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问题：如何删除熊猫数据框的最后一行数据
回答 0
回答 1
回答 2
回答 3
回答 4
回答 5
回答 6

问题：如何删除熊猫数据框的最后一行数据

我认为这应该很简单，但是我尝试了一些想法，但都没有成功：

last_row = len(DF)
DF = DF.drop(DF.index[last_row])  #<-- fail!

我尝试使用负索引，但这也会导致错误。我仍然会误解一些基本知识。

点击查看英文原文

I think this should be simple, but I tried a few ideas and none of them worked:

last_row = len(DF)
DF = DF.drop(DF.index[last_row])  #<-- fail!

I tried using negative indices but that also lead to errors. I must still be misunderstanding something basic.

回答 0

要删除最后n行：

df.drop(df.tail(n).index,inplace=True) # drop last n rows

同样，您可以删除前n行：

df.drop(df.head(n).index,inplace=True) # drop first n rows
点击查看英文原文

To drop last n rows:

df.drop(df.tail(n).index,inplace=True) # drop last n rows

By the same vein, you can drop first n rows:

df.drop(df.head(n).index,inplace=True) # drop first n rows

回答 1

DF[:-n]

其中n是要删除的最后行数。

要删除最后一行：

DF = DF[:-1]
点击查看英文原文 
DF[:-n]

where n is the last number of rows to drop.

To drop the last row :

DF = DF[:-1]

回答 2

由于Python中的索引定位是基于0的，因此index与相对应的位置实际上没有元素len(DF)。您需要这样last_row = len(DF) - 1：

In [49]: dfrm
Out[49]: 
          A         B         C
0  0.120064  0.785538  0.465853
1  0.431655  0.436866  0.640136
2  0.445904  0.311565  0.934073
3  0.981609  0.695210  0.911697
4  0.008632  0.629269  0.226454
5  0.577577  0.467475  0.510031
6  0.580909  0.232846  0.271254
7  0.696596  0.362825  0.556433
8  0.738912  0.932779  0.029723
9  0.834706  0.002989  0.333436

[10 rows x 3 columns]

In [50]: dfrm.drop(dfrm.index[len(dfrm)-1])
Out[50]: 
          A         B         C
0  0.120064  0.785538  0.465853
1  0.431655  0.436866  0.640136
2  0.445904  0.311565  0.934073
3  0.981609  0.695210  0.911697
4  0.008632  0.629269  0.226454
5  0.577577  0.467475  0.510031
6  0.580909  0.232846  0.271254
7  0.696596  0.362825  0.556433
8  0.738912  0.932779  0.029723

[9 rows x 3 columns]

但是，编写起来要简单得多DF[:-1]。

点击查看英文原文

Since index positioning in Python is 0-based, there won’t actually be an element in index at the location corresponding to len(DF). You need that to be last_row = len(DF) - 1:

In [49]: dfrm
Out[49]: 
          A         B         C
0  0.120064  0.785538  0.465853
1  0.431655  0.436866  0.640136
2  0.445904  0.311565  0.934073
3  0.981609  0.695210  0.911697
4  0.008632  0.629269  0.226454
5  0.577577  0.467475  0.510031
6  0.580909  0.232846  0.271254
7  0.696596  0.362825  0.556433
8  0.738912  0.932779  0.029723
9  0.834706  0.002989  0.333436

[10 rows x 3 columns]

In [50]: dfrm.drop(dfrm.index[len(dfrm)-1])
Out[50]: 
          A         B         C
0  0.120064  0.785538  0.465853
1  0.431655  0.436866  0.640136
2  0.445904  0.311565  0.934073
3  0.981609  0.695210  0.911697
4  0.008632  0.629269  0.226454
5  0.577577  0.467475  0.510031
6  0.580909  0.232846  0.271254
7  0.696596  0.362825  0.556433
8  0.738912  0.932779  0.029723

[9 rows x 3 columns]

However, it’s much simpler to just write DF[:-1].

回答 3

没有人惊讶地提出这一点：

# To remove last n rows
df.head(-n)

# To remove first n rows
df.tail(-n)

对1000行的DataFrame进行速度测试表明，切片和head/ tail的速度比使用drop以下方法快约6倍：

>>> %timeit df[:-1]
125 µs ± 132 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

>>> %timeit df.head(-1)
129 µs ± 1.18 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

>>> %timeit df.drop(df.tail(1).index)
751 µs ± 20.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
点击查看英文原文

Surprised nobody brought this one up:

# To remove last n rows
df.head(-n)

# To remove first n rows
df.tail(-n)

Running a speed test on a DataFrame of 1000 rows shows that slicing and head/tail are ~6 times faster than using drop:

>>> %timeit df[:-1]
125 µs ± 132 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

>>> %timeit df.head(-1)
129 µs ± 1.18 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

>>> %timeit df.drop(df.tail(1).index)
751 µs ± 20.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

回答 4

stats = pd.read_csv("C:\\py\\programs\\second pandas\\ex.csv")

统计输出：

       A            B          C
0   0.120064    0.785538    0.465853
1   0.431655    0.436866    0.640136
2   0.445904    0.311565    0.934073
3   0.981609    0.695210    0.911697
4   0.008632    0.629269    0.226454
5   0.577577    0.467475    0.510031
6   0.580909    0.232846    0.271254
7   0.696596    0.362825    0.556433
8   0.738912    0.932779    0.029723
9   0.834706    0.002989    0.333436

只是使用 skipfooter=1

skipfooter：int，默认0

文件底部要跳过的行数

stats_2 = pd.read_csv("C:\\py\\programs\\second pandas\\ex.csv", skipfooter=1, engine='python')

stats_2的输出

       A          B            C
0   0.120064    0.785538    0.465853
1   0.431655    0.436866    0.640136
2   0.445904    0.311565    0.934073
3   0.981609    0.695210    0.911697
4   0.008632    0.629269    0.226454
5   0.577577    0.467475    0.510031
6   0.580909    0.232846    0.271254
7   0.696596    0.362825    0.556433
8   0.738912    0.932779    0.029723
点击查看英文原文 
stats = pd.read_csv("C:\\py\\programs\\second pandas\\ex.csv")

The Output of stats:

       A            B          C
0   0.120064    0.785538    0.465853
1   0.431655    0.436866    0.640136
2   0.445904    0.311565    0.934073
3   0.981609    0.695210    0.911697
4   0.008632    0.629269    0.226454
5   0.577577    0.467475    0.510031
6   0.580909    0.232846    0.271254
7   0.696596    0.362825    0.556433
8   0.738912    0.932779    0.029723
9   0.834706    0.002989    0.333436

just use skipfooter=1

skipfooter : int, default 0

Number of lines at bottom of file to skip

stats_2 = pd.read_csv("C:\\py\\programs\\second pandas\\ex.csv", skipfooter=1, engine='python')

Output of stats_2

       A          B            C
0   0.120064    0.785538    0.465853
1   0.431655    0.436866    0.640136
2   0.445904    0.311565    0.934073
3   0.981609    0.695210    0.911697
4   0.008632    0.629269    0.226454
5   0.577577    0.467475    0.510031
6   0.580909    0.232846    0.271254
7   0.696596    0.362825    0.556433
8   0.738912    0.932779    0.029723

回答 5

drop返回一个新数组，这就是为什么它在og post中阻塞了；由于将格式错误的csv文件转换为Dataframe，我对重命名某些列标题和删除某些行有类似的要求，因此在阅读本文后，我使用了：

newList = pd.DataFrame(newList)
newList.columns = ['Area', 'Price']
print(newList)
# newList = newList.drop(0)
# newList = newList.drop(len(newList))
newList = newList[1:-1]
print(newList)

而且效果很好，正如您在上面两行注释中看到的那样，我尝试了drop。（）方法，它可以工作，但不像使用[n：-n]那样简单易懂，希望对您有所帮助，谢谢。

点击查看英文原文

drop returns a new array so that is why it choked in the og post; I had a similar requirement to rename some column headers and deleted some rows due to an ill formed csv file converted to Dataframe, so after reading this post I used:

newList = pd.DataFrame(newList)
newList.columns = ['Area', 'Price']
print(newList)
# newList = newList.drop(0)
# newList = newList.drop(len(newList))
newList = newList[1:-1]
print(newList)

and it worked great, as you can see with the two commented out lines above I tried the drop.() method and it work but not as kool and readable as using [n:-n], hope that helps someone, thanks.

回答 6

对于具有多索引（例如“股票”和“日期”）且希望删除每个股票的最后一行而不只是最后一个股票的最后一行的更复杂的数据框，解决方案如下：

# To remove last n rows
df = df.groupby(level='Stock').apply(lambda x: x.head(-1)).reset_index(0, drop=True)

# To remove first n rows
df = df.groupby(level='Stock').apply(lambda x: x.tail(-1)).reset_index(0, drop=True)

由于会groupby()为Multi-Index添加一个附加级别，因此我们只需在末尾使用将其删除reset_index()。生成的df与操作之前保持相同的Multi-Index类型。

点击查看英文原文

For more complex DataFrames that have a Multi-Index (say “Stock” and “Date”) and one wants to remove the last row for each Stock not just the last row of the last Stock, then the solution reads:

# To remove last n rows
df = df.groupby(level='Stock').apply(lambda x: x.head(-1)).reset_index(0, drop=True)

# To remove first n rows
df = df.groupby(level='Stock').apply(lambda x: x.tail(-1)).reset_index(0, drop=True)

As the groupby() is adding an additional level to the Multi-Index we just drop it at the end using reset_index(). The resulting df keeps the same type of Multi-Index as before the operation.

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