计算3个不同字符串在多个列中出现的次数,并将此计数放入新列中。

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英文:

Counting the number of times 3 different strings appear over numerous columns and putting this count in a new column

问题

I've translated the code part for you as requested:

df['Received_money'] = df[['Job1', 'Job2']].apply(lambda row: len(row[row == 'Yes']), axis=1)

If you have any more translation requests or need assistance with anything else, please feel free to ask.

英文:

I'm trying to create a new column that tallys up the number of times someone was paid for a job - regardless of if it was all of the money or just some of the money. So for each row, if it says "yes" or "partial" or "paid" in the job columns then I want a count of this in the new column.

My actual data has 15 different job columns that I want to "sum" across.

So before looks like:

Name Job1 Job2
tom Yes No
nick Partial Yes
juli No No

And I'd like afterwards to look like:

Name Job1 Job2 Received_money
tom Yes No 1
nick Partial Yes 2
juli No No 0

Current code

df['Received_money'] = df[['Job1', 'Job2']].apply(lambda row: len(row[row == 'Yes']), axis=1)
This is my current code and it partially does what I want. It adds up the number of times it says "Yes" in the columns listed. But:

  1. I can't figure out how to expand this to include "== 'partial'" and "== 'paid'", and how to get it to give 1 point (so to speak) for each time these occur
  2. Is there any other way of entering in all 15 of my column names instead of [['Job1', 'Job2', 'Job3', 'Job4', 'Job5'....'Job15' ]]

(Example data)

import pandas as pd
  
# initialize list of lists
data = [['tom', "Yes", "No"], ['nick', "Partial", "Yes"], ['juli', "No", "No"]]
  
# Create the pandas DataFrame
df = pd.DataFrame(data, columns=['Name', 'Job1', 'Job2'])

Thank you!

答案1

得分: 1

你的方法看起来没问题。只需像这样添加其他过滤器:

df['Received_money'] = df[['Job1', 'Job2']].apply(lambda row: len(row[row == 'Yes']) + len(row[row == 'Partial']), axis=1)
英文:

Your approach seems okay. Just add the other filters like this:

df['Received_money'] = df[['Job1', 'Job2']].apply(lambda row: len(row[row == 'Yes']) +len(row[row == 'Partial']), axis=1)

答案2

得分: 1

我添加了2个额外的工作列,只是为了测试。这应该满足您的要求。

data = [['tom', "Yes", "No", "Partial", "Paid"], ['nick', "Partial", "Yes"], ['juli', "No", "No", "Partial", "Paid"]]

df = pd.DataFrame(data, columns=['Name', 'Job1', 'Job2', 'Job3', 'Job4'])

job_cols = ['Job1', 'Job2', 'Job3', 'Job4']

paid_values = ['Yes', 'Paid', 'Partial']

df['Received_money'] = df[job_cols].apply(lambda row: len([r for r in row if r in paid_values]), axis=1)

print(df)
英文:

I added 2 more Job columns just for testing. This should satisfy your requirement.

data = [['tom', "Yes", "No","Partial","Paid"], ['nick', "Partial", "Yes"], ['juli', "No", "No","Partial","Paid"]]

df = pd.DataFrame(data, columns=['Name', 'Job1', 'Job2','Job3','Job4'])

job_cols =['Job1','Job2','Job3','Job4']

paid_values = ['Yes','Paid','Partial']

df['Received_money'] = df[job_cols].apply(lambda row : len([r for r in row if r in paid_values]),axis=1)

print(df)

答案3

得分: 1

不要使用 apply,您可以轻松向量化此操作:

df['Received_money'] = df.filter(like='Job').isin(['Yes', 'Partial']).sum(axis=1)

或者,如果 Job 列不是以字面意义上的 "Job" 开头:

cols = ['Job1', 'Job2']
df['Received_money'] = df[cols].isin(['Yes', 'Partial']).sum(axis=1)

输出结果:

   Name     Job1 Job2  Received_money
0   tom      Yes   No               1
1  nick  Partial  Yes               2
2  juli       No   No               0
英文:

Don't use apply, you can easily vectorize this:

df['Received_money'] = df.filter(like='Job').isin(['Yes', 'Partial']).sum(axis=1)

Or, if the Job columns don't start with a literal "Job":

cols = ['Job1', 'Job2']
df['Received_money'] = df[cols].isin(['Yes', 'Partial']).sum(axis=1)

Output:

   Name     Job1 Job2  Received_money
0   tom      Yes   No               1
1  nick  Partial  Yes               2
2  juli       No   No               0

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  • 本文由 发表于 2023年8月4日 01:06:25
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