我有以下索引DataFrame命名列和行不连续的数字:
a b c d
2 0.671399 0.101208 -0.181532 0.241273
3 0.446172 -0.243316 0.051767 1.577318
5 0.614758 0.075793 -0.451460 -0.012493
我想添加一个新列，'e'，到现有的数据帧，并不想改变数据帧中的任何东西(即，新列始终具有与DataFrame相同的长度)。
0 -0.335485
1 -1.166658
2 -0.385571
dtype: float64
如何将列e添加到上面的例子中?
当前回答
简单明了的:
df.loc[:, 'NewCol'] = 'New_Val'
例子:
df = pd.DataFrame(data=np.random.randn(20, 4), columns=['A', 'B', 'C', 'D'])
df
A B C D
0 -0.761269 0.477348 1.170614 0.752714
1 1.217250 -0.930860 -0.769324 -0.408642
2 -0.619679 -1.227659 -0.259135 1.700294
3 -0.147354 0.778707 0.479145 2.284143
4 -0.529529 0.000571 0.913779 1.395894
5 2.592400 0.637253 1.441096 -0.631468
6 0.757178 0.240012 -0.553820 1.177202
7 -0.986128 -1.313843 0.788589 -0.707836
8 0.606985 -2.232903 -1.358107 -2.855494
9 -0.692013 0.671866 1.179466 -1.180351
10 -1.093707 -0.530600 0.182926 -1.296494
11 -0.143273 -0.503199 -1.328728 0.610552
12 -0.923110 -1.365890 -1.366202 -1.185999
13 -2.026832 0.273593 -0.440426 -0.627423
14 -0.054503 -0.788866 -0.228088 -0.404783
15 0.955298 -1.430019 1.434071 -0.088215
16 -0.227946 0.047462 0.373573 -0.111675
17 1.627912 0.043611 1.743403 -0.012714
18 0.693458 0.144327 0.329500 -0.655045
19 0.104425 0.037412 0.450598 -0.923387
df.drop([3, 5, 8, 10, 18], inplace=True)
df
A B C D
0 -0.761269 0.477348 1.170614 0.752714
1 1.217250 -0.930860 -0.769324 -0.408642
2 -0.619679 -1.227659 -0.259135 1.700294
4 -0.529529 0.000571 0.913779 1.395894
6 0.757178 0.240012 -0.553820 1.177202
7 -0.986128 -1.313843 0.788589 -0.707836
9 -0.692013 0.671866 1.179466 -1.180351
11 -0.143273 -0.503199 -1.328728 0.610552
12 -0.923110 -1.365890 -1.366202 -1.185999
13 -2.026832 0.273593 -0.440426 -0.627423
14 -0.054503 -0.788866 -0.228088 -0.404783
15 0.955298 -1.430019 1.434071 -0.088215
16 -0.227946 0.047462 0.373573 -0.111675
17 1.627912 0.043611 1.743403 -0.012714
19 0.104425 0.037412 0.450598 -0.923387
df.loc[:, 'NewCol'] = 0
df
A B C D NewCol
0 -0.761269 0.477348 1.170614 0.752714 0
1 1.217250 -0.930860 -0.769324 -0.408642 0
2 -0.619679 -1.227659 -0.259135 1.700294 0
4 -0.529529 0.000571 0.913779 1.395894 0
6 0.757178 0.240012 -0.553820 1.177202 0
7 -0.986128 -1.313843 0.788589 -0.707836 0
9 -0.692013 0.671866 1.179466 -1.180351 0
11 -0.143273 -0.503199 -1.328728 0.610552 0
12 -0.923110 -1.365890 -1.366202 -1.185999 0
13 -2.026832 0.273593 -0.440426 -0.627423 0
14 -0.054503 -0.788866 -0.228088 -0.404783 0
15 0.955298 -1.430019 1.434071 -0.088215 0
16 -0.227946 0.047462 0.373573 -0.111675 0
17 1.627912 0.043611 1.743403 -0.012714 0
19 0.104425 0.037412 0.450598 -0.923387 0
其他回答
让我补充一下，就像hum3一样，.loc没有解决SettingWithCopyWarning，我不得不求助于df.insert()。在我的例子中，假阳性是由“假”链索引dict['a']['e']生成的，其中'e'是新列，dict['a']是来自字典的数据框架。
还请注意，如果您知道自己在做什么，您可以使用切换警告 pd.options.mode。chained_assignment =无 然后用这里给出的另一个解。
为了完整起见-使用DataFrame.eval()方法的另一个解决方案:
数据:
In [44]: e
Out[44]:
0 1.225506
1 -1.033944
2 -0.498953
3 -0.373332
4 0.615030
5 -0.622436
dtype: float64
In [45]: df1
Out[45]:
a b c d
0 -0.634222 -0.103264 0.745069 0.801288
4 0.782387 -0.090279 0.757662 -0.602408
5 -0.117456 2.124496 1.057301 0.765466
7 0.767532 0.104304 -0.586850 1.051297
8 -0.103272 0.958334 1.163092 1.182315
9 -0.616254 0.296678 -0.112027 0.679112
解决方案:
In [46]: df1.eval("e = @e.values", inplace=True)
In [47]: df1
Out[47]:
a b c d e
0 -0.634222 -0.103264 0.745069 0.801288 1.225506
4 0.782387 -0.090279 0.757662 -0.602408 -1.033944
5 -0.117456 2.124496 1.057301 0.765466 -0.498953
7 0.767532 0.104304 -0.586850 1.051297 -0.373332
8 -0.103272 0.958334 1.163092 1.182315 0.615030
9 -0.616254 0.296678 -0.112027 0.679112 -0.622436
我得到了可怕的SettingWithCopyWarning，它没有通过使用iloc语法修复。我的DataFrame是由read_sql从ODBC源创建的。根据上面low - tech的建议，以下方法对我来说是有效的:
df.insert(len(df.columns), 'e', pd.Series(np.random.randn(sLength), index=df.index))
This worked fine to insert the column at the end. I don't know if it is the most efficient, but I don't like warning messages. I think there is a better solution, but I can't find it, and I think it depends on some aspect of the index. Note. That this only works once and will give an error message if trying to overwrite and existing column. Note As above and from 0.16.0 assign is the best solution. See documentation http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.assign.html#pandas.DataFrame.assign Works well for data flow type where you don't overwrite your intermediate values.
直接通过NumPy这样做将是最有效的:
df1['e'] = np.random.randn(sLength)
注意我最初(非常老)的建议是使用map(这要慢得多):
df1['e'] = df1['a'].map(lambda x: np.random.random())
向pandas数据框架插入新列的4种方法
using simple assignment, insert(), assign() and Concat() methods.
import pandas as pd
df = pd.DataFrame({
'col_a':[True, False, False],
'col_b': [1, 2, 3],
})
print(df)
col_a col_b
0 True 1
1 False 2
2 False 3
使用简单赋值
ser = pd.Series(['a', 'b', 'c'], index=[0, 1, 2])
print(ser)
0 a
1 b
2 c
dtype: object
df['col_c'] = pd.Series(['a', 'b', 'c'], index=[1, 2, 3])
print(df)
col_a col_b col_c
0 True 1 NaN
1 False 2 a
2 False 3 b
使用分配()
e = pd.Series([1.0, 3.0, 2.0], index=[0, 2, 1])
ser = pd.Series(['a', 'b', 'c'], index=[0, 1, 2])
df.assign(colC=s.values, colB=e.values)
col_a col_b col_c
0 True 1.0 a
1 False 3.0 b
2 False 2.0 c
使用insert ()
df.insert(len(df.columns), 'col_c', ser.values)
print(df)
col_a col_b col_c
0 True 1 a
1 False 2 b
2 False 3 c
使用concat ()
ser = pd.Series(['a', 'b', 'c'], index=[10, 20, 30])
df = pd.concat([df, ser.rename('colC')], axis=1)
print(df)
col_a col_b col_c
0 True 1.0 NaN
1 False 2.0 NaN
2 False 3.0 NaN
10 NaN NaN a
20 NaN NaN b
30 NaN NaN c
