使用Openpyxl将工作表从一个工作簿复制到另一个工作表
我有大量的EXCEL文件（即200）我想复制一个特定的工作表从一个工作簿到另一个。 我做了一些调查，我找不到用Openpyxl做的一个方法
这是迄今为止我所开发的代码
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 有什么build议么？
干杯
您不能使用copy_worksheet()在工作簿之间进行复制，因为它依赖于各个工作簿之间可能会有所不同的全局常量。 唯一可行的安全可靠的方法是逐行逐行进行。
您可能想阅读有关此function的讨论
我有一个类似的要求，将来自多个工作簿的数据整理到一个工作簿中。 由于在openpyxl中没有内置的方法。
我创build了下面的脚本来为我完成这个工作。
注意：在我的用例中，所有的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')
我刚刚发现了这个问题。 如上所述，一个很好的解决方法是修改原来的内存中的wb ，并用另一个名字保存。 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')