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Parse a list of dictionaries with apply/lambda

Time:09-21

I have a huge dataframe in which a certain column has a list of dictionaries (it is the school history of several people). So, what I'm trying to do is parsing this data to a new dataframe (because the relation is going to be 1 person to many schools).

However, my first option was to loop over the dataframe with itertuples(). Too slow!

Each list looks like this:

list_of_dicts = {
    0: '[]',
    1: "[{'name': 'USA Health', 'subject': 'Residency, Internal Medicine, 2006 - 2009'}, {'name': 'Ross University School of Medicine', 'subject': 'Class of 2005'}]",
    2: "[{'name': 'Physicians Medical Center Carraway', 'subject': 'Residency, Surgery, 1957 - 1960'}, {'name': 'Physicians Medical Center Carraway', 'subject': 'Internship, Transitional Year, 1954 - 1955'}, {'name': 'University of Alabama School of Medicine', 'subject': 'Class of 1954'}]"
}

df_dict = pd.DataFrame.from_dict(list_of_dicts, orient='index', columns=['school_history'])

What I thought about, was to have a function and them apply it to the dataframe:

def parse_item(row):
    eval_dict = eval(row)[0]
    school_df = pd.DataFrame.from_dict(eval_dict, orient='index').T
    return school_df

df['column'].apply(lambda x: parse_item(x))

However, I'm not able to figure out how to generate a dataframe bigger than original (due to situations of multiple schools to one person). Any ideas?

From those 3 rows, the idea is to have this dataframe (that has 5 rows from 2 rows): enter image description here

CodePudding user response:

Iterate over the column to convert each string into a python list using ast.literal_eval(); the result is a nested list, which can be flattened inside the same comprehension.

N.B. Converting the column to a list (via tolist()) first sees some performance gain.

from ast import literal_eval
result = pd.DataFrame([x 
                       for row in df_dict['school_history'].tolist() 
                       for x in literal_eval(row)])

To retain the original index, instead of iterating over the column as a list, iterate over the zip object created by calling items() method on it. This returns (index, value) tuples where the indexes may be attached to the final output values.

ind, data = zip(*[(i, x) 
                  for i, row in df_dict['school_history'].items() 
                  for x in ast.literal_eval(row)]);
result = pd.DataFrame(data, index=ind)

res

CodePudding user response:

This does the trick using your sample data (thanks for the performance tip in comments):

list_df = df_dict.school_history.map(ast.literal_eval)
exploded = list_df[list_df.str.len() > 0].explode()
final = pd.DataFrame(list(exploded), index=exploded.index)

This produces the following:

In [54]: final
Out[54]:
                                       name                                     subject
1                                USA Health   Residency, Internal Medicine, 2006 - 2009
1        Ross University School of Medicine                               Class of 2005
2        Physicians Medical Center Carraway             Residency, Surgery, 1957 - 1960
2        Physicians Medical Center Carraway  Internship, Transitional Year, 1954 - 1955
2  University of Alabama School of Medicine                               Class of 1954

This will probably not be super fast given the amount of data, but parsing a dictionary of strings with nested objects inside will probably be pretty slow no matter what. You're probably better off parsing the file upstream first, then converting to pandas.

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