python pandas申请循环和groupby功能

时间:2013-11-21 07:55:37

标签: python pandas iteration grouping

我是python的新手,我不熟悉pandas中的groupby函数 我修改了下面的代码,它可以很好地创建一个pandas数据帧

i=['J,Smith,200 G Ct,',
'E,Johnson,200 G Ct,',
'A,Johnson,200 G Ct,',
'M,Simpson,63 F Wy,',
'L,Diablo,60 N Blvd,',
'H,Simpson,63 F Wy,',
'B,Simpson,63 F Wy,']

dbn=[]
dba=[]

for z,g in groupby(
    sorted([l.split(',')for l in i],
    key=lambda x:x[1:]),
    lambda x:x[2:]
):

 l=list(g);r=len(l);Address=','.join(z);o=l[0]
 if r>2:
    dbn.append('The '+o[1]+" Family,")
    dba.append(Address)
 elif r>1:
    dbn.append(o[0]+" and "+l[1][0]+", "+o[1]+",")
    dba.append(Address)
 else:
    dbn.append(o[0]+" "+o[1])
#   print','.join(o),
    dba.append(Address)

Hdf=pd.DataFrame({'Address':dba,'Name':dbn})
print Hdf

      Address                 Name
0  60 N Blvd,             L Diablo
1   200 G Ct,    E and A, Johnson,
2    63 F Wy,  The Simpson Family,
3   200 G Ct,              J Smith

如果我使用pandas数据帧而不是原始csv数据,如何修改for循环以产生相同的结果?

df=pd.DataFrame({'Name':['J','E','A','M','L','H','B'],
'Lastname':['Smith','Johnson','Johnson','Simpson','Diablo','Simpson','Simpson'],
'Address':['200 G Ct','200 G Ct','200 G Ct','63 F Wy','60 N Blvd','63 F Wy','63 F Wy']})

1 个答案:

答案 0 :(得分:1)

带循环/生成器的版本:

首先,我们按Lastname, Address

创建辅助函数和组数据
def helper(k, g):
    r = len(g)
    address, lastname = k
    if r > 2:
        lastname = 'The {} Family'.format(lastname)
    elif r > 1:
        lastname = ' and '.join(g['Name']) + ', ' + lastname
    else:
        lastname = g['Name'].squeeze() + ' ' + lastname
    return (address, lastname)

grouped = df.groupby(['Address', 'Lastname'])

然后创建具有应用于每个组的辅助函数的生成器:

vals = (helper(k, g) for k, g in grouped)

然后从中创建生成的DataFrame:

pd.DataFrame(vals, columns=['Address','Name'])

     Address                Name
0   200 G Ct    E and A, Johnson
1   200 G Ct             J Smith
2  60 N Blvd            L Diablo
3    63 F Wy  The Simpson Family

更多矢量化版本:

Lastname, Address分组数据,然后生成长度为group的新DataFrame,字符串包含两个连接的名字:

grouped = df.groupby(['Address', 'Lastname'])
res = grouped.apply(lambda x: pd.Series({'Len': len(x), 'Names': ' and '.join(x['Name'][:2])})).reset_index()

     Address Lastname  Len    Names
0   200 G Ct  Johnson    2  E and A
1   200 G Ct    Smith    1        J
2  60 N Blvd   Diablo    1        L
3    63 F Wy  Simpson    3  M and H

现在只需应用常规的pandas转换并删除不必要的列:

res.ix[res['Len'] > 2, 'Lastname'] = 'The ' + res['Lastname'] + ' Family'
res.ix[res['Len'] == 2, 'Lastname'] = res['Names'] + ', ' + res['Lastname']
res.ix[res['Len'] < 2, 'Lastname'] = res['Names'] + ' ' + res['Lastname']
del res['Len']
del res['Names']

     Address            Lastname
0   200 G Ct    E and A, Johnson
1   200 G Ct             J Smith
2  60 N Blvd            L Diablo
3    63 F Wy  The Simpson Family