繁体   English   中英

如何提取指定列值组合重复的数据帧的行?

[英]How to extract the rows of a dataframe where a combination of specified column values are duplicated?

假设我有以下数据框:

import pandas as pd
data = {'Year':[2018, 2018, 2018, 2018, 2018, 2018, 2018, 2018],
        'Month':[1,1,1,2,2,3,3,3],
        'ID':['A', 'A', 'B', 'A', 'B', 'A', 'B', 'B'],
        'Fruit':['Apple', 'Banana', 'Apple', 'Pear', 'Mango', 'Banana', 'Apple', 'Mango']}
df = pd.DataFrame(data, columns=['Year', 'Month', 'ID', 'Fruit'])
df = df.astype(str)
df

在此处输入图片说明

我想提取重复的“年”、“月”和“ID”的组合。 因此,使用上述数据框,预期结果是此数据框:

在此处输入图片说明

我这样做的方法是先做一个groupby来计算YearMonthID的组合出现的次数:

df2 = df.groupby(['Year', 'Month'])['ID'].value_counts().to_frame(name = 'Count').reset_index()
df2 = df2[df2.Count>1]
df2

在此处输入图片说明

然后,我的想法是遍历 groupby 数据框中的YearMonthID组合,并将与原始数据框中的组合匹配的那些行提取到一个新的数据框中:

df_new = pd.DataFrame(columns=df.columns, index=range(sum(df2.Count)))

count = 0
for i in df2.index:
    temp = df[(df.ID==df2.ID[i]) & (df.Year==df2.Year[i]) & (df.Month==df2.Month[i])]
    temp.reset_index(drop=True, inplace=True)
    for j in range(len(temp)):
        df_new.iloc[count] = temp.iloc[j]
        count+=1
df_new

但这会产生以下错误:

---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-38-7f2d95d71270> in <module>()
      6     temp.reset_index(drop=True, inplace=True)
      7     for j in range(len(temp)):
----> 8         df_new.iloc[count] = temp.iloc[j]
      9         count+=1
     10 df_new

c:\users\h473\appdata\local\programs\python\python35\lib\site-packages\pandas\core\indexing.py in __setitem__(self, key, value)
    187         else:
    188             key = com.apply_if_callable(key, self.obj)
--> 189         indexer = self._get_setitem_indexer(key)
    190         self._setitem_with_indexer(indexer, value)
    191 

c:\users\h473\appdata\local\programs\python\python35\lib\site-packages\pandas\core\indexing.py in _get_setitem_indexer(self, key)
    173 
    174         try:
--> 175             return self._convert_to_indexer(key, is_setter=True)
    176         except TypeError as e:
    177 

c:\users\h473\appdata\local\programs\python\python35\lib\site-packages\pandas\core\indexing.py in _convert_to_indexer(self, obj, axis, is_setter)
   2245 
   2246         try:
-> 2247             self._validate_key(obj, axis)
   2248             return obj
   2249         except ValueError:

c:\users\h473\appdata\local\programs\python\python35\lib\site-packages\pandas\core\indexing.py in _validate_key(self, key, axis)
   2068             return
   2069         elif is_integer(key):
-> 2070             self._validate_integer(key, axis)
   2071         elif isinstance(key, tuple):
   2072             # a tuple should already have been caught by this point

c:\users\h473\appdata\local\programs\python\python35\lib\site-packages\pandas\core\indexing.py in _validate_integer(self, key, axis)
   2137         len_axis = len(self.obj._get_axis(axis))
   2138         if key >= len_axis or key < -len_axis:
-> 2139             raise IndexError("single positional indexer is out-of-bounds")
   2140 
   2141     def _getitem_tuple(self, tup):

IndexError: single positional indexer is out-of-bounds

有什么错误? 我无法弄清楚。

当我将for循环的内容更改为以下内容时,错误消失了,这会产生所需的结果:

for j in range(len(temp)):
    df_new.ID[count] = temp.ID[j]
    df_new.Year[count] = temp.Year[j]
    df_new.Month[count] = temp.Month[j]
    df_new.Fruit[count] = temp.Fruit[j]
    count+=1

但这是一种繁琐的解决方法,涉及为原始数据帧中的n列中的每一列编写n行。

使用GroupBy.transform与任何列和计算由GroupBy.size的系列与原来一样大小相同,通过这样可以过滤boolean indexing

df1 = df[df.groupby(['Year','Month','ID'])['ID'].transform('size') > 1]

或者,如果小DataFrame或性能不重要,请使用DataFrameGroupBy.filter

df1 = df.groupby(['Year','Month','ID']).filter(lambda x: len(x) > 1)

print (df1)

   Year  Month ID   Fruit
0  2018      1  A   Apple
1  2018      1  A  Banana
6  2018      3  B   Apple
7  2018      3  B   Mango

您可以使用参数keep=False duplicated的方法来选择所有重复项:

df[df.duplicated(subset=['Year', 'Month', 'ID'], keep=False)]

输出:

   Year Month ID   Fruit
0  2018     1  A   Apple
1  2018     1  A  Banana
6  2018     3  B   Apple
7  2018     3  B   Mango

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM