我有一个商店名称的数据框,我正在尝试标准化。在这里测试的小样本:
import pandas as pd
df = pd.DataFrame({'store': pd.Series(['McDonalds', 'Lidls', 'Lidl New York 123', 'KFC', 'Lidi Berlin', 'Wallmart LA 90210', 'Aldi', 'London Lidl', 'Aldi627', 'mcdonaldsabc123', 'Mcdonald_s', 'McDonalds12345', 'McDonalds5555', 'McDonalds888', 'Aldi123', 'KFC-786', 'KFC-908', 'McDonalds511', 'GerALDInes Shop'],dtype='object',index=pd.RangeIndex(start=0, stop=19, step=1)), 'standard': pd.Series([pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan, pd.np.nan],dtype='float64',index=pd.RangeIndex(start=0, stop=19, step=1))}, index=pd.RangeIndex(start=0, stop=19, step=1))
store standard
0 McDonalds NaN
1 Lidls NaN
2 Lidl New York 123 NaN
3 KFC NaN
4 Lidi Berlin NaN
5 Wallmart LA 90210 NaN
6 Aldi NaN
7 London Lidl NaN
8 Aldi627 NaN
9 mcdonaldsabc123 NaN
10 Mcdonald_s NaN
11 McDonalds12345 NaN
12 McDonalds5555 NaN
13 McDonalds888 NaN
14 Aldi123 NaN
15 KFC-786 NaN
16 KFC-908 NaN
17 McDonalds511 NaN
18 GerALDInes Shop NaN
我设置了一个正则表达式字典来搜索字符串,并将商店名称的标准化版本插入到列standard
中。这适用于此小数据帧:
# set up the dictionary
regex_dict = {
"McDonalds": r'(mcdonalds|mcdonald_s)',
"Lidl" : r'(lidl|lidi)',
"Wallmart":r'wallmart',
"KFC": r'KFC',
"Aldi":r'(baldib|baldid+)'
}
# loop through dictionary, using str.replace
for regname, regex_formula in regex_dict.items():
df.loc[df['store'].str.contains(regex_formula,na=False,flags=re.I), 'standard'] = regname
print(df)
store standard
0 McDonalds McDonalds
1 Lidls Lidl
2 Lidl New York 123 Lidl
3 KFC KFC
4 Lidi Berlin Lidl
5 Wallmart LA 90210 Wallmart
6 Aldi Aldi
7 London Lidl Lidl
8 Aldi627 Aldi
9 mcdonaldsabc123 McDonalds
10 Mcdonald_s McDonalds
11 McDonalds12345 McDonalds
12 McDonalds5555 McDonalds
13 McDonalds888 McDonalds
14 Aldi123 Aldi
15 KFC-786 KFC
16 KFC-908 KFC
17 McDonalds511 McDonalds
18 GerALDInes Shop NaN
问题是我有大约六百万行要标准化,正则表达式字典比这里显示的要大得多。(许多不同的商店名称,有些拼写错误等(
我想做的是在每个循环中,只对尚未标准化的行使用str.contains
,而忽略已标准化的行。这个想法是减少每个循环的搜索空间,从而减少整体处理时间。
我已经测试了按standard
列的索引,仅在standard
Nan
的行上执行str.contains
,但它不会导致任何真正的加速。在应用str.contains
之前,仍然需要时间来弄清楚哪些行Nan
。
以下是我试图减少每个循环的处理时间的方法:
for regname, regex_formula in regex_dict.items():
# only apply str.contains to rows where standard == NAN
df.loc[df['standard'].isnull() & df['store'].str.contains(regex_formula,na=False,flags=re.I), 'standard'] = regname
这有效..但是在我的600万行上使用它对速度没有真正的影响。
是否有可能在 600 万行的数据帧上加快速度?
另一种方法是先提取组,然后像下面一样替换,您的循环方法仍然更好。
我们需要稍微改变一下regex_dict,
regex_dict = {
r'mcdonalds|mcdonald_s':"McDonalds",
r'lidl|lidi':"Lidl",
r'wallmart': "Wallmart",
r'kfc':"KFC" ,
r'aldi|aldi':"Aldi"
}
df.str.extract(r'('+ '|'.join(regex_dict.keys())+')',expand=False).replace(regex_dict,regex=True)
0 McDonalds
1 Lidl
2 Lidl
3 KFC
4 Lidl
我设法用这个减少了 40% 所需的时间。我能做的最好的事情
我创建了一个名为fixed_df
的空数据帧来追加新的标准化行,然后在每个循环结束时删除原始数据帧中的相同行。随着每个商店的标准化,每个循环的搜索空间都会减少,并且fixed_df
的大小随着每个循环的增加而增加。最后,fixed_df
应该拥有所有原始行,现在已标准化,原始 df 应为空。
# create empty df to store new results
fixed_df = pd.DataFrame()
# loop through dictionary
for regname, regex_formula in regex_dict.items():
# search for regex formula, add standardized name into standard column
df.loc[df['term_location'].str.contains(regex_formula,na=False,flags=re.I), 'standard'] = regname
# get index of where names were fixed
ind = df[df['standard']==regname].index
# append fixed data to new df
fixed_df.append(df[df.index.isin(ind)].copy())
# remove processed stuff from original df
df = df[~df.index.isin(ind)].copy()