Nested for loop with if in one line for creating new column in dataframe

Question

I have a dataframe "result" and want to create a new column called "type". The value in "type" will be the item value of a dict if the column "Particulars" in the dataframe contains value of the key.

dict_classify={'key1': 'content1', 
           'key2':'content2'
          }

result['type']=[dict_classify[key] if key.lower() in i.lower() else np.nan 
                for key in dict_classify.keys() 
                for i in result['Particulars']]

It returns the error "Length of values (5200) does not match the length of index (1040)". Any idea what I did wrong?

The following is what I want to achieve in a normal for loop. Can I make it into one line?

lst_type=[]

for i in result['Particulars']:
    for key in dict_classify:
        temp=np.nan
        if key.lower() in i.lower():
            temp=dict_classify[key]
            break
    
    lst_type.append(temp)


result['type']=lst_type

Asked By: mogcai

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Source

Answer 1

It sounds like each entry i of result["Particulars"] is either a key from dict_classify in which case we want the corresponding entry of result["type"] to be dict_classify[i], or is not a key in which case we want the corresponding entry to be NaN. If that’s the case, then you should have something like

result['type'] = [dict_classify.get(i,np.nan) for i in result['Particulars']]

The same result could more efficiently be attained with

result['type'] = result['Particulars'].apply(lambda i: dict_classify.get(i,np.nan))

Answered By: Ben Grossmann

Answer 2

result[‘type’] = [dict_classify.get(i,np.nan) for i in result[‘Particulars’]]

The same result could more efficiently be attained with

result[‘type’] = result[‘Particulars’].apply(lambda i: dict_classify.get(i,np.nan))

Answered By: Madhu Sena

Answer 3

The most straightforward way is probably to iterate through the dictionary using loc to find cells that contain each key:

for key, value in dict_classify.items():
    result.loc[result["Particulars"].str.contains(key), "type"] = value

You could also use a regex to identify the matched keys (like this answer). We can then use replace to get the values corresponding to each key.

regex = "(" + "|".join(dict_classify) + ")"
result["type"] = result["Particulars"].str.extract(regex).replace(dict_classify)

(You could of course condense this to one line if you really want to.)

Answered By: Stuart

Nested for loop with if in one line for creating new column in dataframe

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