输出：

Question

df1 = pd.DataFrame({'Chromosome': ['1A','1A','1A','1A','1A'],
              'Marker': ['M1','M2','M3','M4','M5'],
             'Position': [0,1.2,3.5,6,7.3]})
df2 = pd.DataFrame({'Chromosome': ['1A','1A','1A','1A','1A','1B','1B','1B'],
              'Marker': ['M1','M2','M3','M4','M5','mk1','mk2','mk3'],
             'Position': [0,1.2,3.5,6,7.3,0,2.3,3.2]})
#Expected result for df1
#'1A 5 M1 1.2 M2 2.3 M3 2.5 M4 1.3 M5'

#Expected result for df2
#'1A 5 M1 1.2 M2 2.3 M3 2.5 M4 1.3 M5'
#'1B 3 mk1 2.3 mk2 0.9 mk3' 


#My function for computing intermarker distance
def position_interval(df):
    df.loc[:,'diffPos'] = round(df['Position'].diff(),1).shift(-1)

a = []
i = 0
while i < df.shape[0]:#omit the last index
    info = df['Marker'][i]+' '+str(round(df['diffPos'][i],1))
    #print(info)
    a.append(info)
    i +=1
#print(a)
a.insert(0,str(len(df['Marker'])))
a.insert(0,df['Chromosome'][0])
new_info = ' '.join(a).replace(' nan','')#removing the last ' nan'
#print(new_info)    
return new_info

将功能应用于df1的效果很好：

position_interval(df1)

但是我不确定如何应用于每个grouby对象：

position_interval(df2)

Answer 1

由于该函数需要'Chromosome'键，因此必须将as_index=False参数放在groupby中：

df2.groupby('Chromosome', as_index=False).apply(position_interval)

这将引发异常，因为未为“ 1B”组找到索引0。

在函数中用iloc替换Series切片将解决此问题：

def position_interval(df): 
    df.loc[:,'diffPos'] = round(df['Position'].diff(),1).shift(-1)
    a = []
    i = 0
    while i < df.shape[0]:#omit the last index
        info = df['Marker'].iloc[0]+' '+str(round(df['diffPos'].iloc[i],1))
        #print(info)
        a.append(info)
        i +=1
    #print(a)
    a.insert(0,str(len(df['Marker'])))
    a.insert(0,df['Chromosome'].iloc[0])
    new_info = ' '.join(a).replace(' nan','')#removing the last ' nan'
    #print(new_info)    
    return new_info

输出：

1A 5 M1 1.2 M1 2.3 M1 2.5 M1 1.3 M1
1B 3 mk1 2.3 mk1 0.9 mk1

替代：

可以遍历groupby对象：

for i, sub_df in f2.groupby('Chromosome',as_index=False):
    print(position_interval(sub_df))

将自定义功能应用于熊猫groupby对象

1 个答案:

输出：

替代：