Question

我有一个包含大量英文单词的list类型的单词列表。

我还有一个数据框，如下所示： -

    FileName        PageNo     LineNo   GOODS_DESC  
1   17743633 - 1    TM000002    69      Abuj Cen Le 
31  17743633 - 1    TM000007    126     Mr USD  
33  17743633 - 1    TM000008    22      TABLEAU EMBALLAGE
34  17743633 - 1    TM000008    24      LISA e EMBALV
46  17743633 - 1    TM000008    143     Cen 
47  17743633 - 1    TM000008    146     A Gl
50  17743633 - 1    TM000009    121     Ppvv Tn Ppvv In 
51  17743633 - 1    TM000009    129     SPECIFY
52  17743633 - 1    TM000009    136     Decrp G 
58  17743633 - 1    TM000009    97      Je ugn  
60  17743633 - 1    TM000009    108     De Veel 
61  17743633 - 1    TM000014    44      TYRE CHIPS SHREDDED TYRES   
63  17743633 - 1    TM000014    48      TYRE CHIPS SHREDDED TYRES

我想只在＃GOANDS_DESC＆＃39;中保留这些字词。单词列表中的列。

我想要的输出是： -

    FileName        PageNo     LineNo   GOODS_DESC  
1   17743633 - 1    TM000002    69      NaN
31  17743633 - 1    TM000007    126     Mr USD  
33  17743633 - 1    TM000008    22      TABLEAU
34  17743633 - 1    TM000008    24      LISA  
46  17743633 - 1    TM000008    143     NaN 
47  17743633 - 1    TM000008    146     NaN
50  17743633 - 1    TM000009    121     NaN 
51  17743633 - 1    TM000009    129     SPECIFY
52  17743633 - 1    TM000009    136     NaN
58  17743633 - 1    TM000009    97      NaN 
60  17743633 - 1    TM000009    108     NaN
61  17743633 - 1    TM000014    44      TYRE CHIPS SHREDDED TYRES   
63  17743633 - 1    TM000014    48      TYRE CHIPS SHREDDED TYRES

我的方法也提供输出，但我使用列表并且速度很慢。我想快点。

for rows in df.itertuples():
    a = []
    flat_list = []
    a.append(rows.GOODS_DESC)
    flat_list = [item.strip() for sublist in a for item in sublist.split(' ') if item.strip()]
    flat_list = list(sorted(set(flat_list), key=flat_list.index))
    flat_list = [i for i in flat_list if i.lower() in word_list]
    if(not flat_list):
        df.drop(rows.Index,inplace=True)
        continue
    s=' '.join(flat_list)
    df.loc[rows.Index,'GOODS_DESC']=s

df['GOODS_DESC'] = df['GOODS_DESC'].str.upper()

Answer 1

你的逻辑似乎过于复杂。您可以对pd.Series.apply使用单个列表理解。我建议，如下所示，使用set进行O（1）查询，str.casefold匹配字符串，无论情况如何。

s = pd.Series(['Abuj Cen Le', 'Mr USD', 'TABLEAU EMBALLAGE', 'LISA e EMBALV'])

word_set = {i.casefold() for i in ['Mr', 'USD', 'TABLEAU', 'LISA']}

def apply_filter(x):
    out = ' '.join([i for i in x.split() if i.casefold() in word_set])
    return out if out else np.nan

res = s.apply(apply_filter)

print(res)

0        NaN
1     Mr USD
2    TABLEAU
3       LISA
dtype: object

Answer 2

您可以使用merge和join来执行您想要的操作。首先做一些准备：

#input file, you can have any number of columns in it
df_input = pd.DataFrame({'col1':range(1,5),
                         'GOODS_DESC':['Abuj Cen Le', 'Mr USD', 
                                       'TABLEAU EMBALLAGE', 'LISA e EMBALV']})
# DF from the list of words
df_word = pd.DataFrame({'Word':[word.lower() for word in word_list]})
# create a df_stack with each word of your sentences as a row keeping indexes for join later
df_stack = (df_input['GOODS_DESC'].str.split(' ',expand=True).stack().reset_index())
# Column with same name as df_word and lowercase for merge after
df_stack['Word'] = df_stack[0].str.lower()

现在您可以使用merge然后使用join：

df_join = df_stack.merge(df_word).groupby('level_0')[0].apply(lambda x: ' '.join(x))
df_output = df_input.join(df_join)
# get ride
df_output = df_output.drop('GOODS_DESC',1).rename(columns={0:'GOODS_DESC'})

它看起来像很多行，但merge和join是有效的，所以我希望它足够快。

根据单词列表细化数据帧

2 个答案: