Pandas loop from leftmost column and change values by dictinoary-CodePudding

I have the following dictionary and dataframe:

val_dict = {
    'key1': ['val1', 'val2', 'val3'], 
    'key2': ['val4', 'val5']
}

df = pd.DataFrame(data={'val5': [True, False, False], 
                        'val2': [False, True, False], 
                        'val3': [True, True, False], 
                        'val1': [True, False, True], 
                        'val4': [True, True, False],
                        'val6': [False, False, True]},
                        index=pd.Series([1, 2, 3], name='index'))

index	val5	val2	val3	val1	val4	val6
1	True	False	True	True	True	False
2	False	True	True	False	True	False
3	False	False	False	True	False	True

How do I go through the dataframe from the left so that if the column is True, other columns in the val_dict values turn to False?

index	val5	val2	val3	val1	val4	val6
1	True	False	True	FALSE	FALSE	False
2	False	True	FALSE	False	True	False
3	False	False	False	True	False	True

For example, index 1 has val5 as True, so val4 switches to False because they are both assigned to the same val_dict key. Similarly, val2 is False but val3 is True, so val1 gets turned to False. Note that it should skip over val6.

I tried converting df to a dictionary with df.to_dict('index') to work with two dictionaries. However, dictionaries are unordered and the order of the columns is important, so I thought it might make the code buggy.

CodePudding user response：

One way is with a combination of assign and mask:

# either val2 or val3 can be True:
com = df.filter(['val2', 'val3']).sum(1).ge(1)

           # val2 is the leftmost, so start with that
(df.assign(**df.filter(['val1', 'val3']).mask(df.val2, False))
           # next is the combination of val2 and val3
   .assign(val1 = lambda df: df.val1.mask(com, False), 
           val4 = lambda df: df.val4.mask(df.val5, False))
)
Out[84]: 
        val5   val2   val3   val1   val4   val6
index                                          
1       True  False   True  False  False  False
2      False   True  False  False   True  False
3      False  False  False   True  False   True

Note that val6 is untouched, so the values remain the same.

CodePudding user response：

Here's what I have with trying to convert to a dictionary:

def section_filter(df, section_dict):
    result = {}
    for index, vals in df.to_dict('index').items():
        lst = []
        for val in section_dict.values():
            lst.append({k:v for k, v in vals.items() if k in val})
        for k, v in vals.items():
            if k not in [m for mi in section_dict.values() for m in mi]:
                lst.append({k: v})
        for l in lst:
            for i in l:
                if l[i]:
                    l.update({k:False for k in l.keys()})
                    l[i] = True
                    break 
        result[index] = {k: v for d in lst for k, v in d.items()}
    return pd.DataFrame.from_dict(result, orient='index', columns=df.columns)

print(df)
print()
print(section_filter(df, val_dict))

        val5   val2   val3   val1   val4   val6
index                                          
1       True  False   True   True   True  False
2      False   True   True  False   True  False
3      False  False  False   True  False   True

    val5   val2   val3   val1   val4   val6
1   True  False   True  False  False  False
2  False   True  False  False   True  False
3  False  False  False   True  False   True