how to check for match for every letter in row string

user_13131038 0
Viewed 202

I have a subset of the dataframe

df = pd.DataFrame(
    {
            'id': ['1001','1002','1003','1004','1005','1006','1007','1008','1009','1010'],
            'colA': ['H','L B','L H','L B','L S B','B','B S L','L B S','L S B','L S B'],
            'colB': ['H','L|B','H|L','H|L','L|S|B','L|S|B','L|S|B','L|S|B','L|S','L']
    }
    )

I'm doing row level comparison for this dataframe. I want to check whether all the letters in row['colA'] match with all the letters in row['colB'], regardless of what order they appear and ignoring the | in colB. This is the logic for the function, but it doesn't work as intended and how do I update it to ignore |

def match_or_not(df):
    for index,row in df.iterrows():
        if row['colA'] == row['colB']:
            print ("Match for "+str(row['id']))
        else:
            print ("Not match for "+str(row['id']))

I need help to update the condition following the if keyword in the above function, how can I write to get desire output. The cases for which it should match and not match are shown in the picture:

enter image description here

pandas dataframe python
3 Answers

The following should work:

def match_or_not(df):
    for index,row in df.iterrows():
        #First make a list out of the values, then compare the sorted values
        if sorted(row['colA'].split(" ")) == (sorted(row['colB'].split("|"))):
            print ("Match for "+str(row['id']))
        else:
            print ("Not match for "+str(row['id']))
user_106879070

You could make use of the "split()" method for strings, as well as sets, like this:

def match_or_not(df):
    for index,row in df.iterrows():
        a_set = set(row['colA'].split(" "))
        b_set = set(row['colB'].split("|"))
        if a_set == b_set:
            print ("Match for "+str(row['id']))
        else:
            print ("Not match for "+str(row['id']))

split() creates a list of substrings using a given delimiter (in your case a space and a |).

set() makes that list a set. Sets are easy to compare, as order does not matter within them. However this solution does come with a potential drawback, which is not reflected in your example data:

set("L L".split(" "))

This splits the string L L which creates a list of two strings 'L'. However, a sets elements are unique, so it gets converted to the set {'L'} (mind only one L.)

So if your data is expected to contain such double strings and their occurrence is intended to match by amount, this solution does not suffice.

user_34584840

You can use df.apply without df.iterrows():

>>> df.apply(lambda row: f"Match for {row['id']}" if set(row['colA'].replace(" ", "")) == set(row['colB'].replace("|", "")) else f"No match for {row['id']}", axis=1)
0       Match for 1001
1       Match for 1002
2       Match for 1003
3    No match for 1004
4       Match for 1005
5    No match for 1006
6       Match for 1007
7       Match for 1008
8    No match for 1009
9    No match for 1010

The set(row['colA'].replace(" ", "")) == set(row['colB'].replace("|", "")) condition makes sure all chars but spaces from colA exist in colB (other than | chars, all the spaces and pipes are removed before converting to a set).

user_38329700
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