Question

我有一个数据框

df = pd.DataFrame({'id':['one','one','two','two','three','three','three'],
                   'type':['current','saving','current','current','current','saving','credit']})

我想计算只有'当前'的id数应该是这样的：

only_currnt_id_list = ['two']

Answer 1

我认为你需要：

L = df.groupby('id') \
      .filter(lambda x: (x['type'] == 'current').all() and 
                        (x['type'] == 'current').sum() == 1)['id'].tolist()
print (L)

['two']

编辑：

df = pd.DataFrame({'id':['one','one','two','three','three','three'],'type':['current','current','current','current','saving','credit']})
print (df)
      id     type
0    one  current
1    one  current
2    two  current
3  three  current
4  three   saving
5  three   credit

L = df.groupby('id') \
      .filter(lambda x: (x['type'] == 'current').all() and 
                        (x['type'] == 'current').sum() == 1)['id'].tolist()
print (L)
['two']

L = df.groupby('id') \
      .filter(lambda x: (x['type'] == 'current').all())['id'].unique().tolist()
print (L)
['one', 'two']

Answer 2

使用pd.crosstab

尝试此操作

df=pd.crosstab(df.id,df.type)
df.loc[df.sum(1)==df.current,].index.values[0]

Out[1065]: 'two'

或者您可以使用groupby和nunique

df['unique']=df.groupby('id')['type'].transform('nunique')

df.loc[(df.unique==1)&(df.type=='current'),:].id.unique().tolist()


Out[1085]: ['two']

Answer 3

不使用纯Pandas，但您可以使用set所有ID和ID之间的type != 'current'差异：

>>> set(df["id"]) - set(df["id"][df["type"] != "current"])
{2}

Python groupby结果计数频率

3 个答案: