How Can I Efficiently Concatenate Multiple CSV Files into a Single Pandas DataFrame and Track Data Provenance?

Concatenating Multiple CSV Files into a Single DataFrame

Problem Statement

To efficiently combine multiple CSV files into a unified DataFrame, a concise and reliable solution is sought. However, a hurdle has been encountered within the concatenation loop.

Solution

To resolve the issue and successfully concatenate the CSV files, the following comprehensive code snippet can be employed:

import os
import pandas as pd
from pathlib import Path

path = r'C:\DRO\DCL_rawdata_files'
all_files = Path(path).glob('*.csv')

df = pd.concat((pd.read_csv(f) for f in all_files), ignore_index=True)

This code utilizes a generator expression to read each CSV file individually, and then concatenates them into a single DataFrame. The ignore_index parameter ensures that the concatenated DataFrame has continuous row indices.

Adding Information to Identify Data Provenance

In certain scenarios, it may be beneficial to add a column to the concatenated DataFrame indicating the source file of each row. This can be achieved using one of the following approaches:

Option 1: Add Filename as a New Column

dfs = []
for f in all_files:
    data = pd.read_csv(f)
    data['file'] = f.stem
    dfs.append(data)

df = pd.concat(dfs, ignore_index=True)

Option 2: Add Generic File Source as a New Column

dfs = []
for i, f in enumerate(all_files):
    data = pd.read_csv(f)
    data['file'] = f'File {i}'
    dfs.append(data)

df = pd.concat(dfs, ignore_index=True)

Option 3: Add File Source Using List Comprehension

dfs = [pd.read_csv(f) for f in all_files]
df = pd.concat(dfs, ignore_index=True)
df['Source'] = np.repeat([f'S{i}' for i in range(len(dfs))], [len(df) for df in dfs])

Option 4: Single-Line Solution with .assign()

df = pd.concat((pd.read_csv(f).assign(filename=f.stem) for f in all_files), ignore_index=True)

By implementing one of these options, the concatenated DataFrame will be annotated with information to trace the origin of each row.

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