如何检查一个列中的值是否存在于另一个列中，当查询的列有许多值时？

huangapple
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英文:

How to check if a value in one column is in other column when the queried column have many values?

问题

  1. 0    True
  2. 1    False
  3. 2    True
  4. 3    True
  5. 4    True
  6. 5    False
  7. 6    False
  8. 7    False
  9. Name: variant, dtype: bool
英文:

Question

How to check if a value in one column is in other column when the queried column have many values?

The minimal reproducible example

  1. df1 = pd.DataFrame({'patient': ['patient1', 'patient1', 'patient1','patient2', 'patient2', 'patient3','patient3','patient4'], 
  2.                    'gene':['TYR','TYR','TYR','TYR','TYR','TYR','TYR','TYR'],
  3.                    'variant': ['buu', 'luu', 'stm','lol', 'bla', 'buu', 'lol','buu'],
  4.                     'genotype': ['buu,luu,hola', 'gulu,melon', 'melon,stm','melon,buu,lol', 'bla', 'het', 'het','het']})
  6. print(df1)
  8.     patient gene variant       genotype
  9. 0  patient1  TYR     buu   buu,luu,hola
  10. 1  patient1  TYR     luu     gulu,melon
  11. 2  patient1  TYR     stm      melon,stm
  12. 3  patient2  TYR     lol  melon,buu,lol
  13. 4  patient2  TYR     bla            bla
  14. 5  patient3  TYR     buu            het
  15. 6  patient3  TYR     lol            het
  16. 7  patient4  TYR     buu            het

What I have tried

  1. df1.variant.isin(df1.genotype)
  3. 0    False
  4. 1    False
  5. 2    False
  6. 3    False
  7. 4     True
  8. 5    False
  9. 6    False
  10. 7    False
  11. Name: variant, dtype: bool

This does not work. The expected result would be:

  1. 0    True
  2. 1    False
  3. 2    True
  4. 3    True
  5. 4    True
  6. 5    False
  7. 6    False
  8. 7    False
  9. Name: variant, dtype: bool

I don't know how many different values the column genotype has. This vary a lot from 1 to 20

答案1

得分: 1

你可以使用 DataFrame.apply + str.split：

  1. print(df1.apply(lambda x: x['variant'] in x['genotype'].split(','), axis=1))

打印结果：

  1. 0     True
  2. 1    False
  3. 2     True
  4. 3     True
  5. 4     True
  6. 5    False
  7. 6    False
  8. 7    False
  9. dtype: bool
英文:

You can use DataFrame.apply + str.split:

  1. print(df1.apply(lambda x: x['variant'] in x['genotype'].split(','), axis=1))

Prints:

  1. 0     True
  2. 1    False
  3. 2     True
  4. 3     True
  5. 4     True
  6. 5    False
  7. 6    False
  8. 7    False
  9. dtype: bool

答案2

得分: 1

With a listcomp:

  1. [var in gen for var, gen in zip(df1["variant"], df1["genotype"])]

Output:

  1. # with the Series constructor pd.Series(...)
  3. 0     True
  4. 1    False
  5. 2     True
  6. 3     True
  7. 4     True
  8. 5    False
  9. 6    False
  10. 7    False
  11. dtype: bool
英文:

With a listcomp :

  1. [var in gen for var,gen in zip(df1["variant"], df1["genotype"])]

Output :

  1. # with the Series constructor pd.Series(...)
  3. 0     True
  4. 1    False
  5. 2     True
  6. 3     True
  7. 4     True
  8. 5    False
  9. 6    False
  10. 7    False
  11. dtype: bool

答案3

得分: 0

你需要创建一个简单的 function 并使用 apply 来迭代所有行。

  1. def check_variant(df):
  2.     return True if df['genotype'].find(df['variant']) != -1 else False

和触发：

  1. df1.apply(check_variant, axis=1)

结果：

  1. 0     True
  2. 1    False
  3. 2     True
  4. 3     True
  5. 4     True
  6. 5    False
  7. 6    False
  8. 7    False
英文:

You need to create a simple function and use apply to iterate it through all rows.

  1. def check_variant(df):
  2.     
  3.     return True if df['genotype'].find(df['variant']) != -1 else False

and Trigger:

  1. df1.apply(check_variant, axis=1)

Results:

  1. 0     True
  2. 1    False
  3. 2     True
  4. 3     True
  5. 4     True
  6. 5    False
  7. 6    False

7 False

huangapple
  • 本文由 发表于 2023年4月11日 01:29:23
  • 转载请务必保留本文链接：https://go.coder-hub.com/75979268.html
  • pandas
  • python
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