【问题标题】:How to split merged cells in excel workbook using python如何使用python拆分excel工作簿中的合并单元格
【发布时间】:2019-03-29 21:01:20
【问题描述】:

有没有办法使用 python 拆分/取消合并 Excel 工作簿中的单元格?我想要的解释如下-

结果应该是一个包含以下条目的新 excel 文件 -

我使用 xlrd 为所有合并列复制相同字符串的解决方案如下 -

[注意:“formatted_info = True”标志尚未在我使用的 xlrd 中实现,因此我无法直接获取合并单元格的列表。我不应该在设置上升级 xlrd。]

def xlsx_to_dict():
    workbook = xlrd.open_workbook(xlsfile)
    worksheet_names = workbook.sheet_names()
    for worksheet_name in worksheet_names:
        worksheet = workbook.sheet_by_name(worksheet_name)
        num_rows = worksheet.nrows - 1
        num_cells = worksheet.ncols - 1
        curr_row = -1
        header_row = worksheet.row(0)
        columns = []
        for cell in range(len(header_row)):
            value = worksheet.cell_value(0, cell)
            columns.append(value)

        cities = []

        for row in range(1,num_rows):
            value = worksheet.cell_value(row,0)
            type = worksheet.cell_type(row,0)
            if  not value == "":
                cities.append(value)

        names = []
        for row in range(1,num_rows):
            value = worksheet.cell_value(row,1)
            type = worksheet.cell_type(row,1)
            if  not value == "":
                names.append(value)

            current_city = cities[0]
            result_dict = {}
            for curr_row in range(1,num_rows):
                row = worksheet.row(curr_row)
                curr_cell = -1
                curr_name = names[0]
                while curr_cell < num_cells:
                    curr_cell += 1
                    cell_value = worksheet.cell_value(curr_row, curr_cell)
                    if cell_value in cities and curr_cell == 0:
                        current_city = cell_value
                        if not result_dict.has_key(current_city):
                            result_dict[current_city] = {}
                        continue
                    if cell_value == "" and curr_cell == 0:
                        continue
                    if cell_value in names and curr_cell == 1:
                        curr_name = cell_value
                        if not result_dict[current_city].has_key(curr_name):
                            result_dict[current_city][curr_name] = {}
                        continue
                    if cell_value == "" and curr_cell == 1:
                        continue
                    try:
                        result_dict[current_city][curr_name]['Phone'].append(cell_Value)
                    except:
                        result_dict[current_city][curr_name]['Phone'] = [cell_value]

上面的函数将返回python字典如下-

{ 'New York' : { 'Tom' : [92929292, 33929] }, ........}

然后我将遍历目录并编写新的excel。

但是,我想要一些通用的拆分合并单元格的方法。

【问题讨论】:

  • 请分享您到目前为止尝试过的内容吗?否则人们会一直反对它

标签: python


【解决方案1】:

此函数获取“真实”单元格值,即如果坐标位于合并单元格内的任意位置,则为合并单元格的值。

def unmergedValue(rowx,colx,thesheet):
    for crange in thesheet.merged_cells:
        rlo, rhi, clo, chi = crange
        if rowx in xrange(rlo, rhi):
            if colx in xrange(clo, chi):
                return thesheet.cell_value(rlo,clo)
    #if you reached this point, it's not in any merged cells
    return thesheet.cell_value(rowx,colx)

大致基于http://www.lexicon.net/sjmachin/xlrd.html#xlrd.Sheet.merged_cells-attribute

非常低效,但对于小型电子表格应该是可以接受的。

【讨论】:

    【解决方案2】:

    如果你的文件中间没有空单元格,这可能会有所帮助,读取文件,做一些工作,重写它。

    def read_merged_xls(file_contents):
        book = xlrd.open_workbook(file_contents=file_contents)
        data = []
        sheet = book.sheet_by_index(0)
        for rx in range(sheet.nrows):
            line = [] 
            for ry in range(sheet.ncols):
                cell = sheet.cell_value(rx,ry)
                if not cell:
                    cell = data[-1][ry] if data else ''
                line.append(cell)
            data.append(line)
        return data
    

    【讨论】:

      【解决方案3】:
      import xlrd
      import xlsxwriter
      import numpy as np
      import pandas as pd
      def rep(l,i):
          j= i
          while(j>=0):
              if not l[j-1] == u'':
                  return l[j-1]
              else:
                  j = j-1
      def write_df2xlsx(df,filename):
          # Create a Pandas Excel writer using XlsxWriter as the engine.
          writer = pd.ExcelWriter(filename,engine='xlsxwriter')
      
          # Convert the dataframe to an XlsxWriter Excel object.
          df.to_excel(writer, sheet_name='Sheet1', index = False)
      
          # Close the Pandas Excel writer and output the Excel file.
          writer.save()
      
      def csv_from_excel(filename):
      
          wb = xlrd.open_workbook(filename)
          worksheet_names = wb.sheet_names()
          for worksheet_name in worksheet_names:
              sh = wb.sheet_by_name(worksheet_name)
              #To find the headers/column names of the xlsx file
      
              header_index = 0
              for i in range(sh.nrows):
                  if(len(filter(lambda x: not (x.value == xlrd.empty_cell.value), sh.row(i))) == len(sh.row(i))):
                      header_row = sh.row(i)
                      header_index = i
                      break
              columns = []
              for cell in range(len(header_row)):
                  value = sh.cell_value(header_index, cell)
                  columns.append(value)
              rows = []
              for rownum in range(header_index+1,sh.nrows):
                  rows.append(sh.row_values(rownum))
              data = pd.DataFrame(rows,columns = columns)
              cols = [col for col in data.columns if u'' in list(data[col])]
              res = []
              for col in cols:
                  t_list = list(data[col])
                  res.append(map(lambda x,y: rep(list(data[col]),y[0]) if x == u'' else x,t_list,enumerate(t_list)))
              for (col,r) in zip(cols,res):
                  data[col] = pd.core.series.Series(r)
              write_df2xlsx(data,'ResultFile.xlsx')   
      

      【讨论】:

      • 欢迎来到 StackOverflow。在发布代码作为答案时,最好附上简短的解释。这是How to Answer的指南
      猜你喜欢
      • 1970-01-01
      • 1970-01-01
      • 1970-01-01
      • 2016-11-21
      • 1970-01-01
      • 1970-01-01
      • 1970-01-01
      • 1970-01-01
      • 2018-05-24
      相关资源
      最近更新 更多