Lost in Pandas - Part 1

Data Science

Nodar Okroshiashvili


April 10, 2020


Many have been told and written about Pandas and its capabilities. I could not imagine data scientist or data analyst who had not heard about Pandas or had not used it at least once. We all use it. Every day, every week. It does not matter how many times. It’s a great tool. I use it all the time when I want to do data analysis, either it is simple calculations or complex data transformations, and it surprises me. Pandas is so simple in its form. However, imagine, how much you can do with some simple method chaining.

Saying all of these, this blog aims to share my experience and amazment with Pandas. This series is not meant for beginner users and will not be short in length. These series will be based on my experience and I will try to give a detailed explanation for every step from problem definition to solving it. That’s enough for now. Let get down to business.

Problem Statement

We have data. This data comes from the HR department of the company. The data contains two columns, company name, and information about its employees. Each row of the employee information column is a list of lists. The lists inside the outer list can be duplicated. It also can have duplicate values, and inner lists have at most two values.

Disclaimer: Any name, phone, email, and the title is a pure coincidence. Data is random and fake.

Here is our data.

from collections import defaultdict

import pandas as pd
import numpy as np
data = {
    "company_name": ["A", "B", "C", "D", "E", "F"],
    "info": [
        [["Name", "David Jones"], ["Title", "CEO"], ["Phone", "207-685-1626"], ["Email", "djones@example.org"]],
            ["Name", "Kate Brown"],
            ["Title", "Senior Lawyer"],
            ["Phone", "316-978-7791"],
            ["Email", "Kate.Brown@example.edu"],
            ["Name", "Darin White"],
            ["Title", "Associate Vice President"],
            ["Phone", "316-978-3887"],
            ["Email", "Darin.White@example.edu"],
            ["Name", "Scott Lamb"],
            ["Title", "Actuary"],
            ["Phone", "316-978-3804"],
            ["Email", "scott.lamb@example.edu"],
            ["Name", "Scott Lamb"],
            ["Title", "Senior Officer"],
            ["Title", "Application Developer"],
            ["Title", "Blockchain Architect"],
            ["Title", "Director of External Affairs"],
            ["Name", "Scott"],
            ["Name", "Scott"],
            ["Title", "Director of Medicine"],
            ["Title", "Product Owner"],
            ["Name", "Mike"],
            ["Title", "Domain Expert"],
            ["Title", "Growth Hacker"],
            ["Title", "Engineering Head"],
            ["Title", "Event Manager"],
            ["Name", "Joe"],
            ["Name", "Mike"],
            ["Title", "Fundraising"],
            ["Title", "VP of Customers"],
            ["Name", "Mike"],
            ["Title", "Venture Capital Analyst"],
            ["Title", "UX Designer"],
            ["Name", "Mike"],
            ["Name", "Susan"],
            ["Name", "Bryan"],
            ["Name", "Mia"],
            ["Title", "Songwriter"],
            ["Name", "Rose Smith Rose Smith"],
            ["Title", "Vice President"],
            ["Title", "Finance and Operations Head"],
            ["Phone", "316-978-3810"],
            ["Email", "rose.smith@example.edu"],
            ["Name", "Rose Smith"],
            ["Title", "Foundation"],
            ["Name", "Susan"],
            ["Title", "Foundation"],
            ["Title", "Accountant"],
            ["Title", "Accountant"],
            ["Title", "Executive"],
            ["Title", "Director"],
            ["Title", "Executive"],
            ["Name", "Ray"],
            ["Title", "Strategic Planning"],
            ["Title", "Financial Analyst"],
            ["Title", "Foundation"],
            ["Title", "Foundation"],
            ["Name", "Susan"],
            ["Title", "member of the board"],
            ["Title", "board of directors"],
            ["Title", "president"],
            ["Title", "board of directors"],
            ["Name", "Carl Clark"],
            ["Title", "Chief "],
            ["Title", "Operating Officer"],
            ["Title", "PhD"],
            ["Phone", "413-534-2745"],
            ["Email", "Clark_Carl@example.com"],
            ["Title", "Board Member"],
            ["Name", "Taylor Garcia"],
            ["Phone", "307-733-2164"],
            ["Phone", "307-733-4568"],
            ["Email", "Garcia@example.org"],

Let convert this dictionary into Pandas DataFrame and see what data we have.

df = pd.DataFrame(data)

company_name info
0 A [[Name, David Jones], [Title, CEO], [Phone, 20...
1 B [[Name, Kate Brown], [Title, Senior Lawyer], [...
2 C [[Name, Scott Lamb], [Title, Actuary], [Phone,...
3 D [[Name, Rose Smith Rose Smith], [Title, Vice P...
4 E [[Name, Carl Clark], [Title, Chief ], [Title, ...

We see that the first column seems okay, but the second one not. Here we have one big list containing smaller two-element lists.

[['Name', 'Scott Lamb'],
 ['Title', 'Actuary'],
 ['Phone', '316-978-3804'],
 ['Email', 'scott.lamb@example.edu'],
 ['Name', 'Scott Lamb'],
 ['Title', 'Senior Officer'],
 ['Title', 'Application Developer'],
 ['Title', 'Blockchain Architect'],
 ['Title', 'Director of External Affairs'],
 ['Name', 'Scott'],
 ['Name', 'Scott'],
 ['Title', 'Director of Medicine'],
 ['Title', 'Product Owner'],
 ['Name', 'Mike'],
 ['Title', 'Domain Expert'],
 ['Title', 'Growth Hacker'],
 ['Title', 'Engineering Head'],
 ['Title', 'Event Manager'],
 ['Name', 'Joe'],
 ['Name', 'Mike'],
 ['Title', 'Fundraising'],
 ['Title', 'VP of Customers'],
 ['Name', 'Mike'],
 ['Title', 'Venture Capital Analyst'],
 ['Title', 'UX Designer'],
 ['Name', 'Mike'],
 ['Name', 'Susan'],
 ['Name', 'Bryan'],
 ['Name', 'Mia'],
 ['Title', 'Songwriter']]

As we figured out the data structure, let define the aim.

We need to unpack lists from the second column and flatten them in tabular format in the way to preserve the order. Meaning that, from the above example, Scott Lamb has to have title Actuary and not other titles are allowed. Long story short, we need to flatten list of list and make proper DataFrame from it.

How I approached this problem

The first idea that came to my mind was to use Pandas DataFrame .explode() method to unpack list of lists, which returned lists containing two elements. After that, I extracted these two elements into two different columns.

df_exploded = df.explode("info")

company_name info
0 A [Name, David Jones]
0 A [Title, CEO]
0 A [Phone, 207-685-1626]
0 A [Email, djones@example.org]
1 B [Name, Kate Brown]
# Add two new columns
df_exploded.loc[:, "tag"] = df_exploded["info"].map(lambda x: x[0])

df_exploded.loc[:, "result"] = df_exploded["info"].map(lambda x: x[1])

company_name info tag result
0 A [Name, David Jones] Name David Jones
0 A [Title, CEO] Title CEO
0 A [Phone, 207-685-1626] Phone 207-685-1626
0 A [Email, djones@example.org] Email djones@example.org
1 B [Name, Kate Brown] Name Kate Brown

Despite unpacking the list of lists, that is not the format I wanted. So, I need to do an extra transformation.

df_exploded_final = (
    df_exploded.groupby(["company_name", "tag"])["result"]
    .apply(lambda x: pd.Series(x.values))
    .drop(["level_1"], axis=1)

tag company_name Email Name Phone Title
0 A djones@example.org David Jones 207-685-1626 CEO
1 B Kate.Brown@example.edu Kate Brown 316-978-7791 Senior Lawyer
2 B Darin.White@example.edu Darin White 316-978-3887 Associate Vice President
3 C scott.lamb@example.edu Scott Lamb 316-978-3804 Actuary
4 C NaN Scott Lamb NaN Senior Officer

It seems we did a good job. However, this approach is prone to errors. Namely, it does not preserve the order of the list values and may assign a different email to a different person. That was not what I need.

So, I decided to use another way to solve this problem. Notably, as the data contained millions of rows, it seemed impossible to be too precise, but I wanted to reduce the error of non-matching cases. To achieve this, I iterated over the values of info column and converted it to dict of lists, where keys are tags (identifiers) and values are actual employee information.

out = []

for x in df["info"].tolist():
    groups = defaultdict(list)
    for g, v in x:

df.loc[:, "new_info"] = out

{'Name': ['David Jones'],
 'Title': ['CEO'],
 'Phone': ['207-685-1626'],
 'Email': ['djones@example.org']}

That’s a step forward. After this, I was interested in counting the values for each key in dicts for each row. I made small changes in the above code and applied it to the new_info column.

out = []

for x in df["new_info"]:
    groups = defaultdict(int)
    for g, v in x.items():
        groups[g] = len(list(filter(None, v)))

df.loc[:, "new_info_stats"] = out

{'Name': 1, 'Title': 1, 'Phone': 1, 'Email': 1}

As we calculated value counts for each dict, now we need to add three helper columns to the dataset for further usage. These helper columns will help to differentiate matching cases and non-matching cases.

df["_max"] = df["new_info_stats"].apply(lambda x: max(x.values()))

df["_min"] = df["new_info_stats"].apply(lambda x: min(x.values()))

df.loc[:, "max_equal_min"] = pd.Series(np.where((df["_max"] == df["_min"]), 1, 0))

The column max_equal_min is a dummy variable and helps us to differentiate matching and non-matching cases. The value 1 indicates we have a matching case and value 0 - non-matching case. According to this column, I split data into two parts. The first only contains matching examples, and the second will have only non-matching cases.

df_first = df[df["max_equal_min"] > 0].reset_index(drop=True)

df_second = df[df["max_equal_min"] == 0].reset_index(drop=True)

The pre-processing of the first DataFrame is over and is ready to flatten. To do so, I iterate over new_info column and transform each row into Pandas DataFrame. After this step, data will be flat.

new_data = []

for j in df_first["new_info"]:

df_first_final_i = (
    pd.concat(new_data, axis=0, sort=False).drop_duplicates().dropna(subset=["Name"]).reset_index(drop=True)

Name Title Phone Email
0 David Jones CEO 207-685-1626 djones@example.org
1 Kate Brown Senior Lawyer 316-978-7791 Kate.Brown@example.edu
2 Darin White Associate Vice President 316-978-3887 Darin.White@example.edu

Woohoo, it works! However, imagine having millions of rows how slow this approach will be. For this reason, I tried another method and found it much faster. Here it is.

df_first_final_ii = (
    df_first["new_info"].apply(pd.Series).apply(lambda x: x.explode()).drop_duplicates().reset_index(drop=True)

Name Title Phone Email
0 David Jones CEO 207-685-1626 djones@example.org
1 Kate Brown Senior Lawyer 316-978-7791 Kate.Brown@example.edu
2 Darin White Associate Vice President 316-978-3887 Darin.White@example.edu

Faster and cleaner solution. But, what about the second DataFrame? It turned out that the above solution did not fit the second DataFrame and gave me an error. The error was ValueError: cannot reindex from a duplicate axis. Before finding the solution for this error, let take a look at the data.

company_name info new_info new_info_stats _max _min max_equal_min
0 C [[Name, Scott Lamb], [Title, Actuary], [Phone,... {'Name': ['Scott Lamb', 'Scott Lamb', 'Scott',... {'Name': 12, 'Title': 16, 'Phone': 1, 'Email': 1} 16 1 0
1 D [[Name, Rose Smith Rose Smith], [Title, Vice P... {'Name': ['Rose Smith Rose Smith', 'Rose Smith... {'Name': 5, 'Title': 17, 'Phone': 1, 'Email': 1} 17 1 0
2 E [[Name, Carl Clark], [Title, Chief ], [Title, ... {'Name': ['Carl Clark'], 'Title': ['Chief ', '... {'Name': 1, 'Title': 3, 'Phone': 1, 'Email': 1} 3 1 0
3 F [[Title, Board Member], [Name, Taylor Garcia],... {'Title': ['Board Member'], 'Name': ['Taylor G... {'Title': 1, 'Name': 1, 'Phone': 2, 'Email': 1} 2 1 0

In the second and third row, we have one Name and three Title and two Phone, respectively, for the new_info_stats column. This may be due to the data entry or extraction reason. Not 100% sure that this is the case, but the likelihood is very high. So, we have to handle this problem properly. One solution is to concatenate strings for the values of Title and Phone keys.

def process_info(record: dict) -> dict:
    if (
        len(record.keys()) == 4
        and len(record.get("Name")) == 1
        and len(record.get("Title")) > 1
        and len(record.get("Email")) == 1
        and len(record.get("Phone")) == 1
        record["Title"] = [" ".join(record.get("Title"))]

    elif (
        len(record.keys()) == 4
        and len(record.get("Name")) == 1
        and len(record.get("Title")) == 1
        and len(record.get("Email")) > 1
        and len(record.get("Phone")) == 1
        record["Email"] = [",".join(record.get("Email"))]

    elif (
        len(record.keys()) == 4
        and len(record.get("Name")) == 1
        and len(record.get("Title")) == 1
        and len(record.get("Email")) == 1
        and len(record.get("Phone")) > 1
        record["Phone"] = [",".join(record.get("Phone"))]

    return record

df_second["new_info"] = df_second["new_info"].apply(process_info)

This is a simple logic to check if we are correctly concatenating string. After applying this function, the second DataFrame is ready to flatten. As I mentioned above the good old method did not give me the desired result for this case and then I came up to the following:

def flatten(df, column):
    data = []
    for i in df[column]:
        data.append(pd.DataFrame(dict([(k, pd.Series(v)) for k, v in i.items()])))

    new_df = (
        pd.concat(data, axis=0, sort=False)
        .drop_duplicates(subset=["Name", "Title", "Phone", "Email"])

    return new_df

df_second_final = flatten(df_second, "new_info")

Name Title Phone Email
0 Scott Lamb Actuary 316-978-3804 scott.lamb@example.edu
1 Scott Lamb Senior Officer NaN NaN
2 Scott Application Developer NaN NaN
3 Scott Blockchain Architect NaN NaN
4 Mike Director of External Affairs NaN NaN

Applied this function to the second DataFrame flattened it, and combining first and second DataFrames will give the final result.


To sum up, that was only a tiny part of this data pre-processing process. However, it was a great journey and lots to learn. What do you think? Did you find a more elegant solution? Please share it in the comments.