  1973.561204   NaN
16     7  (270, 275]  3500.619163   NaN
17     8    (45, 50]  4234.621801   NaN
18     9  (255, 260]  4267.575087   NaN
19    10  (320, 325]  3031.733130   NaN
20     1  (235, 240]  3137.832272   NaN
21     2  (330, 335]  4113.654195   NaN
22     3  (265, 270]  3060.886390   NaN
23     4  (290, 295]  2836.105371   NaN
24     5  (255, 260]  2756.894839   NaN
..   ...         ...          ...   ...
75     6  (325, 330]  2471.775169   NaN
76     7    (70, 75]  4463.964881   NaN
77     8  (110, 115]  5681.124294   NaN
78     9  (135, 140]  2500.650717   NaN
79    10  (225, 230]  2936.364153   NaN
80     1  (280, 285]  1138.591459   NaN
81     2  (250, 255]  3121.142300   NaN
82     3  (150, 155]  2991.257906   NaN
83     4  (160, 165]  3078.156743   NaN
84     5  (130, 135]  4335.076559   NaN
85     6    (85, 90]  4970.471290   NaN
86     7  (335, 340]  3207.906304   NaN
87     8  (350, 355]  3605.474926   NaN
88     9  (125, 130]  4922.963220   NaN
89    10    (60, 65]  3121.061944   NaN
90     1  (105, 110]  3092.191627   NaN
91     2      (0, 5]  3693.602055   2.0
92     3  (195, 200]  2291.508096   NaN
93     4    (40, 45]  4628.409801   NaN
94     5  (215, 220]  3327.321452   NaN
95     6  (110, 115]  4347.471046   NaN
96     7  (110, 115]  4494.707840   NaN
97     8  (110, 115]  3545.460851   NaN
98     9    (55, 60]  2831.042251   NaN
99    10    (30, 35]  3705.225870   NaN

[100 rows x 4 columns]

但它会变慢，看起来有点麻烦

Answer 2

使用列表理解：

bin_filt = ['(350, 355]', '(355, 360]', '(0, 5]', '(5, 10]']
# Creates a column 'col1'
df['col1'] = [1 for i in range(df.shape[0]) if df['Unit'][i] in [1, 3, 5] and df['bin_dir'][i] in bin_filt else 0]
# Creates a column 'col2'
df['col2'] = [2 for i in range(df.shape[0]) if df['Unit'][i] in [2, 4, 9] and df['bin_dir'][i] in bin_filt else 0]
# You can replace the 'else' statement a t the end of the list comprehension to put the value you want instead

根据其他列中的多个条件创建列值

2 个答案: