Batch Generate Word Docs from Excel with Python: zipfile & python-docx Solutions

Task Requirement

Given an Excel file containing target data, each row's values need to replace specific placeholders in a Word document and be saved individually.

Task Breakdown

The Word template is a .doc file; the placeholders are English strings highlighted in red within the main body.

The Excel file is well‑structured, so column names must match the placeholders in the Word document.

Solution 1: Using python-docx to Read Word

Problems Encountered

Incorrect installation command; the correct package is python-docx. python-docx cannot manipulate .doc files; the file must be converted to .docx first, which is done via win32com.

Text replacement fails when runs are split by formatting or language input, causing underlines to disappear. The workaround is to replace text at the run level and ensure placeholders are unique (e.g., #ColumnName#).

After solving these issues, the following script performs the conversion and replacement:

from copy import deepcopy
from pathlib import Path
from win32com import client as wc  # pip install pypiwin32
from docx import Document  # pip install python-docx
import pandas as pd

# Convert .doc to .docx if needed
def doctransform2docx(doc_path):
    docx_path = doc_path + 'x'
    suffix = doc_path.split('.')[-1]
    assert 'doc' in suffix, 'Input is not a Word document!'
    if suffix == 'docx':
        return Document(doc_path)
    word = wc.Dispatch('Word.Application')
    doc = word.Documents.Open(doc_path)
    doc.SaveAs2(docx_path, 16)  # 16 = wdFormatXMLDocument
    doc.Close()
    word.Quit()
    return Document(docx_path)

# Replace placeholders in a docx Document object
def replace_docx(name, values, wordfile, path_name='Company'):
    wordfile_copy = deepcopy(wordfile)
    for col_name, value in zip(name, values):
        if col_name == 'Company':
            path_name = str(value)
        for paragraph in wordfile_copy.paragraphs:
            for run in paragraph.runs:
                run.text = run.text.replace(col_name, str(value))
    wordfile_copy.save(f'{save_folder}/{path_name}.docx')

if __name__ == '__main__':
    doc_path = r"D:\solve_path\单位.doc"
    excel_path = r"D:\solve_path\信息.xls"
    save_folder = Path('D:/docx_save')
    save_folder.mkdir(parents=True, exist_ok=True)
    data = pd.read_excel(excel_path)
    wordfile = doctransform2docx(doc_path)
    data.apply(lambda x: replace_docx(x.index, x.values, wordfile), axis=1)

Even after successful replacement, the original checkboxes disappeared, indicating formatting loss.

Solution 2: Using zipfile to Edit the Underlying XML

Since a .docx file is essentially a ZIP archive of XML files, the workflow is:

Convert .doc to .docx via win32com (same as before).

Unzip the .docx and read word/document.xml.

Replace placeholders directly in the XML string.

Re‑zip the folder back into a .docx file.

Key pitfalls:

Use zipfile.zlib.DEFLATED to preserve the original folder hierarchy.

When writing files into the archive, provide the relative path (no absolute paths).

from shutil import rmtree
import zipfile
from copy import deepcopy
from pathlib import Path
from win32com import client as wc  # pip install pypiwin32
import pandas as pd

# Convert .doc to .docx (same as before)
def doctransform2docx(doc_path):
    docx_path = doc_path + 'x'
    suffix = doc_path.split('.')[-1]
    assert 'doc' in suffix, 'Input is not a Word document!'
    if suffix == 'docx':
        return Path(doc_path)
    word = wc.Dispatch('Word.Application')
    doc = word.Documents.Open(doc_path)
    doc.SaveAs2(docx_path, 16)
    doc.Close()
    word.Quit()
    return Path(docx_path)

# Unzip .docx and read document.xml
def docx_unzip(docx_path):
    docx_path = Path(docx_path)
    unzip_path = docx_path.with_name(docx_path.stem)
    with zipfile.ZipFile(docx_path, 'r') as f:
        for file in f.namelist():
            f.extract(file, path=unzip_path)
    xml_path = unzip_path.joinpath('word/document.xml')
    xml_file = xml_path.read_text(encoding='utf-8')
    return unzip_path, xml_path, xml_file

# Re‑zip the folder into a .docx
def docx_zipped(docx_path, zipped_path):
    docx_path = Path(docx_path)
    with zipfile.ZipFile(zipped_path, 'w', zipfile.zlib.DEFLATED) as f:
        for file in docx_path.glob('**/*.*'):
            f.write(file, file.as_posix().replace(docx_path.as_posix() + '/', ''))

# Remove temporary folder
def remove_folder(path):
    path = Path(path)
    if path.exists():
        rmtree(path)
    else:
        raise "System cannot find the specified file"

# Replace placeholders in the XML string and save
def replace_docx(name, values, xml_file, xml_path, unzip_path, path_name='Company'):
    xml_path = Path(xml_path)
    xml_file_copy = deepcopy(xml_file)
    for col_name, value in zip(name, values):
        if col_name == 'Company':
            path_name = str(value)
        xml_file_copy = xml_file_copy.replace(col_name, str(value))
    xml_path.write_text(xml_file_copy, encoding='utf-8')
    docx_zipped(unzip_path, f'{save_folder}/{path_name}.docx')

if __name__ == '__main__':
    doc_path = r"D:\solve_path\单位.doc"
    excel_path = r"D:\solve_path\信息.xls"
    save_folder = Path('D:/docx_save')
    save_folder.mkdir(parents=True, exist_ok=True)
    data = pd.read_excel(excel_path)
    docx_path = doctransform2docx(doc_path)
    unzip_path, xml_path, xml_file = docx_unzip(docx_path)
    data.apply(lambda x: replace_docx(x.index, x.values, xml_file, xml_path, unzip_path), axis=1)
    remove_folder(unzip_path)

After re‑zipping, the checkboxes and underlines are preserved.

Conclusion

Through trial and error, a method was found that replaces strings in a Word document without altering its original formatting. While this solution works, there may be more elegant approaches.