也可能：

out = df.loc[ df["col1"].ne(3) | df["col1"].duplicated(keep="last") ]

输出：

col1  col2
0     0     0
1     1    11
2     2    21
3     3    31
4     3    32
5     3    33
7     4    41
8     5    51
9     6    61

您可以用途：

>>> df.drop(df['col1'].where(df['col1'].eq(3)).last_valid_index())
   col1  col2
0     0     0
1     1    11
2     2    21
3     3    31
4     3    32
5     3    33
7     4    41
8     5    51
9     6    61

您可以用途：

# get the last index of 3 in col1
idx = df.loc[::-1, 'col1'].eq(3).idxmax()

# if there was no 3 in col1, this would give a false positive
# idxmax would return the last non-3 instead
# ensure that we drop the correct row
if df.loc[idx, 'col1'] == 3:
    df = df.drop(idx)

• 注意：如果您已经知道3在col 1中，那么简单的df = df.drop(idx)就足够了。*

输出：

col1  col2
0     0     0
1     1    11
2     2    21
3     3    31
4     3    32
5     3    33
7     4    41
8     5    51
9     6    61

计时比较

Dataframe 大小从8到33 M行。所有答案的时间都是相似的，除了Chrysophylaxs的时间更快（rhug 123和Corralian/LaurentB的时间分别对于大/小 Dataframe 更慢）。

下面是另一个解决方案：

df.loc[df['col1'].iloc[::-1].ne(3).rank(method = 'first').ne(1)]

输出：

col1  col2
0     0     0
1     1    11
2     2    21
3     3    31
4     3    32
5     3    33
7     4    41
8     5    51
9     6    61

df.drop(index=df[df['col1'].eq(3)].index[-1:], axis=0)

col1  col2
0     0     0
1     1    11
2     2    21
3     3    31
4     3    32
5     3    33
7     4    41
8     5    51
9     6    61

编辑：
筛选最后一个3的索引并将其删除：

df1 = df.drop(df.index[df['col1'].eq(3)][-1])

如果可能，始终存在值3，您可以找到索引并删除它：

df1 = df.drop((df['col1'].iloc[::-1] == 3).idxmax())

在numpy里也是这样：

df1 = df.drop(np.argwhere(df['col1'].to_numpy() == 3)[-1])

计时非常相似：

#Last value of 1M is 3
np.random.seed(100)
df = pd.DataFrame({'col1': np.random.randint(100, size=1000000)})
df.loc[len(df), 'col1'] = 3
#print (df)

In [261]: %timeit df.loc[ df["col1"].ne(3) | df["col1"].duplicated(keep="last") ]
43.8 ms ± 479 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [262]: %timeit df.drop(df[df['col1'].eq(3)].index[-1])
44.4 ms ± 120 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [263]: %timeit df.drop(df[df['col1'].eq(3)].index[-1:])
44.8 ms ± 490 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [264]: %timeit df.drop((df['col1'].iloc[::-1] == 3).idxmax())
44.6 ms ± 1.62 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [265]: %timeit df.drop(np.argwhere(df['col1'].to_numpy() == 3)[-1])
43.3 ms ± 422 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [266]: %timeit df.drop(df['col1'].where(df['col1'].eq(3)).last_valid_index())
64.3 ms ± 11.3 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [267]: %timeit df.loc[df['col1'].iloc[::-1].ne(3).rank(method = 'first').ne(1)]
168 ms ± 2.59 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)