Copy Cell Style Openpyxl

Python Excel copy cell style from one to another openpyxl

Shortly after typing this out, I found the solution.

from copy import copy
wb = openpyxl.load_workbook('C:\\path...\\')
sheet = wb["Name of the worksheet"]

sheet["E1"].font = copy(sheet["D1"].font)
sheet["E1"].border = copy(sheet["D1"].border)
sheet["E1"].fill = copy(sheet["D1"].fill)
sheet["E1"].number_format = copy(sheet["D1"].number_format)
sheet["E1"].protection = copy(sheet["D1"].protection)
sheet["E1"].alignment = copy(sheet["D1"].alignment)

The only thing left to do would be to do this in a loop. That would be achievable by doing something like

for i in range(....): 
    sheet["E" + str(i)].font= copy(sheet["D" +str(i)].font) 
    etc. 

Copying styles from a range to another range?

Your workload are to copy the six styles, e.g.

new_third_col_cell.font = copy(third_col_cell.font)

Try to copy only the style reference, instead of assigning new styles, for instance

target_cell._style = copy(source_cell._style)

You can save ~10 %, per Style, if not all Styles are are used in all cells.

Copy only used styles, for instance:

if source_cell._style.fontId: target_cell.font = copy(s_cell.font)
...

Besides this, consider to slim your code, for instance:

def copy_style(s_cell, t_cell):
    ...

def merge_cells(ws, row, offset, columns_456):
    ...

def main(recommended, plans):
    column_456 = [4,5,6]

    for a in xrange(1, len(plans)):
        offset = 3 * a

        # Create blank templates for plans
        for x in xrange(5, 549):
            if x != 42:
              for column in column_456:
                  cell = recommended.cell(row=x, column=column)
                  if cell.has_style:
                      copy_style(cell, recommended.cell(row=x, column=column+offset))

            if (x >= 6 and x <= 33) or x == 36 or x == 41 or x == 44:
                merge_cells(recommended, x, offset, column_456)

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