Introduction

In this blog post, we will explore how to create an indicator based on complex combinations in a pandas DataFrame. The problem involves creating an indicator that depends on the value of two columns (cd1 and cd2) and the latest timestamp associated with each combination. We will discuss different approaches to solving this problem and provide code examples.

Problem Statement

We have a table like this:

We need to create an indicator that is 1 if cd1 is either 50 or 600 and cd2 is not zero, otherwise it’s 0. However, for the same clm_id, if multiple occurrences of cd1 occur, we need to see which one is the latest and what are their combinations.

Initial Solution

The initial solution attempts to solve this problem by converting 'tmstp' to a datetime object so comparisons can be made:

df['tmstp'] = pd.to_datetime(df['tmstp'])

Then, group the DataFrame by clm_id:

gb = df.groupby('clm_id')

Keep a list of named tuples for the results:

import collections
results = []
Result = collections.namedtuple('Result',['clm_id','indicator'])

Iterate over the clm_id groups; for each clm_id split by cd1; find the most recent cd1 and determine if its cd2 is not zero; check if either the 50 or 600 subgroup is True; store the result:

for clm_id,group in gb:
    cd1_grp = group.groupby('cd1')
    # Start with the indicators set to False
    ind = {50:False,600:False}
    for cd1,subgroup in cd1_grp:
        if cd1 not in (50,600):
            continue
        most_recent = subgroup.loc[subgroup['tmstp'].idxmax()]
        ind[cd1] = most_recent.cd2 != 0
    indicator = (ind[50] or ind[600]) * 1
    results.append(Result(clm_id,indicator))

Improved Solution

A better solution is to use grouping with two columns. This approach avoids the need for nested loops and makes the code more efficient:

gb1 = df.groupby(['clm_id','cd1'])
d = {}
for (clm_id,cd1),group in gb1:
    if clm_id not in d:
        d[clm_id]=0
    if cd1 not in (50,600):
        continue
    most_recent = group.loc[group['tmstp'].idxmax()]
    d[clm_id] = d[clm_id] or int(most_recent.cd2 != 0)

print(d)

This solution uses a dictionary to store the results and only updates it when necessary, making it more efficient.

Using dict.fromkeys instead of d[clm_id] not in d:

The dictionary can be made with d = dict.fromkeys(df['clm_id'].unique(),0), which would make the conditional statement unnecessary. This might affect performance, but it is an interesting alternative solution.

Conclusion

In this blog post, we explored how to create an indicator based on complex combinations in a pandas DataFrame. We discussed different approaches and provided code examples for each approach. The improved solution using grouping with two columns is more efficient than the initial solution. Additionally, we explored the use of dict.fromkeys instead of d[clm_id] not in d:.

Last modified on 2023-07-05