熊猫：如何获取一列的所有值，其中另一列的值是特定值

Question

I have a dataframe which contains a sample_id and mutation: Each sample contains several mutations

sample_id    mutation
sample1      mutation_A
sample1      mutation_B
sample1      mutation_D

sample2      mutation_C
sample2      mutation_D

sample3      mutation_A
sample3      mutation_B
sample3      mutation_C

I want to be able to obtain the values where say, mutation_C exists. However I want to get all the results out for that sample -

df.loc[(df[mutation] == 'mutation_C')]

returns:

sample_id    mutation
sample2      mutation_C

How do I get the rest of sample2 mutation data, so:

sample_id    mutation
sample2      mutation_C
sample2      mutation_D

I have been trying to use grouopby but can't figure out how to obtain all the results

Answer 1

First filter all samples and then filter again by isin :

a = df.loc[df['mutation'] == 'mutation_C', 'sample_id']
df = df[df['sample_id'].isin(a)]
print (a)

3    sample2
7    sample3
Name: sample_id, dtype: object

df = df[df['sample_id'].isin(a)]
print (df)
  sample_id    mutation
3   sample2  mutation_C
4   sample2  mutation_D
5   sample3  mutation_A
6   sample3  mutation_B
7   sample3  mutation_C

Answer 2

Assuming you have other data, a neater idea would be to set the index the way you are after. (I've added a dummy column with df['value'] = 1 )

>>> a = df.set_index(['mutation', 'sample_id'])
>>> a.sort_index()
                      value
mutation   sample_id       
mutation_A sample1        1
           sample3        1
mutation_B sample1        1
           sample3        1
mutation_C sample2        1
           sample3        1
mutation_D sample1        1
           sample2        1
>>> a.loc['mutation_C']
               value
sample_id       
sample2        1
sample3        1

If you really need the sample_ids as a list then you could do:

>>> a.loc['mutation_C'].index.tolist()
['sample2', 'sample3']

Not what you asked but perhaps another useful view:

>>> df.pivot_table(values='value', index='sample_id', columns='mutation')
mutation   mutation_A  mutation_B  mutation_C  mutation_D
sample_id                                                
sample1           1.0         1.0         NaN         1.0
sample2           NaN         NaN         1.0         1.0
sample3           1.0         1.0         1.0         NaN