我知道删除数据框的列应该像以下一样简单：
df.drop(df.columns[1], axis=1)按索引删除
或dr.dropna(axis=1, how='any')根据其是否包含NaN来删除。
但这些都不适用于我的数据框架，我不确定这是因为格式问题或数据类型问题，还是对这些命令的误用或误解。
这是我的数据框：
fish_frame after append new_column: 0 1 2 3 4 \
2 GBE COD NaN NaN 600 NaN
3 GBW COD NaN 11,189 NaN NaN
4 GOM COD NaN 0 NaN Package Deal - $40,753.69
5 POLLOCK NaN NaN 1,103 NaN
6 WHAKE NaN NaN 12 NaN
7 GBE HADDOCK NaN 10,730 NaN NaN
8 GBW HADDOCK NaN 64,147 NaN NaN
9 GOM HADDOCK NaN 0 NaN NaN
10 REDFISH NaN NaN 0 NaN
11 WITCH FLOUNDER NaN 370 NaN NaN
12 PLAICE NaN NaN 622 NaN
13 GB WINTER FLOUNDER 54,315 NaN NaN NaN
14 GOM WINTER FLOUNDER 653 NaN NaN NaN
15 SNEMA WINTER FLOUNDER 14,601 NaN NaN NaN
16 GB YELLOWTAIL NaN 1,663 NaN NaN
17 SNEMA YELLOWTAIL NaN 1,370 NaN NaN
18 CCGOM YELLOWTAIL 1,812 NaN NaN NaN
6 package_deal_column Package_Price new_column
2 NaN Package Deal - $40,753.69 None 600
3 NaN Package Deal - $40,753.69 None 11,1890
4 None Package Deal - $40,753.69 None 0
5 NaN Package Deal - $40,753.69 None 1,103
6 NaN Package Deal - $40,753.69 None 12
7 NaN Package Deal - $40,753.69 None 10,7300
8 NaN Package Deal - $40,753.69 None 64,1470
9 NaN Package Deal - $40,753.69 None 0
10 NaN Package Deal - $40,753.69 None 0
11 NaN Package Deal - $40,753.69 None 3700
12 NaN Package Deal - $40,753.69 None 622
13 None Package Deal - $40,753.69 None 54,31500
14 None Package Deal - $40,753.69 None 65300
15 None Package Deal - $40,753.69 None 14,60100
16 NaN Package Deal - $40,753.69 None 1,6630
17 NaN Package Deal - $40,753.69 None 1,3700
18 None Package Deal - $40,753.69 None 1,81200
然后我有以下几行代码：
fish_frame.drop(fish_frame.columns[1], axis=1)
fish_frame.drop(fish_frame.columns[2], axis=1)
fish_frame.drop(fish_frame.columns[3], axis=1)
fish_frame.drop(fish_frame.columns[4:5], axis=1)
#del fish_frame[4:5] #doesn't work, "TypeError: slice(4, 5, None) is an invalid key"
del fish_frame['Package_Price']
fish_frame.dropna(axis=1, how='any')
然后我再次打印出数据框，结果如下：
NEW fish_frame: 0 1 2 3 4 \
2 GBE COD NaN NaN 600 NaN
3 GBW COD NaN 11,189 NaN NaN
4 GOM COD NaN 0 NaN Package Deal - $40,753.69
5 POLLOCK NaN NaN 1,103 NaN
6 WHAKE NaN NaN 12 NaN
7 GBE HADDOCK NaN 10,730 NaN NaN
8 GBW HADDOCK NaN 64,147 NaN NaN
9 GOM HADDOCK NaN 0 NaN NaN
10 REDFISH NaN NaN 0 NaN
11 WITCH FLOUNDER NaN 370 NaN NaN
12 PLAICE NaN NaN 622 NaN
13 GB WINTER FLOUNDER 54,315 NaN NaN NaN
14 GOM WINTER FLOUNDER 653 NaN NaN NaN
15 SNEMA WINTER FLOUNDER 14,601 NaN NaN NaN
16 GB YELLOWTAIL NaN 1,663 NaN NaN
17 SNEMA YELLOWTAIL NaN 1,370 NaN NaN
18 CCGOM YELLOWTAIL 1,812 NaN NaN NaN
6 package_deal_column new_column
2 NaN Package Deal - $40,753.69 600
3 NaN Package Deal - $40,753.69 11,1890
4 None Package Deal - $40,753.69 0
5 NaN Package Deal - $40,753.69 1,103
6 NaN Package Deal - $40,753.69 12
7 NaN Package Deal - $40,753.69 10,7300
8 NaN Package Deal - $40,753.69 64,1470
9 NaN Package Deal - $40,753.69 0
10 NaN Package Deal - $40,753.69 0
11 NaN Package Deal - $40,753.69 3700
12 NaN Package Deal - $40,753.69 622
13 None Package Deal - $40,753.69 54,31500
14 None Package Deal - $40,753.69 65300
15 None Package Deal - $40,753.69 14,60100
16 NaN Package Deal - $40,753.69 1,6630
17 NaN Package Deal - $40,753.69 1,3700
18 None Package Deal - $40,753.69 1,81200
NaN丢弃工作和索引丢弃都不起作用。只有特定的drop[column name]命令有效，但我不能对此脚本的每次迭代都这样做。
我很困惑，我希望这不是一个非常愚蠢的错误。
此外，我自己并不完全理解这些信息，但打印fish_frame.info()会产生：
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 17 entries, 2 to 18
Data columns (total 8 columns):
0 17 non-null object
1 4 non-null object
2 8 non-null object
3 5 non-null object
4 1 non-null object
6 0 non-null object
package_deal_column 17 non-null object
new_column 17 non-null object
dtypes: object(8)
memory usage: 586.0+ bytes
任何有助于解决此问题的人都将不胜感激。
答案 0 :(得分：5)
如果没有错误我从输出中看不到，您只是忘记使用inplace参数：
df.drop(df.columns[1], axis=1, inplace=True)
答案 1 :(得分：2)
以下是一些替代方案：
<强>设定：
df = pd.DataFrame(np.random.rand(3,5), columns=list('abcde'))
In [57]: cols_to_drop = ['b', 'd']
In [63]: df
Out[63]:
a b c d e
0 0.758670 0.734007 0.027711 0.614674 0.955711
1 0.833110 0.242010 0.922831 0.165401 0.546079
2 0.414916 0.949050 0.608527 0.018036 0.230343
选项1：
df = df[df.columns.drop(col_to_drop)]
选项2：
df = df[df.columns.difference(cols_to_drop)]
选项3：
df = df.loc[:, ~df.columns.isin(cols_to_drop)]
所有回报：
a c e
0 0.758670 0.027711 0.955711
1 0.833110 0.922831 0.546079
2 0.414916 0.608527 0.230343
答案 2 :(得分：0)
如果您尝试使用NaN删除列，则下面的代码就足够了。好吧，我自己尝试了一下，而且效果很好。
df = df.dropna(axis = 1)