Python/Pandas:如何将字符串列表与DataFrame列匹配 [英] Python/Pandas: How to Match List of Strings with a DataFrame column
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问题描述
我想比较两列-Description和Employer.我想查看在Description列中是否找到Employer中的任何关键字.我已经将Employer列分解为单词并转换为列表.现在，我想看看这些单词中的任何一个是否在相应的Description列中.
I want to compare two columnn -- Description and Employer. I want to see if any keywords in Employer are found in the Description column. I have broken the Employer column down to words and converted to a list. Now I want to see if any of those words are in the corresponding Description column.
样本输入:
print(df.head(25))
Date Description Amount AutoNumber \
0 3/17/2015 WW120 TFR?FR xxx8690 140.00 49246
2 3/13/2015 JX154 TFR?FR xxx8690 150.00 49246
5 3/6/2015 CANSEL SURVEY E PAY 1182.08 49246
9 3/2/2015 UE200 TFR?FR xxx8690 180.00 49246
10 2/27/2015 JH401 TFR?FR xxx8690 400.00 49246
11 2/27/2015 CANSEL SURVEY E PAY 555.62 49246
12 2/25/2015 HU204 TFR?FR xxx8690 200.00 49246
13 2/23/2015 UQ263 TFR?FR xxx8690 102.00 49246
14 2/23/2015 UT460 TFR?FR xxx8690 200.00 49246
15 2/20/2015 CANSEL SURVEY E PAY 1222.05 49246
17 2/17/2015 UO414 TFR?FR xxx8690 250.00 49246
19 2/11/2015 HI540 TFR?FR xxx8690 130.00 49246
20 2/11/2015 HQ010 TFR?FR xxx8690 177.00 49246
21 2/10/2015 WU455 TFR?FR xxx8690 200.00 49246
22 2/6/2015 JJ500 TFR?FR xxx8690 301.00 49246
23 2/6/2015 CANSEL SURVEY E PAY 1182.08 49246
24 2/5/2015 IR453 TFR?FR xxx8690 168.56 49246
26 2/2/2015 RQ574 TFR?FR xxx8690 500.00 49246
27 2/2/2015 UT022 TFR?FR xxx8690 850.00 49246
28 12/31/2014 HU521 TFR?FR xxx8690 950.17 49246
Employer
0 Cansel Survey Equipment
2 Cansel Survey Equipment
5 Cansel Survey Equipment
9 Cansel Survey Equipment
10 Cansel Survey Equipment
11 Cansel Survey Equipment
12 Cansel Survey Equipment
13 Cansel Survey Equipment
14 Cansel Survey Equipment
15 Cansel Survey Equipment
17 Cansel Survey Equipment
19 Cansel Survey Equipment
20 Cansel Survey Equipment
21 Cansel Survey Equipment
22 Cansel Survey Equipment
23 Cansel Survey Equipment
24 Cansel Survey Equipment
26 Cansel Survey Equipment
27 Cansel Survey Equipment
28 Cansel Survey Equipment
我尝试了类似的方法，但似乎不起作用.:
I tried something like this, but it doesn't seem to work.:
df['Text_Search'] = df['Employer'].apply(lambda x: x.split(" "))
df['Match'] = np.where(df['Description'].str.contains("|".join(df['Text_Search'])), "Yes", "No")
我想要的输出如下所示:
My desired output would be as shown below:
Date Description Amount AutoNumber \
0 3/17/2015 WW120 TFR?FR xxx8690 140.00 49246
2 3/13/2015 JX154 TFR?FR xxx8690 150.00 49246
5 3/6/2015 CANSEL SURVEY E PAY 1182.08 49246
9 3/2/2015 UE200 TFR?FR xxx8690 180.00 49246
10 2/27/2015 JH401 TFR?FR xxx8690 400.00 49246
11 2/27/2015 CANSEL SURVEY E PAY 555.62 49246
12 2/25/2015 HU204 TFR?FR xxx8690 200.00 49246
13 2/23/2015 UQ263 TFR?FR xxx8690 102.00 49246
14 2/23/2015 UT460 TFR?FR xxx8690 200.00 49246
15 2/20/2015 CANSEL SURVEY E PAY 1222.05 49246
17 2/17/2015 UO414 TFR?FR xxx8690 250.00 49246
19 2/11/2015 HI540 TFR?FR xxx8690 130.00 49246
20 2/11/2015 HQ010 TFR?FR xxx8690 177.00 49246
21 2/10/2015 WU455 TFR?FR xxx8690 200.00 49246
22 2/6/2015 JJ500 TFR?FR xxx8690 301.00 49246
23 2/6/2015 CANSEL SURVEY E PAY 1182.08 49246
24 2/5/2015 IR453 TFR?FR xxx8690 168.56 49246
26 2/2/2015 RQ574 TFR?FR xxx8690 500.00 49246
27 2/2/2015 UT022 TFR?FR xxx8690 850.00 49246
28 12/31/2014 HU521 TFR?FR xxx8690 950.17 49246
29 12/30/2014 WZ553 TFR?FR xxx8690 200.00 49246
32 12/29/2014 JW173 TFR?FR xxx8690 300.00 49246
33 12/24/2014 CANSEL SURVEY E PAY 1219.21 49246
34 12/24/2014 CANSEL SURVEY E PAY 434.84 49246
36 12/23/2014 WT002 TFR?FR xxx8690 160.00 49246
Employer Text_Search Match
0 Cansel Survey Equipment [Cansel, Survey, Equipment] No
2 Cansel Survey Equipment [Cansel, Survey, Equipment] No
5 Cansel Survey Equipment [Cansel, Survey, Equipment] Yes
9 Cansel Survey Equipment [Cansel, Survey, Equipment] No
10 Cansel Survey Equipment [Cansel, Survey, Equipment] No
11 Cansel Survey Equipment [Cansel, Survey, Equipment] Yes
12 Cansel Survey Equipment [Cansel, Survey, Equipment] No
13 Cansel Survey Equipment [Cansel, Survey, Equipment] No
14 Cansel Survey Equipment [Cansel, Survey, Equipment] No
15 Cansel Survey Equipment [Cansel, Survey, Equipment] Yes
17 Cansel Survey Equipment [Cansel, Survey, Equipment] No
19 Cansel Survey Equipment [Cansel, Survey, Equipment] No
20 Cansel Survey Equipment [Cansel, Survey, Equipment] No
21 Cansel Survey Equipment [Cansel, Survey, Equipment] No
22 Cansel Survey Equipment [Cansel, Survey, Equipment] No
23 Cansel Survey Equipment [Cansel, Survey, Equipment] Yes
24 Cansel Survey Equipment [Cansel, Survey, Equipment] No
26 Cansel Survey Equipment [Cansel, Survey, Equipment] No
27 Cansel Survey Equipment [Cansel, Survey, Equipment] No
28 Cansel Survey Equipment [Cansel, Survey, Equipment] No
29 Cansel Survey Equipment [Cansel, Survey, Equipment] No
32 Cansel Survey Equipment [Cansel, Survey, Equipment] No
33 Cansel Survey Equipment [Cansel, Survey, Equipment] Yes
34 Cansel Survey Equipment [Cansel, Survey, Equipment] Yes
36 Cansel Survey Equipment [Cansel, Survey, Equipment] No
推荐答案
以下是使用单个search_func的可读解决方案:
Here is a readable solution using an individual search_func:
def search_func(row):
matches = [test_value in row["Description"].lower()
for test_value in row["Text_Search"]]
if any(matches):
return "Yes"
else:
return "No"
然后按行应用此函数:
# create example data
df = pd.DataFrame({"Description": ["CANSEL SURVEY E PAY", "JX154 TFR?FR xxx8690"],
"Employer": ["Cansel Survey Equipment", "Cansel Survey Equipment"]})
print(df)
Description Employer
0 CANSEL SURVEY E PAY Cansel Survey Equipment
1 JX154 TFR?FR xxx8690 Cansel Survey Equipment
# create text searches and match column
df["Text_Search"] = df["Employer"].str.lower().str.split()
df["Match"] = df.apply(search_func, axis=1)
# show result
print(df)
Description Employer Text_Search Match
0 CANSEL SURVEY E PAY Cansel Survey Equipment [cansel, survey, equipment] Yes
1 JX154 TFR?FR xxx8690 Cansel Survey Equipment [cansel, survey, equipment] No
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