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<p>我有一个商店名称的数据框架,我正试图将其标准化。此处要测试的小样本:</p>
<pre><code>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
</code></pre>
<p>我设置了一个regex字典来搜索字符串,并将商店名称的标准化版本插入到<code>standard</code>列中。这适用于这个小数据帧:</p>
<pre><code># 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
</code></pre>
<p>问题是我有大约600万行需要标准化,其中一个regex字典比这里显示的要大得多。(许多不同的店名有一些拼写错误等)</p>
<p>我想做的是在每个循环中,只对<strong>未标准化的行使用<code>str.contains</code>,而忽略已标准化的行。其思想是减少每个循环的搜索空间,从而减少总体处理时间</p>
<p>我已经通过<code>standard</code>列测试了索引,只对<code>standard</code>为<code>Nan</code>的行执行<code>str.contains</code>,但这不会导致任何实际的加速。在应用<code>str.contains</code>之前,仍然需要时间来确定哪些行是<code>Nan</code></p>
<p>以下是我试图减少每个循环的处理时间的内容:</p>
<pre><code>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
</code></pre>
<p>这很有效。。但是在我的600万行中使用这个并没有真正的速度差异</p>
<p>在一个600万行的数据帧上,是否有可能加快速度</p>