首页 > 解决方案 > Pandas 删除后重复索引

问题描述

我得到:“ValueError:索引包含重复的条目,无法重塑”

我正在使用的数据非常庞大,我无法提供样本数据,也无法用较小的数据集复制错误。我试图用虚拟数据生成副本以复制我的原始帧,但由于某种神秘的原因,该代码仅适用于虚拟数据,而不适用于我的真实数据。这就是我所知道的我正在使用的形状。


df.shape

>> (6820, 26) 

df.duplicated()

>> 0       False
>> 1       False
>> 2       False
>>        ...  
>> 6818    False
>> 6819    False
>> Length: 6820, dtype: bool

现在我想找出哪些行是重复的。

df[df.duplicated(keep=False)]

>> 0 rows × 26 columns

只是为了确保我删除所有重复项并只保留第一个:

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

这是我收到 ValueError 的时候:

df2 = df.melt('Release')\
        .assign(variable = lambda x: x.variable.map({'Created Date':1,'Finished Date':-1}))\
        .pivot('value','Release','variable').fillna(0)\
        .rename(columns = lambda c: f'{c} netmov' )


---> 33         .pivot('value','Release','variable').fillna(0)\
ValueError: Index contains duplicate entries, cannot reshape

从进一步调查来看,似乎不是重复的行,而是索引。我尝试使用 df.reset_index() 重置索引,但它会引发相同的 ValueError。

编辑:

我可以提供应该复制我正在使用的框架的虚拟数据(只需少几列不需要)

df = pd.DataFrame({'name': ["Peter", "Anna", "Anna", "Peter", "Simon", "Johan", "Nils", "Oskar", "Peter"]
                  , 'Deposits': ["2019-03-07", "2019-03-08", "2019-03-12", "2019-03-12", "2019-03-14", "2019-03-07", "2019-03-08", "2016-03-07", "2019-03-07"]
                  , 'Withdrawals': ["2019-03-11", "2019-03-19", "2019-05-22", "2019-10-31", "2019-04-05", "2019-03-11", "NaN", "2017-03-06", "2019-03-11"]})

df.duplicated()

0    False
1    False
2    False
.....
7    False
8     True
dtype: bool

df = df.drop_duplicates(keep='first')
df2 = df.melt('name')\
        .assign(variable = lambda x: x.variable.map({'Deposits':1,'Withdrawals':-1}))\
        .pivot('value','name','variable').fillna(0)\
        .rename(columns = lambda c: f'{c} netmov' )

df2 = pd.concat([df2,df2.cumsum().rename(columns = lambda c: c.split()[0] + ' balance')], axis = 1)\
        .sort_index(axis=1)


print(df2.head())

name        Anna balance  Anna netmov  Johan balance  Johan netmov  \
value                                                                
2016-03-07           0.0          0.0            0.0           0.0   
2017-03-06           0.0          0.0            0.0           0.0   
2019-03-07           0.0          0.0            1.0           1.0   
2019-03-08           1.0          1.0            1.0           0.0   
2019-03-11           1.0          0.0            0.0          -1.0

即使 DataFrame 中有重复项,这也会顺利运行。

最好我也不想删除重复项,因为“安娜”一天内可能进行了 4 次存款和 4 次取款,所以我想计算所有这些。

我正在使用的数据框:


df = df.drop_duplicates().reset_index(drop=True)
df = df.drop(['id'], axis=1)
df

Output:

        name    Deposits     Withdrawals
0       Anna    2020-07-31   NaN
1       Peter   2020-07-30   NaN
2       Simon   2020-07-30   NaN
3       Simon   2020-07-29   NaN
4       Simon   2020-07-29   NaN
... ... ... ...
6154    Peter   2014-01-22  2014-02-03
6155    Peter   2014-01-22  2014-01-29
6156    Peter   2014-01-22  2014-01-24
6157    Peter   2014-01-21  2014-01-29
6158    Peter   2014-01-15  2014-02-03
6159 rows × 3 columns

更新:向社区大喊帮助我解决这个问题。

这解决了这个问题:

df.Deposits = pd.to_datetime(df.Deposits)
df.Withdrawals = pd.to_datetime(df.Withdrawals)

df2 = (
    df.melt('name') 
    .assign(variable = lambda x: x.variable.map({'Deposits':1,'Withdrawals':-1}))
    .dropna(subset=['value']) # you need this for cases like Nils's Withdrawal
    )
df2 = df2.groupby(['value', 'name']).sum().unstack(fill_value=0).droplevel(0, axis=1)


df2 = (
    pd.concat([df2, df2.cumsum()], keys=['netmov', 'balance'], axis=1)
     notice how concat has the functionality you want for naming columns
     and is a better idea to have netmov/balance in a separate level
     in case you want to groupby or .loc later on
    .reorder_levels([1, 0], axis=1).sort_index(axis=1)
    )

偶然发现下一个问题,与此无关。当将此 DataFrame 转换为 json 时,它会出于某种原因将日期转换为另一种格式。

data = df2.to_json()
print(data)

{
    "Peter":
    {
        "1389744000000": 0,
        "1390262400000": 0,
        "1390348800000": 0,
        "1390521600000": 0,
    .....
    .....
    }
}

总是别的东西,呵呵..尽管为帮助欢呼,我几乎可以触及球门线。

标签: python-3.xpandasduplicatespivot

解决方案


当一个名称在完全相同的存款/取款日期(因此重复)有多个移动时,似乎会出现问题。Dataframe.pivot方法不能处理重复的索引,它只是不是为此而设计的。为了您的分析目的.pivot_table,主要区别在于这个可以应用聚合函数来处理重复索引(在这种情况下为总和)。https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.pivot_table.html

我个人倾向于使用.groupby任何此类问题,因为它不仅提供了通过 df 中的任何列组合进行分组的功能,还可以包括外生序列、计算、索引或自我或其他索引的级别、掩码等.

所以我的代码是:

df.Deposits = pd.to_datetime(df.Deposits)
df.Withdrawals = pd.to_datetime(df.Withdrawals) # this parsing probably happens in read_csv
df2 = (
    df.melt('name') 
    .assign(variable = lambda x: x.variable.map({'Deposits':1, 'Withdrawals':-1}))
    # use lambda if you must
    # replace on 'variable' after creating df2 would also work
    # and is probably faster for larger dfs
    .dropna(subset=['value']) # you need this for cases like Nils's Withdrawal
    )
df2 = df2.groupby(['value', 'name']).sum().unstack(fill_value=0).droplevel(0, axis=1)
df2 = (
    pd.concat([df2, df2.cumsum()], keys=['netmov', 'balance'], axis=1)
    # notice how concat has the functionality you want for naming columns
    # and is a better idea to have netmov/balance in a separate level
    # in case you want to groupby or .loc later on
    .reorder_levels([1, 0], axis=1).sort_index(axis=1)
    )

输出

name          Anna          Johan           Nils  ...  Oskar   Peter          Simon
           balance netmov balance netmov balance  ... netmov balance netmov balance netmov
value                                             ...
2016-03-07       0      0       0      0       0  ...      1       0      0       0      0
2017-03-06       0      0       0      0       0  ...     -1       0      0       0      0
2019-03-07       0      0       1      1       0  ...      0       2      2       0      0
2019-03-08       1      1       1      0       1  ...      0       2      0       0      0
2019-03-11       1      0       0     -1       1  ...      0       0     -2       0      0
2019-03-12       2      1       0      0       1  ...      0       1      1       0      0
2019-03-14       2      0       0      0       1  ...      0       1      0       1      1
2019-03-19       1     -1       0      0       1  ...      0       1      0       1      0
2019-04-05       1      0       0      0       1  ...      0       1      0       0     -1
2019-05-22       0     -1       0      0       1  ...      0       1      0       0      0
2019-10-31       0      0       0      0       1  ...      0       0     -1       0      0

推荐阅读