我如何在python中使用熊猫获得所有重复项的列表?

我有一份可能有出口问题的商品清单。我想获得重复项目的列表，以便我可以手动比较它们。当我尝试使用熊猫复制方法时，它只返回第一个副本。有没有办法得到所有的副本，而不只是第一个?

我的数据集的一小部分是这样的:

ID,ENROLLMENT_DATE,TRAINER_MANAGING,TRAINER_OPERATOR,FIRST_VISIT_DATE
1536D,12-Feb-12,"06DA1B3-Lebanon NH",,15-Feb-12
F15D,18-May-12,"06405B2-Lebanon NH",,25-Jul-12
8096,8-Aug-12,"0643D38-Hanover NH","0643D38-Hanover NH",25-Jun-12
A036,1-Apr-12,"06CB8CF-Hanover NH","06CB8CF-Hanover NH",9-Aug-12
8944,19-Feb-12,"06D26AD-Hanover NH",,4-Feb-12
1004E,8-Jun-12,"06388B2-Lebanon NH",,24-Dec-11
11795,3-Jul-12,"0649597-White River VT","0649597-White River VT",30-Mar-12
30D7,11-Nov-12,"06D95A3-Hanover NH","06D95A3-Hanover NH",30-Nov-11
3AE2,21-Feb-12,"06405B2-Lebanon NH",,26-Oct-12
B0FE,17-Feb-12,"06D1B9D-Hartland VT",,16-Feb-12
127A1,11-Dec-11,"064456E-Hanover NH","064456E-Hanover NH",11-Nov-12
161FF,20-Feb-12,"0643D38-Hanover NH","0643D38-Hanover NH",3-Jul-12
A036,30-Nov-11,"063B208-Randolph VT","063B208-Randolph VT",
475B,25-Sep-12,"06D26AD-Hanover NH",,5-Nov-12
151A3,7-Mar-12,"06388B2-Lebanon NH",,16-Nov-12
CA62,3-Jan-12,,,
D31B,18-Dec-11,"06405B2-Lebanon NH",,9-Jan-12
20F5,8-Jul-12,"0669C50-Randolph VT",,3-Feb-12
8096,19-Dec-11,"0649597-White River VT","0649597-White River VT",9-Apr-12
14E48,1-Aug-12,"06D3206-Hanover NH",,
177F8,20-Aug-12,"063B208-Randolph VT","063B208-Randolph VT",5-May-12
553E,11-Oct-12,"06D95A3-Hanover NH","06D95A3-Hanover NH",8-Mar-12
12D5F,18-Jul-12,"0649597-White River VT","0649597-White River VT",2-Nov-12
C6DC,13-Apr-12,"06388B2-Lebanon NH",,
11795,27-Feb-12,"0643D38-Hanover NH","0643D38-Hanover NH",19-Jun-12
17B43,11-Aug-12,,,22-Oct-12
A036,11-Aug-12,"06D3206-Hanover NH",,19-Jun-12

我的代码现在是这样的:

df_bigdata_duplicates = df_bigdata[df_bigdata.duplicated(cols='ID')]

有两个重复的项目。但是，当我使用上面的代码时，我只得到第一项。在API引用中，我看到了如何获得最后一项，但我想拥有所有这些项，这样我就可以直观地检查它们，看看为什么我得到了差异。因此，在这个示例中，我希望获得所有三个A036条目和11795条目以及任何其他重复条目，而不仅仅是第一个条目。任何帮助都非常感激。

当前回答

方法#1:打印所有ID是重复ID之一的行:

>>> import pandas as pd
>>> df = pd.read_csv("dup.csv")
>>> ids = df["ID"]
>>> df[ids.isin(ids[ids.duplicated()])].sort_values("ID")
       ID ENROLLMENT_DATE        TRAINER_MANAGING        TRAINER_OPERATOR FIRST_VISIT_DATE
24  11795       27-Feb-12      0643D38-Hanover NH      0643D38-Hanover NH        19-Jun-12
6   11795        3-Jul-12  0649597-White River VT  0649597-White River VT        30-Mar-12
18   8096       19-Dec-11  0649597-White River VT  0649597-White River VT         9-Apr-12
2    8096        8-Aug-12      0643D38-Hanover NH      0643D38-Hanover NH        25-Jun-12
12   A036       30-Nov-11     063B208-Randolph VT     063B208-Randolph VT              NaN
3    A036        1-Apr-12      06CB8CF-Hanover NH      06CB8CF-Hanover NH         9-Aug-12
26   A036       11-Aug-12      06D3206-Hanover NH                     NaN        19-Jun-12

但我想不出一个好办法来防止重复这么多次id。我更喜欢方法#2:在ID上groupby。

>>> pd.concat(g for _, g in df.groupby("ID") if len(g) > 1)
       ID ENROLLMENT_DATE        TRAINER_MANAGING        TRAINER_OPERATOR FIRST_VISIT_DATE
6   11795        3-Jul-12  0649597-White River VT  0649597-White River VT        30-Mar-12
24  11795       27-Feb-12      0643D38-Hanover NH      0643D38-Hanover NH        19-Jun-12
2    8096        8-Aug-12      0643D38-Hanover NH      0643D38-Hanover NH        25-Jun-12
18   8096       19-Dec-11  0649597-White River VT  0649597-White River VT         9-Apr-12
3    A036        1-Apr-12      06CB8CF-Hanover NH      06CB8CF-Hanover NH         9-Aug-12
12   A036       30-Nov-11     063B208-Randolph VT     063B208-Randolph VT              NaN
26   A036       11-Aug-12      06D3206-Hanover NH                     NaN        19-Jun-12

2013-02-02 01:01:09

其他回答

df[df['ID'].duplicated() == True]

这对我很有效

2018-11-19 15:06:27

这段代码提供给你一个数据帧，指示一行在数据帧中是否有重复:

df2 = df1.duplicated()

这段代码消除了重复，只保留了一个实例:

df3 = df1.drop_duplicates(keep="first")

Df3将是一个由唯一项(行)组成的数据帧。

2022-08-23 10:26:25

方法#1:打印所有ID是重复ID之一的行:

>>> import pandas as pd
>>> df = pd.read_csv("dup.csv")
>>> ids = df["ID"]
>>> df[ids.isin(ids[ids.duplicated()])].sort_values("ID")
       ID ENROLLMENT_DATE        TRAINER_MANAGING        TRAINER_OPERATOR FIRST_VISIT_DATE
24  11795       27-Feb-12      0643D38-Hanover NH      0643D38-Hanover NH        19-Jun-12
6   11795        3-Jul-12  0649597-White River VT  0649597-White River VT        30-Mar-12
18   8096       19-Dec-11  0649597-White River VT  0649597-White River VT         9-Apr-12
2    8096        8-Aug-12      0643D38-Hanover NH      0643D38-Hanover NH        25-Jun-12
12   A036       30-Nov-11     063B208-Randolph VT     063B208-Randolph VT              NaN
3    A036        1-Apr-12      06CB8CF-Hanover NH      06CB8CF-Hanover NH         9-Aug-12
26   A036       11-Aug-12      06D3206-Hanover NH                     NaN        19-Jun-12

但我想不出一个好办法来防止重复这么多次id。我更喜欢方法#2:在ID上groupby。

>>> pd.concat(g for _, g in df.groupby("ID") if len(g) > 1)
       ID ENROLLMENT_DATE        TRAINER_MANAGING        TRAINER_OPERATOR FIRST_VISIT_DATE
6   11795        3-Jul-12  0649597-White River VT  0649597-White River VT        30-Mar-12
24  11795       27-Feb-12      0643D38-Hanover NH      0643D38-Hanover NH        19-Jun-12
2    8096        8-Aug-12      0643D38-Hanover NH      0643D38-Hanover NH        25-Jun-12
18   8096       19-Dec-11  0649597-White River VT  0649597-White River VT         9-Apr-12
3    A036        1-Apr-12      06CB8CF-Hanover NH      06CB8CF-Hanover NH         9-Aug-12
12   A036       30-Nov-11     063B208-Randolph VT     063B208-Randolph VT              NaN
26   A036       11-Aug-12      06D3206-Hanover NH                     NaN        19-Jun-12

2013-02-02 01:01:09

使用元素逻辑或并将pandas duplicate方法的take_last参数设置为True和False，您可以从您的数据帧中获得一个包含所有副本的集合。

df_bigdata_duplicates = 
    df_bigdata[df_bigdata.duplicated(cols='ID', take_last=False) |
               df_bigdata.duplicated(cols='ID', take_last=True)
              ]

2015-06-17 13:38:55

你可以用:

df[df.duplicated(['ID'])==True].sort_values('ID')

复制的行及其所有列值的索引loc#

def dup_rows_index(df):
  dup = df[df.duplicated()]
  print('Duplicated index loc:',dup[dup == True ].index.tolist())
  return dup

2019-04-09 07:53:45

我如何在python中使用熊猫获得所有重复项的列表?

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