如何使用 openpyxl 将工作表从一个工作簿复制到另一个工作簿?

问题描述 投票:0回答:11

我有大量 EXCEL 文件(即 200 个),我想将一个特定工作表从一个工作簿复制到另一个工作簿。我做了一些调查,但找不到使用 Openpyxl 实现这一点的方法

这是我迄今为止开发的代码

def copy_sheet_to_different_EXCEL(path_EXCEL_read,Sheet_name_to_copy,path_EXCEL_Save,Sheet_new_name):
''' Function used to copy one EXCEL sheet into another file.
    
    def path_EXCEL_read,Sheet_name_to_copy,path_EXCEL_Save,Sheet_new_name
    
Input data:
    1.) path_EXCEL_read: the location of the EXCEL file along with the name where the information is going to be saved
    2.) Sheet_name_to_copy= The name of the EXCEL sheet to copy
    3.) path_EXCEL_Save: The path of the EXCEL file where the sheet is going to be copied
    3.) Sheet_new_name: The name of the new EXCEL sheet
    
Output data:
    1.) Status= If 0, everything went OK. If 1, one error occurred.

Version History:
1.0 (2017-02-20): Initial version.

'''
status=0

if(path_EXCEL_read.endswith('.xls')==1): 
    print('ERROR - EXCEL xls file format is not supported by openpyxl. Please, convert the file to an XLSX format')
    status=1
    return status
    
try:
   wb = openpyxl.load_workbook(path_EXCEL_read,read_only=True)
except:
    print('ERROR - EXCEL file does not exist in the following location:\n  {0}'.format(path_EXCEL_read))
    status=1
    return status

Sheet_names=wb.get_sheet_names()    # We copare against the sheet name we would like to cpy

if ((Sheet_name_to_copy in Sheet_names)==0):
    print('ERROR - EXCEL sheet does not exist'.format(Sheet_name_to_copy))
    status=1
    return status   

# We checking if the destination file exists

if (os.path.exists(path_EXCEL_Save)==1):
    #If true, file exist so we open it
    
    if(path_EXCEL_Save.endswith('.xls')==1): 
        print('ERROR - Destination EXCEL xls file format is not supported by openpyxl. Please, convert the file to an XLSX format')
        status=1
    return status
    
    try:
        wdestiny = openpyxl.load_workbook(path_EXCEL_Save)
    except:
        print('ERROR - Destination EXCEL file does not exist in the following location:\n  {0}'.format(path_EXCEL_read))
        status=1
    return status

    #we check if the destination sheet exists. If so, we will delete it
    
    destination_list_sheets = wdestiny.get_sheet_names()
    
    if((Sheet_new_name in destination_list_sheets) ==True):
        print('WARNING - Sheet "{0}" exists in: {1}. It will be deleted!'.format(Sheet_new_name,path_EXCEL_Save))
        wdestiny.remove_sheet(Sheet_new_name) 

else:
    wdestiny=openpyxl.Workbook()
# We copy the Excel sheet
    
try:
    sheet_to_copy = wb.get_sheet_by_name(Sheet_name_to_copy) 
    target = wdestiny.copy_worksheet(sheet_to_copy)
    target.title=Sheet_new_name
except:
    print('ERROR - Could not copy the EXCEL sheet. Check the file')
    status=1
    return status

try:
    wdestiny.save(path_EXCEL_Save)
except:
    print('ERROR - Could not save the EXCEL sheet. Check the file permissions')
    status=1
    return status

#Program finishes
return status
python excel copy openpyxl worksheet
11个回答
23
投票

我也有同样的问题。对我来说,风格、格式和布局非常重要。此外,我不想复制公式,而只想复制(公式的)值。经过大量的尝试、错误和 stackoverflow 之后,我想出了以下函数。它可能看起来有点吓人,但代码将一张工作表从一个 Excel 文件复制到另一个(可能是现有文件),同时保留:

  1. 文字的字体和颜色
  2. 单元格填充颜色
  3. 合并单元格
  4. 评论和超链接
  5. 单元格值的格式
  6. 每行每列的宽度
  7. 行列是否隐藏
  8. 冻结行

当您想要从多个工作簿中收集工作表并将它们绑定到一本工作簿中时,此功能非常有用。我复制了大部分属性,但可能还有更多。在这种情况下,您可以使用此脚本作为起点来添加更多内容。

###############
## Copy a sheet with style, format, layout, ect. from one Excel file to another Excel file
## Please add the ..path\\+\\file..  and  ..sheet_name.. according to your desire.

import openpyxl
from copy import copy

def copy_sheet(source_sheet, target_sheet):
    copy_cells(source_sheet, target_sheet)  # copy all the cel values and styles
    copy_sheet_attributes(source_sheet, target_sheet)


def copy_sheet_attributes(source_sheet, target_sheet):
    target_sheet.sheet_format = copy(source_sheet.sheet_format)
    target_sheet.sheet_properties = copy(source_sheet.sheet_properties)
    target_sheet.merged_cells = copy(source_sheet.merged_cells)
    target_sheet.page_margins = copy(source_sheet.page_margins)
    target_sheet.freeze_panes = copy(source_sheet.freeze_panes)

    # set row dimensions
    # So you cannot copy the row_dimensions attribute. Does not work (because of meta data in the attribute I think). So we copy every row's row_dimensions. That seems to work.
    for rn in range(len(source_sheet.row_dimensions)):
        target_sheet.row_dimensions[rn] = copy(source_sheet.row_dimensions[rn])

    if source_sheet.sheet_format.defaultColWidth is None:
        print('Unable to copy default column wide')
    else:
        target_sheet.sheet_format.defaultColWidth = copy(source_sheet.sheet_format.defaultColWidth)

    # set specific column width and hidden property
    # we cannot copy the entire column_dimensions attribute so we copy selected attributes
    for key, value in source_sheet.column_dimensions.items():
        target_sheet.column_dimensions[key].min = copy(source_sheet.column_dimensions[key].min)   # Excel actually groups multiple columns under 1 key. Use the min max attribute to also group the columns in the targetSheet
        target_sheet.column_dimensions[key].max = copy(source_sheet.column_dimensions[key].max)  # https://stackoverflow.com/questions/36417278/openpyxl-can-not-read-consecutive-hidden-columns discussed the issue. Note that this is also the case for the width, not onl;y the hidden property
        target_sheet.column_dimensions[key].width = copy(source_sheet.column_dimensions[key].width) # set width for every column
        target_sheet.column_dimensions[key].hidden = copy(source_sheet.column_dimensions[key].hidden)


def copy_cells(source_sheet, target_sheet):
    for (row, col), source_cell in source_sheet._cells.items():
        target_cell = target_sheet.cell(column=col, row=row)

        target_cell._value = source_cell._value
        target_cell.data_type = source_cell.data_type

        if source_cell.has_style:
            target_cell.font = copy(source_cell.font)
            target_cell.border = copy(source_cell.border)
            target_cell.fill = copy(source_cell.fill)
            target_cell.number_format = copy(source_cell.number_format)
            target_cell.protection = copy(source_cell.protection)
            target_cell.alignment = copy(source_cell.alignment)

        if source_cell.hyperlink:
            target_cell._hyperlink = copy(source_cell.hyperlink)

        if source_cell.comment:
            target_cell.comment = copy(source_cell.comment)


wb_target = openpyxl.Workbook()
target_sheet = wb_target.create_sheet(..sheet_name..)

wb_source = openpyxl.load_workbook(..path\\+\\file_name.., data_only=True)
source_sheet = wb_source[..sheet_name..]

copy_sheet(source_sheet, target_sheet)

if 'Sheet' in wb_target.sheetnames:  # remove default sheet
    wb_target.remove(wb_target['Sheet'])

wb_target.save('out.xlsx')

9
投票

我找到了一种方法来玩弄它

import openpyxl

xl1 = openpyxl.load_workbook('workbook1.xlsx')
# sheet you want to copy
s = openpyxl.load_workbook('workbook2.xlsx').active
s._parent = xl1
xl1._add_sheet(s)
xl1.save('some_path/name.xlsx')

8
投票

您不能使用

copy_worksheet()
在工作簿之间进行复制,因为它取决于工作簿之间可能有所不同的全局常量。唯一安全可靠的方法是逐行、逐个单元地进行。

您可能想阅读有关此功能的讨论


4
投票

为了提高速度,我在打开工作簿时使用

data_only
read_only
属性。而且
iter_rows()
也非常快。

@Oscar 的出色答案需要进行一些更改以支持 ReadOnlyWorksheet 和 EmptyCell

# Copy a sheet with style, format, layout, ect. from one Excel file to another Excel file
# Please add the ..path\\+\\file..  and  ..sheet_name.. according to your desire.
import openpyxl
from copy import copy


def copy_sheet(source_sheet, target_sheet):
    copy_cells(source_sheet, target_sheet)  # copy all the cel values and styles
    copy_sheet_attributes(source_sheet, target_sheet)


def copy_sheet_attributes(source_sheet, target_sheet):
    if isinstance(source_sheet, openpyxl.worksheet._read_only.ReadOnlyWorksheet):
        return
    target_sheet.sheet_format = copy(source_sheet.sheet_format)
    target_sheet.sheet_properties = copy(source_sheet.sheet_properties)
    target_sheet.merged_cells = copy(source_sheet.merged_cells)
    target_sheet.page_margins = copy(source_sheet.page_margins)
    target_sheet.freeze_panes = copy(source_sheet.freeze_panes)

    # set row dimensions
    # So you cannot copy the row_dimensions attribute. Does not work (because of meta data in the attribute I think). So we copy every row's row_dimensions. That seems to work.
    for rn in range(len(source_sheet.row_dimensions)):
        target_sheet.row_dimensions[rn] = copy(source_sheet.row_dimensions[rn])

    if source_sheet.sheet_format.defaultColWidth is None:
        print('Unable to copy default column wide')
    else:
        target_sheet.sheet_format.defaultColWidth = copy(source_sheet.sheet_format.defaultColWidth)

    # set specific column width and hidden property
    # we cannot copy the entire column_dimensions attribute so we copy selected attributes
    for key, value in source_sheet.column_dimensions.items():
        target_sheet.column_dimensions[key].min = copy(source_sheet.column_dimensions[key].min)   # Excel actually groups multiple columns under 1 key. Use the min max attribute to also group the columns in the targetSheet
        target_sheet.column_dimensions[key].max = copy(source_sheet.column_dimensions[key].max)  # https://stackoverflow.com/questions/36417278/openpyxl-can-not-read-consecutive-hidden-columns discussed the issue. Note that this is also the case for the width, not onl;y the hidden property
        target_sheet.column_dimensions[key].width = copy(source_sheet.column_dimensions[key].width) # set width for every column
        target_sheet.column_dimensions[key].hidden = copy(source_sheet.column_dimensions[key].hidden)


def copy_cells(source_sheet, target_sheet):
    for r, row in enumerate(source_sheet.iter_rows()):
        for c, cell in enumerate(row):
            source_cell = cell
            if isinstance(source_cell, openpyxl.cell.read_only.EmptyCell):
                continue
            target_cell = target_sheet.cell(column=c+1, row=r+1)

            target_cell._value = source_cell._value
            target_cell.data_type = source_cell.data_type

            if source_cell.has_style:
                target_cell.font = copy(source_cell.font)
                target_cell.border = copy(source_cell.border)
                target_cell.fill = copy(source_cell.fill)
                target_cell.number_format = copy(source_cell.number_format)
                target_cell.protection = copy(source_cell.protection)
                target_cell.alignment = copy(source_cell.alignment)

            if not isinstance(source_cell, openpyxl.cell.ReadOnlyCell) and source_cell.hyperlink:
                target_cell._hyperlink = copy(source_cell.hyperlink)

            if not isinstance(source_cell, openpyxl.cell.ReadOnlyCell) and source_cell.comment:
                target_cell.comment = copy(source_cell.comment)

使用类似

    wb = Workbook()
    
    wb_source = load_workbook(filename, data_only=True, read_only=True)
    for sheetname in wb_source.sheetnames:
        source_sheet = wb_source[sheetname]
        ws = wb.create_sheet("Orig_" + sheetname)
        copy_sheet(source_sheet, ws)

    wb.save(new_filename)

3
投票

我有类似的要求,将多个工作簿中的数据整理到一本工作簿中。因为 openpyxl 中没有可用的内置方法。

我创建了以下脚本来为我完成这项工作。

注意:在我的用例中,所有工作簿都包含相同格式的数据。

from openpyxl import load_workbook
import os


# The below method is used to read data from an active worksheet and store it in memory.
def reader(file):
    global path
    abs_file = os.path.join(path, file)
    wb_sheet = load_workbook(abs_file).active
    rows = []
    # min_row is set to 2, to ignore the first row which contains the headers
    for row in wb_sheet.iter_rows(min_row=2):
        row_data = []
        for cell in row:
            row_data.append(cell.value)
        # custom column data I am adding, not needed for typical use cases
        row_data.append(file[17:-6])
        # Creating a list of lists, where each list contain a typical row's data
        rows.append(row_data)
    return rows


if __name__ == '__main__':
    # Folder in which my source excel sheets are present
    path = r'C:\Users\tom\Desktop\Qt'
    # To get the list of excel files
    files = os.listdir(path)
    for file in files:
        rows = reader(file)
        # below mentioned file name should be already created
        book = load_workbook('new.xlsx')
        sheet = book.active
        for row in rows:
            sheet.append(row)
        book.save('new.xlsx')

3
投票

我的解决方法是这样的:

您有一个模板文件,假设它是“template.xlsx”。 您打开它,根据需要对其进行更改,将其另存为新文件,然后关闭该文件。 根据需要重复。只要确保在测试/乱搞时保留原始模板的副本即可。


2
投票

我刚刚发现这个问题。正如here提到的,一个好的解决方法可以是修改内存中的原始

wb
,然后用另一个名称保存它。例如:

import openpyxl

# your starting wb with 2 Sheets: Sheet1 and Sheet2
wb = openpyxl.load_workbook('old.xlsx')

sheets = wb.sheetnames # ['Sheet1', 'Sheet2']

for s in sheets:

    if s != 'Sheet2':
        sheet_name = wb.get_sheet_by_name(s)
        wb.remove_sheet(sheet_name)

# your final wb with just Sheet1
wb.save('new.xlsx')

0
投票

我使用的解决方法是将当前工作表另存为 pandas 数据框并将其加载到您需要的 Excel 工作簿中


0
投票

其实可以用非常简单的方法来完成! 只需要3步:

  1. 使用 load_workbook 打开文件

    wb = load_workbook('File_1.xlsx')

  2. 选择您要复制的工作表

    ws = wb.active

  3. 使用新文件的名称来保存文件

    wb.save('New_file.xlsx')

此代码会将第一个文件(File_1.xlsx)的工作表保存到第二个文件(New_file.xlsx)。


0
投票

通过使用 openpyxl - 边框复制未成功。 就我而言 - 使用 xlwings 取得了成功。它在操作系统中打开 Excel,将选项卡复制到其他 Excel,保存,重命名并关闭。

import openpyxl, os
import xlwings as xw

def copy_tab(file_old, tab_source, file_new, tab_destination):
    delete_tab = False
    if not os.path.exists(file_new):
        wb_target = openpyxl.Workbook()
        wb_target.save(file_new)
        delete_tab = True

    wb = xw.Book(file_old)
    app = wb.app
    app.visible = False
    sht = wb.sheets[tab_source]
    new_wb = xw.Book(file_new)
    new_app = new_wb.app
    new_app.visible = False
    sht.api.Copy(None, After=new_wb.sheets[-1].api)
    if delete_tab:
        new_wb.sheets['Sheet'].delete()
    wb.close()
    for sheet in new_wb.sheets:
        if tab_destination in sheet.name:
            sheet.delete()
    new_wb.sheets[tab_source].name = tab_destination
    new_wb.save()
    new_wb.close()

if __name__ == "__main__":
    file_old = r"C:\file_old.xlsx"
    file_new = r"C:\file_new.xlsx"

    copy_tab(file_old, "sheet_old", file_new, "sheet_new")

0
投票

Oscar 的解决方案
中使用
deepcopy
而不是循环 source_sheet.row_dimensions 将消除 LibreOffice 错误:“无法完全加载数据,因为超出了每张纸的最大行数。

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