SCo*_*ool 5 python regex pandas
我有一个我试图标准化的商店名称数据框。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
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我设置了一个正则表达式字典来搜索一个字符串,并将商店名称的标准化版本插入到列中standard。这适用于这个小数据框:
# set up the dictionary
regex_dict = {
"McDonalds": r'(mcdonalds|mcdonald_s)',
"Lidl" : r'(lidl|lidi)',
"Wallmart":r'wallmart',
"KFC": r'KFC',
"Aldi":r'(\baldi\b|\baldi\d+)'
}
# 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
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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,有一些拼写错误等)
我想这样做是在每个循环,只能使用str.contains对已行没有被标准化,而忽略已经标准化的行。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。
我已经按standard列测试了索引,只str.contains在standardis 的行上执行Nan,但它不会导致任何真正的加速。Nan在应用之前确定哪些行仍然需要时间str.contains。
这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
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这有效..但是在我的全部 600 万行上使用它对速度没有真正的影响。
甚至有可能在 600 万行的数据帧上加快速度吗?
我使用它成功地将所需时间减少了 40%。我能做的最好的
我创建一个空数据帧,调用它fixed_df来附加新的标准化行,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每个循环的大小都会增加。最后,fixed_df应该拥有所有原始行,现在已标准化,并且原始 df 应该为空。
# 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()
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