Python Pandas: Apply Function to Split Column and Generate Multiple New Columns

Here's the breakdown:

1. Import pandas:

   import pandas as pd
   

2. Create a sample DataFrame:

   data = {'text_col': ['apple banana', 'cherry orange']}
   df = pd.DataFrame(data)
   

3. Define a function to process the column:

   This function will take a single string value from the column and return a list of values representing the new columns.

   def split_and_extract(text):
       fruits = text.split()  # Split the string into a list of words
       length = len(fruits)  # Get the number of words
       has_apple = 'apple' in fruits  # Check if "apple" is present
       return length, has_apple  # Return a list of values
   

4. Apply the function using apply:

   new_cols = df['text_col'].apply(split_and_extract)
   

5. Unpack the results and create new columns:

   Since split_and_extract returns a list, we need to unpack it and assign them as separate columns. You can use list unpacking or column selection with indexing:

   Option 1: List unpacking

   df['num_words'], df['has_apple'] = zip(*new_cols)
   

   Option 2: Column selection with indexing

   df['num_words'] = new_cols[0]
   df['has_apple'] = new_cols[1]
   

Complete code:

import pandas as pd

data = {'text_col': ['apple banana', 'cherry orange']}
df = pd.DataFrame(data)

def split_and_extract(text):
    fruits = text.split()
    length = len(fruits)
    has_apple = 'apple' in fruits
    return length, has_apple

new_cols = df['text_col'].apply(split_and_extract)
df['num_words'], df['has_apple'] = zip(*new_cols)

print(df)

Output:

   text_col  num_words  has_apple
0  apple banana         2       True
1  cherry orange         2      False

This approach effectively creates new columns based on the logic defined in your function.

import pandas as pd

# Sample data with an address column
data = {'address': ['123 Main St, Anytown, CA 12345', '456 Elm St, Springfield, IL 67890']}
df = pd.DataFrame(data)

# Function to extract city, state, and zip code
def extract_location(address):
    parts = address.split(',')  # Split on commas
    city = parts[1].strip()  # Extract city (assuming format)
    state = parts[2].split()[0].strip()  # Extract state (assuming format)
    zip_code = parts[2].split()[1].strip()  # Extract zip code (assuming format)
    return city, state, zip_code  # Return a tuple of values

# Apply the function using `apply`
new_cols = df['address'].apply(extract_location)

# Option 1: Using list unpacking (more concise)
df['city'], df['state'], df['zip_code'] = zip(*new_cols)

# Option 2: Using column selection with indexing (more explicit)
# df['city'] = new_cols[0]
# df['state'] = new_cols[1]
# df['zip_code'] = new_cols[2]  # Uncomment if preferred

print(df)

Explanation:

1. Function with specific address format: This extract_location function assumes a specific address format with commas separating city, state, and zip code. You can modify it to handle different formats based on your data.
2. Returning a tuple: The function now returns a tuple containing city, state, and zip code.
3. List unpacking (Option 1): This option directly unpacks the results from apply into three new columns named city, state, and zip_code.
4. Column selection with indexing (Option 2, commented out): This option provides more explicit control by assigning each element of the returned tuple (accessible by index) to its respective column.

This code demonstrates how to adapt the core concept to extract specific information from a column and create new columns with the results. Remember to adjust the extract_location function to match your actual address format.

List comprehension and assignment:

This approach uses a list comprehension to create a list of lists containing the results for each row, then assigns them directly to new columns:

def split_and_extract(text):
    fruits = text.split()
    length = len(fruits)
    has_apple = 'apple' in fruits
    return length, has_apple

data = {'text_col': ['apple banana', 'cherry orange']}
df = pd.DataFrame(data)

num_words = [len(row.split()) for row in df['text_col']]
has_apple = ['apple' in row for row in df['text_col']]

df['num_words'] = num_words
df['has_apple'] = has_apple

print(df)

Vectorized operations with NumPy (if applicable):

If your function can be vectorized using NumPy functions, it can be significantly faster than using apply. This approach works best for element-wise operations on entire columns.

import numpy as np

data = {'text_col': ['apple banana', 'cherry orange']}
df = pd.DataFrame(data)

num_words = np.array([len(row.split()) for row in df['text_col']])
has_apple = np.array(['apple' in row for row in df['text_col']])

df['num_words'] = num_words
df['has_apple'] = has_apple

print(df)

Custom function with column selection:

You can define a custom function that takes the DataFrame and column name as arguments, processes the column, and assigns the results to new columns:

def create_new_columns(df, col_name):
    fruits = df[col_name].str.split()
    df['num_words'] = fruits.str.len()
    df['has_apple'] = fruits.str.contains('apple')

data = {'text_col': ['apple banana', 'cherry orange']}
df = pd.DataFrame(data)

create_new_columns(df, 'text_col')

print(df)

Choosing the right method:

• Use list comprehension or custom function with column selection for simple transformations and better readability.
• Consider vectorized operations if your function can be efficiently vectorized using NumPy for performance gains.
• Generally avoid apply for performance reasons unless the function is complex and not easily vectorized.