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打开特定的excel文件/表格以获取特定的内容单元格位置

[英]Open specific excel file/sheet to get a specific content cell location

I have a list of excel files and their corresponding sheet number. 我有一个excel文件及其对应的工作表编号的列表。 I need python to go to those sheets and find out the cell location for a particular content. 我需要python转到这些工作表,并找到特定内容的单元格位置。 Can someone point out the error in my code? 有人可以指出我的代码中的错误吗? Thanks in advance 提前致谢

import xlrd
value = 'Avg.'
filename = ('C:/002 Av SW W of 065 St SW 2011-Jul-05.xls', 'C:/003 Avenue SW West of 058 Street SW 2012-Jun-23.xls')
sheetnumber = ('505840', '505608')
dictionary = dict(zip(filename, sheetnumber))
for item in dictionary:
    book = xlrd.open_workbook(item)
    sheet = book.sheet_by_name(dictionary[key])
    for row in range(sheet.nrows):
        for column in range(sheet.ncols):
            if sheet.cell(row,column).value == value:
                print row, column

You don't need to make a dictionary. 您无需制作字典。 Iterate over zip(filename, sheetnumber) : 遍历zip(filename, sheetnumber)

for name, sheet_name in zip(filename, sheetnumber):
    book = xlrd.open_workbook(name)
    sheet = book.sheet_by_name(sheet_name)
    for row in range(sheet.nrows):
        for column in range(sheet.ncols):
            if sheet.cell(row,column).value == value:
                print row, column

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