Filter column list based on another column in python-CodePudding

In python, I have a dataset like this below, where column1 and column2 are objects and not strings:

data = {'id':  ['first_value', 'first_value', 'second_value', 'third_value'],
    'column1': [a0, a0 a1, a2, a87],
    'column2': [a0=4, a0=2;a1=8;a7=9, a2=0, a33=9]
    }

And I want to keep on 'column2' the values that:
a. are also in 'column1'.
b. the values that are bigger than zero after the "=" sign on 'column2' (so, for example, 'a2=0' will be removed from the result because is zero (...=0), while a1=8 will stay because it's 8 (...=8))

Therefore in this case my result would be:

data = {'id':  ['first_value', 'first_value'],
        'column1': ['aO', 'a0;a1'],
        'column2': ['a0=4', 'a0=2;a1=8']
        }

I have tried many times but I just cannot find how to :(

Any help is appreciated!

CodePudding user response：

You can try apply on rows

def filter_column(row):
    keys = row['column1'].split(' ')
    dicts = dict([kv.split('=') for kv in row['column2'].split(';')])
    lists = [f'{k}={v}' for k, v in dicts.items()
             if k in keys and int(v) > 0]
    return ';'.join(lists)

df['column3'] = df.apply(filter_column, axis=1)

print(df)

             id column1         column2    column3
0   first_value      a0            a0=4       a0=4
1   first_value   a0 a1  a0=2;a1=8;a7=9  a0=2;a1=8
2  second_value      a2            a2=0
3   third_value     a87           a33=9

------------

print(df[df['column3'].ne('')])

            id column1         column2    column3
0  first_value      a0            a0=4       a0=4
1  first_value   a0 a1  a0=2;a1=8;a7=9  a0=2;a1=8

CodePudding user response：

For your problem, it seems easier to use a loop. Create sets from column1 values, and dictionaries from column2 values; then traverse both together using zip and identify common keys and see if the value under those keys are zero or not:

out = []
for col1, col2 in zip(({*x.split(';')} for x in df['column1']), 
                      (dict([kv.split('=') for kv in x.split(';')]) for x in df['column2'])):
    inner = []
    s = col2.keys() & col1
    while s:
        k = s.pop()
        if col2[k] != '0':
            inner.append(f"{k}={col2[k]}")
    out.append(';'.join(inner))
df['column2'] = out
out = df[df['column2']!='']

Output:

            id column1    column2
0  first_value      a0       a0=4
1  first_value   a0;a1  a0=2;a1=8