在 Pandas DataFrame 中调用 groupby 对象时出错 [英] Error when calling a groupby object inside a Pandas DataFrame

问题描述

我有这个数据框:

    person_code  #CNAE   growth   size 
0           231     32     0.54     32
1           233     43     0.12    333
2           432     32     0.44     21
3           431     56     0.32     23
4           654     89     0.12     89
5           764     32     0.20    211
6           434     32     0.82     90

我需要创建一个名为top3growth"的新列.为此，我需要为每一行检查 df 的 #CNAE 并添加一个额外的列，指出哪些是该 CNAE 增长最高的 3 个人(它将在 df 数据帧中添加一个数据帧).要创建top3dfs"，我正在使用这个 groupby:

I need to create a new column called "top3growth". For that I will need to check df's #CNAE for each row and add an extra column pointing out which are the 3 persons with highest growth for that CNAE (it will add a dataframe inside the df dataframe). To create the "top3dfs" I'm using this groupby:

a=sql2.groupby('#CNAE',group_keys=False).apply(pd.DataFrame.nlargest,n=3,columns='growth')

(这个解决方案来自 这个问题.)

(This solution came out of this question.)

它应该是这样的:

    person_code  #CNAE   growth   size              top3growth ...
0 .         231     32     0.54     32       [df_top3_type_32]
1 .         233     43     0.12    333       [df_top3_type_43]
2 .         432     32     0.44     21       [df_top3_type_32]                     
3 .         431     56     0.32     23       [df_top3_type_56]
4 .         654     89     0.12     89       [df_top3_type_89]
5 .         764     32     0.20    211       [df_top3_type_32]
6 .         434     32     0.82     90       [df_top3_type_32]
...

df_top3_type_32 应如下所示(例如):

df_top3_type_32 should look like this (for example):

     person_code  #CNAE  growth  size
6 .          434    32    0.82    90
0 .          231    32    0.54    32
2 .          432    32    0.44    21

我正在尝试使用以下方法解决我的问题:

I'm trying to solve my problem by using:

df['top3growth']=np.nan
for i in df.index:
    df['top3growth'].loc[i]=a[a['#CNAE'] == df['#CNAE'].loc[i]]

但我得到:

ValueError: Incompatible indexer with DataFrame

有人知道这是怎么回事吗?有没有更有效的方法(不使用 for 循环)?

Does anyone know what's going on? Is there a more efficient way of doing this (not using a for loop)?

推荐答案

有一种方法，将 a 转换为 dict ，然后再映射回来

There is one way, convert a to dict , then map it back

#a=df.groupby('#CNAE',group_keys=False).apply(pd.DataFrame.nlargest,n=3,columns='growth')
df['top3growth']=df['#CNAE'].map(a.groupby('#CNAE').apply(lambda x : x.to_dict()))
df
Out[195]: 
   person_code  #CNAE  growth  size  \
0          231     32    0.54    32   
1          233     43    0.12   333   
2          432     32    0.44    21   
3          431     56    0.32    23   
4          654     89    0.12    89   
5          764     32    0.20   211   
6          434     32    0.82    90   
                                          top3growth  
0  {'person_code': {0: 231, 2: 432, 6: 434}, 'gro...  
1  {'person_code': {1: 233}, 'growth': {1: 0.12},...  
2  {'person_code': {0: 231, 2: 432, 6: 434}, 'gro...  
3  {'person_code': {3: 431}, 'growth': {3: 0.32},...  
4  {'person_code': {4: 654}, 'growth': {4: 0.12},...  
5  {'person_code': {0: 231, 2: 432, 6: 434}, 'gro...  
6  {'person_code': {0: 231, 2: 432, 6: 434}, 'gro...  

创建新列后，如果要将单个单元格转换回数据框

After create your new column , if you want to convert the single cell back to dataframe

pd.DataFrame(df.top3growth[0])
Out[197]: 
   #CNAE  growth  person_code  size
0     32    0.54          231    32
2     32    0.44          432    21
6     32    0.82          434    90

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