我有一个有点大的.xlsx文件--19列,5185行。我想打开文件,读取一列中的所有值,对这些值执行一些操作,然后在同一工作簿中创建一个新列并写出修改后的值。因此,我需要能够在同一个文件中进行读写。
我的原始代码执行了此操作:
def readExcel(doc):
wb = load_workbook(generalpath + exppath + doc)
ws = wb["Sheet1"]
# iterate through the columns to find the correct one
for col in ws.iter_cols(min_row=1, max_row=1):
for mycell in col:
if mycell.value == "PerceivedSound.RESP":
origCol = mycell.column
# get the column letter for the first empty column to output the new values
newCol = utils.get_column_letter(ws.max_column+1)
# iterate through the rows to get the value from the original column,
# do something to that value, and output it in the new column
for myrow in range(2, ws.max_row+1):
myrow = str(myrow)
# do some stuff to make the new value
cleanedResp = doStuff(ws[origCol + myrow].value)
ws[newCol + myrow] = cleanedResp
wb.save(doc)
然而,python在第3853行之后抛出了内存错误,因为工作簿太大了。 openpyxl文档说使用只读模式(https://openpyxl.readthedocs.io/en/latest/optimized.html)来处理大型工作簿。我现在正试图用它;但是,当我添加read_only = True参数时似乎没有办法遍历列:
def readExcel(doc):
wb = load_workbook(generalpath + exppath + doc, read_only=True)
ws = wb["Sheet1"]
for col in ws.iter_cols(min_row=1, max_row=1):
#etc.
python抛出此错误: 属性错误:' ReadOnlyWorksheet'对象没有属性' iter_cols'
如果我将上述代码段中的最后一行更改为:
for col in ws.columns:
python抛出同样的错误: 属性错误:' ReadOnlyWorksheet'对象没有属性'列'
迭代行很好(并且包含在我上面链接的文档中):
for col in ws.rows:
(无错误)
This question询问AttritubeError,但解决方法是删除只读模式,这对我不起作用,因为openpyxl不会以非只读模式读取我的整个工作簿。
那么:我如何遍历大型工作簿中的列?
我还没遇到过这个问题,但是我可以在列中进行迭代:如果所说的工作簿很大,我如何读写同一个工作簿呢?
谢谢!
答案 0 :(得分:2)
如果工作表只有大约100,000个单元格,那么您不应该有任何内存问题。您可能应该进一步调查此事。
iter_cols()
在只读模式下不可用,因为它需要对基础XML文件进行常量且非常低效的重新分析。但是,使用iter_rows()
将行转换为zip
的列相对容易。
def _iter_cols(self, min_col=None, max_col=None, min_row=None,
max_row=None, values_only=False):
yield from zip(*self.iter_rows(
min_row=min_row, max_row=max_row,
min_col=min_col, max_col=max_col, values_only=values_only))
import types
for sheet in workbook:
sheet.iter_cols = types.MethodType(_iter_cols, sheet)
答案 1 :(得分:1)
根据documentation,ReadOnly模式仅支持基于行的读取(未实现列读取)。但这并不难解决:
wb2 = Workbook(write_only=True)
ws2 = wb2.create_sheet()
# find what column I need
colcounter = 0
for row in ws.rows:
for cell in row:
if cell.value == "PerceivedSound.RESP":
break
colcounter += 1
# cells are apparently linked to the parent workbook meta
# this will retain only values; you'll need custom
# row constructor if you want to retain more
row2 = [cell.value for cell in row]
ws2.append(row2) # preserve the first row in the new file
break
for row in ws.rows:
row2 = [cell.value for cell in row]
row2.append(doStuff(row2[colcounter]))
ws2.append(row2) # write a new row to the new wb
wb2.save('newfile.xlsx')
wb.close()
wb2.close()
# copy `newfile.xlsx` to `generalpath + exppath + doc`
# Either using os.system,subprocess.popen, or shutil.copy2()
您将无法写入同一工作簿,但如上所示,您可以打开一个新工作簿(以writeonly模式),写入它,并使用操作系统副本覆盖旧文件。