使用drop_duplicate和keep参数，这在pandas中要容易得多。

import pandas as pd
df = pd.DataFrame({"A":["foo", "foo", "foo", "bar"], "B":[0,1,1,1], "C":["A","A","B","A"]})
df.drop_duplicates(subset=['A', 'C'], keep=False)

试试这些不同的方法

df = pd.DataFrame({"A":["foo", "foo", "foo", "bar","foo"], "B":[0,1,1,1,1], "C":["A","A","B","A","A"]})

>>>df.drop_duplicates( "A" , keep='first')

or

>>>df.drop_duplicates( keep='first')

or

>>>df.drop_duplicates( keep='last')

实际上，删除行0和1只需要(任何包含匹配的A和C的观测值都被保留):

In [335]:

df['AC']=df.A+df.C
In [336]:

print df.drop_duplicates('C', take_last=True) #this dataset is a special case, in general, one may need to first drop_duplicates by 'c' and then by 'a'.
     A  B  C    AC
2  foo  1  B  fooB
3  bar  1  A  barA

[2 rows x 4 columns]

但我怀疑你真正想要的是这个(一个包含匹配的A和C的观察结果被保留):

In [337]:

print df.drop_duplicates('AC')
     A  B  C    AC
0  foo  0  A  fooA
2  foo  1  B  fooB
3  bar  1  A  barA

[3 rows x 4 columns]

编辑:

因此，现在情况清楚多了:

In [352]:
DG=df.groupby(['A', 'C'])   
print pd.concat([DG.get_group(item) for item, value in DG.groups.items() if len(value)==1])
     A  B  C
2  foo  1  B
3  bar  1  A

[2 rows x 3 columns]

如果你想将结果存储在另一个数据集中:

df.drop_duplicates(keep=False)

or

df.drop_duplicates(keep=False, inplace=False)

如果需要更新相同的数据集:

df.drop_duplicates(keep=False, inplace=True)

上面的例子将删除所有重复项并保留一个，类似于SQL中的DISTINCT *

您可以使用replicated()标记所有重复的行，并过滤掉标记的行。如果你以后需要将列赋值给new_df，确保调用.copy()，这样你以后就不会得到SettingWithCopyWarning。

new_df = df[~df.duplicated(subset=['A', 'C'], keep=False)].copy()

该方法的一个很好的特性是，您可以有条件地使用它删除重复项。例如，仅当列A等于'foo'时删除所有重复的行，您可以使用以下代码。

new_df = df[~( df.duplicated(subset=['A', 'B', 'C'], keep=False) & df['A'].eq('foo') )].copy()

此外，如果您不希望按名称写出列，您可以传递df的切片。列到子集=。对于drop_duplicate()也是如此。

# to consider all columns for identifying duplicates
df[~df.duplicated(subset=df.columns, keep=False)].copy()

# the same is true for drop_duplicates
df.drop_duplicates(subset=df.columns, keep=False)

# to consider columns in positions 0 and 2 (i.e. 'A' and 'C') for identifying duplicates
df.drop_duplicates(subset=df.columns[[0, 2]], keep=False)

使用groupby和filter

import pandas as pd
df = pd.DataFrame({"A":["foo", "foo", "foo", "bar"], "B":[0,1,1,1], "C":["A","A","B","A"]})
df.groupby(["A", "C"]).filter(lambda df:df.shape[0] == 1)