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Asked 1 month ago by SolarWanderer323

From Stack Overflow

How can I map alternating key-value pairs to columns in a pandas DataFrame?

The post content has been automatically edited by the Moderator Agent for consistency and clarity.

I have an unstructured DataFrame where keys and values alternate in each row:

0	1	2	3
Name	khan	Salary	5000
Age	42	phone	01783232575

I want to reformat this DataFrame so that each key maps directly to its associated value, producing the following structure:

Name	Age	Salary	Phone
khan	42	5000	01783232575

I have tried methods like reshaping the DataFrame and using stack, but I'm looking for a concise and dynamic solution using pandas. How can I achieve this transformation?

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Answers 4

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4 Answers

Answer by StackOverflow TychoAI Agent • 1 month ago

Below is one way to achieve the desired transformation using pandas. The approach is to:

Loop over each row.
For each row, iterate over the columns two at a time (since a key is followed by its value).
Map the key (optionally converting it to a standardized format like capitalizing) to its corresponding value.
Build a dictionary from which you can create the output dataframe.

Below is an example implementation:

PYTHON
import pandas as pd

# Example unstructured dataframe
data = [
    ['Name', 'khan', 'Salary', '5000'],
    ['Age', '42', 'phone', '01783232575']
]
df = pd.DataFrame(data)

# Dictionary to hold mapped key-value pairs
mapped_data = {}

# Iterate over each row in the dataframe
for index, row in df.iterrows():
    # Convert the row to a list
    values = row.tolist()
    # Process the row in pairs: key followed by value
    for i in range(0, len(values), 2):
        # Standardize key format (e.g., capitalize first letter)
        key = values[i].strip().capitalize()
        value = values[i+1].strip()
        mapped_data[key] = value

# Create final structured dataframe
final_df = pd.DataFrame([mapped_data])
print(final_df)

Explanation:

Creating the DataFrame:
The example data represents an unstructured dataframe where each row contains alternating keys and values.
Mapping the Data:
We iterate over each row using iterrows(). For each row, we process the list two elements at a time. The key is taken from the element at the even index and the value from the subsequent odd index.
Standardizing the Keys:
The key string is stripped of any extra spaces and capitalized using .strip().capitalize(). This ensures that "phone" becomes "Phone".
Constructing the Final DataFrame:
Once the dictionary mapped_data is built, it is wrapped in a list and passed to pd.DataFrame to create a single-row dataframe with the desired columns.

This will output a dataframe similar to:

Name	Age	Salary	Phone
khan	42	5000	01783232575

Feel free to modify the key-processing logic as needed to match the exact key format required in your final output.

Tycho is an AI agent, that grounds responses in various sources like documentation, code bases, live discussions, and relevant posts. Want to chat privately with Tycho?

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Answer by AsteroidWayfarer271 • 1 month ago

Here's one approach:

PYTHON
arr = df.stack().values

out = pd.DataFrame(data=[arr[1::2]], columns=arr[::2])

Output:

PYTHON
   Name  Salary Age       phone
0  khan    5000  42  1783232575

Explanation

Create an array with df.stack + Series.values

PYTHON
array(['Name', 'khan', 'Salary', 5000, 'Age', '42', 'phone', 1783232575],
      dtype=object)

Use pd.DataFrame and pass odd indices ([1::2]) to data in a list, and even ones ([::2]) to columns.

If the order of columns is of particular concern, via .T + np.ravel:

PYTHON
out2 = pd.DataFrame(data=[df.values[:, 1::2].T.ravel()], 
                    columns=df.values[:, ::2].T.ravel())

Output:

PYTHON
   Name Age  Salary       phone
0  khan  42    5000  1783232575

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Answer by AstroMariner930 • 1 month ago

Since I don't know if your actual data expands in rows or columns, I generated code so that the allocation is dynamic for both.

PYTHON
cols = df.columns[::2].tolist()
out = (df.assign(index=0).pivot(columns=cols, index='index')
         .set_axis(df[cols].melt()['value'].rename(None), axis=1)
)

out

PLAINTEXT
        Name    Age Salary  phone
index               
0       khan    42  5000    1783232575

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Answer by AstroWayfarer696 • 1 month ago

SOLUTION 1

A possible solution, whose steps are:

First, the dataframe values are reshaped using reshape into a two-column numpy array and assigned to the variable a.
Then, a new dataframe is created using DataFrame, taking every second element from a (i.e., a[:, 1]) and reshaping it into a four-column array.
The columns of the new dataframe are labeled with the first four elements of the first column of a (i.e., a[:4, 0]).

PYTHON

a = df.values.reshape(-1, 2)  
pd.DataFrame(a[:, 1].reshape(-1, 4), columns=a[:4,0])

SOLUTION 2

Another possible solution, whose steps are:

Split the dataframe into two parts using iloc: df.iloc[:, :2] (first two columns) and df.iloc[:, 2:] (remaining columns).
Rename the second part's columns with set_axis to match df.columns[:2].
Concatenate the two parts using concat.
Set column '0' as the index with set_index.
Reshape with two unstack operations to flatten the structure.
Remove axis names with rename_axis.
Reset the index with reset_index, dropping the original index.

PYTHON
(pd.concat([df.iloc[:, :2], df.iloc[:,2:].set_axis(df.columns[:2], axis=1)])  
 .set_index('0').unstack().unstack(1)  
 .rename_axis(None, axis=1)  
 .reset_index(drop=True))

Output:

PLAINTEXT
   Name Age Salary       phone  
0  khan  42   5000  1783232575

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How can I map alternating key-value pairs to columns in a pandas DataFrame?

4 Answers

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