18

我正在寻找使用 openpyxl 将行插入电子表格的最佳方法。

实际上,我有一个电子表格(Excel 2007),它有一个标题行,后面是(最多)几千行数据。我希望将该行作为实际数据的第一行插入,所以在标题之后。我的理解是 append 函数适用于在文件末尾添加内容。

阅读 openpyxl 和 xlrd(和 xlwt)的文档,除了手动循环内容并插入新工作表(在插入所需的行之后)之外,我找不到任何明确的方法来做到这一点。

鉴于我迄今为止对 Python 的有限经验,我试图了解这是否确实是最好的选择(最 Pythonic!),如果是这样,有人可以提供一个明确的例子。具体来说,我可以使用 openpyxl 读写行还是必须访问单元格?另外我可以(过)写相同的文件(名称)吗?

4

12 回答 12

19

== 根据此处的反馈更新到功能齐全的版本:groups.google.com/forum/#!topic/openpyxl-users/wHGecdQg3Iw。==

正如其他人指出的那样,openpyxl不提供此功能,但我已将Worksheet类扩展如下以实现插入行。希望这对其他人有用。

def insert_rows(self, row_idx, cnt, above=False, copy_style=True, fill_formulae=True):
    """Inserts new (empty) rows into worksheet at specified row index.

    :param row_idx: Row index specifying where to insert new rows.
    :param cnt: Number of rows to insert.
    :param above: Set True to insert rows above specified row index.
    :param copy_style: Set True if new rows should copy style of immediately above row.
    :param fill_formulae: Set True if new rows should take on formula from immediately above row, filled with references new to rows.

    Usage:

    * insert_rows(2, 10, above=True, copy_style=False)

    """
    CELL_RE  = re.compile("(?P<col>\$?[A-Z]+)(?P<row>\$?\d+)")

    row_idx = row_idx - 1 if above else row_idx

    def replace(m):
        row = m.group('row')
        prefix = "$" if row.find("$") != -1 else ""
        row = int(row.replace("$",""))
        row += cnt if row > row_idx else 0
        return m.group('col') + prefix + str(row)

    # First, we shift all cells down cnt rows...
    old_cells = set()
    old_fas   = set()
    new_cells = dict()
    new_fas   = dict()
    for c in self._cells.values():

        old_coor = c.coordinate

        # Shift all references to anything below row_idx
        if c.data_type == Cell.TYPE_FORMULA:
            c.value = CELL_RE.sub(
                replace,
                c.value
            )
            # Here, we need to properly update the formula references to reflect new row indices
            if old_coor in self.formula_attributes and 'ref' in self.formula_attributes[old_coor]:
                self.formula_attributes[old_coor]['ref'] = CELL_RE.sub(
                    replace,
                    self.formula_attributes[old_coor]['ref']
                )

        # Do the magic to set up our actual shift    
        if c.row > row_idx:
            old_coor = c.coordinate
            old_cells.add((c.row,c.col_idx))
            c.row += cnt
            new_cells[(c.row,c.col_idx)] = c
            if old_coor in self.formula_attributes:
                old_fas.add(old_coor)
                fa = self.formula_attributes[old_coor].copy()
                new_fas[c.coordinate] = fa

    for coor in old_cells:
        del self._cells[coor]
    self._cells.update(new_cells)

    for fa in old_fas:
        del self.formula_attributes[fa]
    self.formula_attributes.update(new_fas)

    # Next, we need to shift all the Row Dimensions below our new rows down by cnt...
    for row in range(len(self.row_dimensions)-1+cnt,row_idx+cnt,-1):
        new_rd = copy.copy(self.row_dimensions[row-cnt])
        new_rd.index = row
        self.row_dimensions[row] = new_rd
        del self.row_dimensions[row-cnt]

    # Now, create our new rows, with all the pretty cells
    row_idx += 1
    for row in range(row_idx,row_idx+cnt):
        # Create a Row Dimension for our new row
        new_rd = copy.copy(self.row_dimensions[row-1])
        new_rd.index = row
        self.row_dimensions[row] = new_rd
        for col in range(1,self.max_column):
            col = get_column_letter(col)
            cell = self.cell('%s%d'%(col,row))
            cell.value = None
            source = self.cell('%s%d'%(col,row-1))
            if copy_style:
                cell.number_format = source.number_format
                cell.font      = source.font.copy()
                cell.alignment = source.alignment.copy()
                cell.border    = source.border.copy()
                cell.fill      = source.fill.copy()
            if fill_formulae and source.data_type == Cell.TYPE_FORMULA:
                s_coor = source.coordinate
                if s_coor in self.formula_attributes and 'ref' not in self.formula_attributes[s_coor]:
                    fa = self.formula_attributes[s_coor].copy()
                    self.formula_attributes[cell.coordinate] = fa
                # print("Copying formula from cell %s%d to %s%d"%(col,row-1,col,row))
                cell.value = re.sub(
                    "(\$?[A-Z]{1,3}\$?)%d"%(row - 1),
                    lambda m: m.group(1) + str(row),
                    source.value
                )   
                cell.data_type = Cell.TYPE_FORMULA

    # Check for Merged Cell Ranges that need to be expanded to contain new cells
    for cr_idx, cr in enumerate(self.merged_cell_ranges):
        self.merged_cell_ranges[cr_idx] = CELL_RE.sub(
            replace,
            cr
        )

Worksheet.insert_rows = insert_rows
于 2015-06-15T07:20:20.450 回答
12

添加适用于以下版本的最新版本 v2.5+ 的答案openpyxl

现在有一个insert_rows()and insert_cols()

insert_rows(idx, amount=1)

在 row==idx 之前插入一行或多行

于 2018-03-24T09:21:15.583 回答
10

用我现在用来达到预期结果的代码来回答这个问题。请注意,我在位置 1 处手动插入行,但这应该很容易根据特定需求进行调整。您还可以轻松地对其进行调整以插入多行,并简单地从相关位置开始填充其余数据。

另外,请注意,由于下游依赖性,我们手动指定“Sheet1”中的数据,并且数据被复制到插入工作簿开头的新工作表中,同时将原始工作表重命名为“Sheet1.5” .

编辑:我还(稍后)添加了对 format_code 的更改,以解决此处默认复制操作删除所有格式的问题:new_cell.style.number_format.format_code = 'mm/dd/yyyy'. 我找不到任何可以设置的文档,这更像是一个反复试验的案例!

最后,不要忘记这个例子是在原来的基础上保存。您可以更改适用的保存路径以避免这种情况。

    import openpyxl

    wb = openpyxl.load_workbook(file)
    old_sheet = wb.get_sheet_by_name('Sheet1')
    old_sheet.title = 'Sheet1.5'
    max_row = old_sheet.get_highest_row()
    max_col = old_sheet.get_highest_column()
    wb.create_sheet(0, 'Sheet1')

    new_sheet = wb.get_sheet_by_name('Sheet1')

    # Do the header.
    for col_num in range(0, max_col):
        new_sheet.cell(row=0, column=col_num).value = old_sheet.cell(row=0, column=col_num).value

    # The row to be inserted. We're manually populating each cell.
    new_sheet.cell(row=1, column=0).value = 'DUMMY'
    new_sheet.cell(row=1, column=1).value = 'DUMMY'

    # Now do the rest of it. Note the row offset.
    for row_num in range(1, max_row):
        for col_num in range (0, max_col):
            new_sheet.cell(row = (row_num + 1), column = col_num).value = old_sheet.cell(row = row_num, column = col_num).value

    wb.save(file)
于 2013-06-26T14:31:42.387 回答
5

Openpyxl 工作表在执行行或列级别操作时功能有限。工作表具有的与行/列相关的唯一属性是属性row_dimensionscolumn_dimensions,它们分别为每一行和每一列存储“RowDimensions”和“ColumnDimensions”对象。这些字典也用于类似get_highest_row()和的函数中get_highest_column()

其他一切都在单元格级别上运行,在字典中跟踪 Cell 对象_cells(以及在字典中跟踪它们的样式_styles)。大多数看起来像是在行或列级别上执行任何操作的函数实际上是在一系列单元格上运行的(例如前面提到的append())。

最简单的做法就是您的建议:创建一个新工作表,附加标题行,附加新数据行,附加旧数据行,删除旧工作表,然后将新工作表重命名为旧工作表。这种方法可能出现的问题是行/列维度属性和单元格样式的丢失,除非您也专门复制它们。

或者,您可以创建自己的插入行或列的函数。

我有大量非常简单的工作表,需要从中删除列。由于您要求提供明确的示例,因此我将提供我快速组合在一起的功能:

from openpyxl.cell import get_column_letter

def ws_delete_column(sheet, del_column):

    for row_num in range(1, sheet.get_highest_row()+1):
        for col_num in range(del_column, sheet.get_highest_column()+1):

            coordinate = '%s%s' % (get_column_letter(col_num),
                                   row_num)
            adj_coordinate = '%s%s' % (get_column_letter(col_num + 1),
                                       row_num)

            # Handle Styles.
            # This is important to do if you have any differing
            # 'types' of data being stored, as you may otherwise get
            # an output Worksheet that's got improperly formatted cells.
            # Or worse, an error gets thrown because you tried to copy
            # a string value into a cell that's styled as a date.

            if adj_coordinate in sheet._styles:
                sheet._styles[coordinate] = sheet._styles[adj_coordinate]
                sheet._styles.pop(adj_coordinate, None)
            else:
                sheet._styles.pop(coordinate, None)

            if adj_coordinate in sheet._cells:
                sheet._cells[coordinate] = sheet._cells[adj_coordinate]
                sheet._cells[coordinate].column = get_column_letter(col_num)
                sheet._cells[coordinate].row = row_num
                sheet._cells[coordinate].coordinate = coordinate

                sheet._cells.pop(adj_coordinate, None)
            else:
                sheet._cells.pop(coordinate, None)

        # sheet.garbage_collect()

我将我正在使用的工作表以及我想要删除的列号传递给它,然后它就消失了。我知道这不是您想要的,但我希望这些信息对您有所帮助!

编辑:注意到有人给了另一个投票,并认为我应该更新它。Openpyxl 中的坐标系统在过去几年的某个时候经历了一些变化,coordinate_cell. 这也需要编辑,否则行将留空(而不是删除),Excel 将抛出有关文件问题的错误。这适用于 Openpyxl 2.2.3(未经更高版本测试)

于 2013-06-25T20:04:35.380 回答
5

从 openpyxl 1.5 开始,您现在可以使用 .insert_rows(idx, row_qty)

from openpyxl import load_workbook
wb = load_workbook('excel_template.xlsx')
ws = wb.active
ws.insert_rows(14, 10)

如果您在 Excel 中手动执行此操作,它将不会获取 idx 行的格式。之后您将应用正确的格式,即单元格颜色。

于 2018-08-07T09:39:56.183 回答
3

在 Python 中使用 openpyxl 将行插入 Excel 电子表格

下面的代码可以帮助你: -

import openpyxl

file = "xyz.xlsx"
#loading XL sheet bassed on file name provided by user
book = openpyxl.load_workbook(file)
#opening sheet whose index no is 0
sheet = book.worksheets[0]

#insert_rows(idx, amount=1) Insert row or rows before row==idx, amount will be no of 
#rows you want to add and it's optional
sheet.insert_rows(13)

对于插入列 openpyxl 也有类似的功能 ieinsert_cols(idx, amount=1)

于 2018-10-12T08:47:46.160 回答
2

我编写了一个函数,它可以使用 openpyxl 在电子表格或整个 2D 表中的任何位置插入整行。

函数的每一行都用注释解释,但如果你只想插入一行,只需让你的行等于 [row]。即如果 row = [1,2,3,4,5] 然后将您的输入设置为 [[1,2,3,4,5]]。如果您希望将此行插入电子表格的第一行 (A1),则 Start = [1,1]。

您确实可以覆盖文件名,如底部的示例所示。

def InputList(Start, List): #This function is to input an array/list from a input start point; len(Start) must equal 2, where Start = [1,1] is cell 1A. List must be a two dimensional array; if you wish to input a single row then this can be done where len(List) == 1, e.g. List = [[1,2,3,4]]
    x = 0 #Sets up a veriable to go through List columns
    y = 0 #Sets up a veriable to go through List rows
    l = 0 #Sets up a veriable to count addional columns against Start[1] to allow for column reset on each new row
    for row in List: #For every row in List
        l = 0 #Set additonal columns to zero
        for cell in row: #For every cell in row
            ws.cell(row=Start[0], column=Start[1]).value = List[y][x] #Set value for current cell
            x = x + 1 #Move to next data input (List) column
            Start[1] = Start[1] + 1 #Move to next Excel column
            l = l + 1 #Count addional row length
        y = y + 1 #Move to next Excel row
        Start[0] = Start[0] + 1 #Move to next Excel row
        x = 0 #Move back to first column of input data (ready for next row)
        Start[1] = Start[1] - l #Reset Excel column back to orignal start column, ready to write next row

在第 7 行开头插入单行的示例:

from openpyxl import load_workbook
wb = load_workbook('New3.xlsx')
ws = wb.active

def InputList(Start, List): #This function is to input an array/list from a input start point; len(Start) must equal 2, where Start = [1,1] is cell 1A. List must be a two dimensional array; if you wish to input a single row then this can be done where len(List) == 1, e.g. List = [[1,2,3,4]]
    x = 0 #Sets up a veriable to go through List columns
    y = 0 #Sets up a veriable to go through List rows
    l = 0 #Sets up a veriable to count addional columns against Start[1] to allow for column reset on each new row
    for row in List: #For every row in List
        l = 0 #Set additonal columns to zero
        for cell in row: #For every cell in row
            ws.cell(row=Start[0], column=Start[1]).value = List[y][x] #Set value for current cell
            x = x + 1 #Move to next data input (List) column
            Start[1] = Start[1] + 1 #Move to next Excel column
            l = l + 1 #Count addional row length
        y = y + 1 #Move to next Excel row
        Start[0] = Start[0] + 1 #Move to next Excel row
        x = 0 #Move back to first column of input data (ready for next row)
        Start[1] = Start[1] - l #Reset Excel column back to orignal start column, ready to write next row

test = [[1,2,3,4]]
InputList([7,1], test)

wb.save('New3.xlsx')
于 2020-01-06T18:05:48.263 回答
1

I took Dallas solution and added support for merged cells:

    def insert_rows(self, row_idx, cnt, above=False, copy_style=True, fill_formulae=True):
        skip_list = []
        try:
            idx = row_idx - 1 if above else row_idx
            for (new, old) in zip(range(self.max_row+cnt,idx+cnt,-1),range(self.max_row,idx,-1)):
                for c_idx in range(1,self.max_column):
                  col = self.cell(row=1, column=c_idx).column #get_column_letter(c_idx)
                  print("Copying %s%d to %s%d."%(col,old,col,new))
                  source = self["%s%d"%(col,old)]
                  target = self["%s%d"%(col,new)]
                  if source.coordinate in skip_list:
                      continue

                  if source.coordinate in self.merged_cells:
                      # This is a merged cell
                      for _range in self.merged_cell_ranges:
                          merged_cells_list = [x for x in cells_from_range(_range)][0]
                          if source.coordinate in merged_cells_list:
                              skip_list = merged_cells_list
                              self.unmerge_cells(_range)
                              new_range = re.sub(str(old),str(new),_range)
                              self.merge_cells(new_range)
                              break

                  if source.data_type == Cell.TYPE_FORMULA:
                    target.value = re.sub(
                      "(\$?[A-Z]{1,3})%d"%(old),
                      lambda m: m.group(1) + str(new),
                      source.value
                    )
                  else:
                    target.value = source.value
                  target.number_format = source.number_format
                  target.font   = source.font.copy()
                  target.alignment = source.alignment.copy()
                  target.border = source.border.copy()
                  target.fill   = source.fill.copy()
            idx = idx + 1
            for row in range(idx,idx+cnt):
                for c_idx in range(1,self.max_column):
                  col = self.cell(row=1, column=c_idx).column #get_column_letter(c_idx)
                  #print("Clearing value in cell %s%d"%(col,row))
                  cell = self["%s%d"%(col,row)]
                  cell.value = None
                  source = self["%s%d"%(col,row-1)]
                  if copy_style:
                    cell.number_format = source.number_format
                    cell.font      = source.font.copy()
                    cell.alignment = source.alignment.copy()
                    cell.border    = source.border.copy()
                    cell.fill      = source.fill.copy()
                  if fill_formulae and source.data_type == Cell.TYPE_FORMULA:
                    #print("Copying formula from cell %s%d to %s%d"%(col,row-1,col,row))
                    cell.value = re.sub(
                      "(\$?[A-Z]{1,3})%d"%(row - 1),
                      lambda m: m.group(1) + str(row),
                      source.value
                    )
于 2015-08-09T10:12:55.057 回答
0

这对我有用:

    openpyxl.worksheet.worksheet.Worksheet.insert_rows(wbs,idx=row,amount=2)

在 row==idx 之前插入 2 行

请参阅: http: //openpyxl.readthedocs.io/en/stable/api/openpyxl.worksheet.worksheet.html

于 2018-06-07T21:04:28.900 回答
0

编辑尼克的解决方案,此版本采用起始行、要插入的行数和文件名,并插入必要数量的空白行。

#! python 3

import openpyxl, sys

my_start = int(sys.argv[1])
my_rows = int(sys.argv[2])
str_wb = str(sys.argv[3])

wb = openpyxl.load_workbook(str_wb)
old_sheet = wb.get_sheet_by_name('Sheet')
mcol = old_sheet.max_column
mrow = old_sheet.max_row
old_sheet.title = 'Sheet1.5'
wb.create_sheet(index=0, title='Sheet')

new_sheet = wb.get_sheet_by_name('Sheet')

for row_num in range(1, my_start):
    for col_num in range(1, mcol + 1):
        new_sheet.cell(row = row_num, column = col_num).value = old_sheet.cell(row = row_num, column = col_num).value

for row_num in range(my_start + my_rows, mrow + my_rows):
    for col_num in range(1, mcol + 1):
        new_sheet.cell(row = (row_num + my_rows), column = col_num).value = old_sheet.cell(row = row_num, column = col_num).value

wb.save(str_wb)
于 2016-12-30T19:00:07.347 回答
0

我已经成功地使用了Dalas 的答案,尽管对 openpyxl 3.0.9 做了一些修改。我在这里为其他想知道如何在 2022 年做到这一点的人发布代码。

不同之处在于:

  • 添加导入
  • 更改Cell.TYPE_FORMULATYPE_FORMULA
  • str()使用orint()在需要的地方添加类型转换

我是 Python 新手,因此请随意提出任何更改建议,但这就是我让它工作的方式。

import copy
import re
from openpyxl.utils import get_column_letter
from openpyxl.cell.cell import TYPE_FORMULA

def insert_rows(self, row_idx, cnt, above=False, copy_style=True, 
    fill_formulae=True):
"""Inserts new (empty) rows into worksheet at specified row index.

:param row_idx: Row index specifying where to insert new rows.
:param cnt: Number of rows to insert.
:param above: Set True to insert rows above specified row index.
:param copy_style: Set True if new rows should copy style of immediately above row.
:param fill_formulae: Set True if new rows should take on formula from immediately above row, filled with references new to rows.

Usage:

* insert_rows(2, 10, above=True, copy_style=False)

"""
CELL_RE  = re.compile("(?P<col>\$?[A-Z]+)(?P<row>\$?\d+)")

row_idx = row_idx - 1 if above else row_idx

def replace(m):
    row = m.group('row')
    prefix = "$" if row.find("$") != -1 else ""
    row = int(row.replace("$",""))
    row += cnt if row > row_idx else 0
    return m.group('col') + prefix + str(row)

# First, we shift all cells down cnt rows...
old_cells = set()
old_fas   = set()
new_cells = dict()
new_fas   = dict()
for c in self._cells.values():

    old_coor = c.coordinate

    # Shift all references to anything below row_idx
    if c.data_type == TYPE_FORMULA:
        c.value = CELL_RE.sub(
            replace,
            c.value
        )
        # Here, we need to properly update the formula references to reflect new row indices
        if old_coor in self.formula_attributes and 'ref' in self.formula_attributes[old_coor]:
            self.formula_attributes[old_coor]['ref'] = CELL_RE.sub(
                replace,
                self.formula_attributes[old_coor]['ref']
            )

    # Do the magic to set up our actual shift    
    if c.row > row_idx:
        old_coor = c.coordinate
        old_cells.add((c.row,c.column))
        c.row += cnt
        new_cells[(c.row,c.column)] = c
        if old_coor in self.formula_attributes:
            old_fas.add(old_coor)
            fa = self.formula_attributes[old_coor].copy()
            new_fas[c.coordinate] = fa

for coor in old_cells:
    del self._cells[coor]
self._cells.update(new_cells)

for fa in old_fas:
    del self.formula_attributes[fa]
self.formula_attributes.update(new_fas)

# Next, we need to shift all the Row Dimensions below our new rows down by cnt...
for row in range(len(self.row_dimensions)-1+cnt,row_idx+cnt,-1):
    new_rd = copy.copy(self.row_dimensions[row-cnt])
    new_rd.index = row
    self.row_dimensions[row] = new_rd
    del self.row_dimensions[row-cnt]

# Now, create our new rows, with all the pretty cells
row_idx += 1
for row in range(row_idx,row_idx+cnt):
    # Create a Row Dimension for our new row
    new_rd = copy.copy(self.row_dimensions[row-1])
    new_rd.index = row
    self.row_dimensions[row] = new_rd
    for col in range(1,self.max_column):
        col = get_column_letter(col)
        cell = self[str(col)+str(row)]
        cell.value = None
        source = self[str(col)+str(row-1)]
        if copy_style:
            cell.number_format = source.number_format
            cell.font      = source.font.copy()
            cell.alignment = source.alignment.copy()
            cell.border    = source.border.copy()
            cell.fill      = source.fill.copy()
        if fill_formulae and source.data_type == TYPE_FORMULA:
            s_coor = source.coordinate
            if s_coor in self.formula_attributes and 'ref' not in self.formula_attributes[s_coor]:
                fa = self.formula_attributes[s_coor].copy()
                self.formula_attributes[cell.coordinate] = fa
            # print("Copying formula from cell %s%d to %s%d"%(col,row-1,col,row))
            cell.value = re.sub(
                "(\$?[A-Z]{1,3}\$?)%d"%(row - 1),
                lambda m: m.group(1) + str(row),
                source.value
            )   
            cell.data_type = TYPE_FORMULA

# Check for Merged Cell Ranges that need to be expanded to contain new cells
for cr_idx, cr in enumerate(self.merged_cells.ranges):
    self.merged_cells.ranges[cr_idx] = CELL_RE.sub(
        replace,
        str(cr)
    )
于 2022-02-20T15:23:42.533 回答
-1

不幸的是,没有更好的方法来读取文件,并使用像 xlwt 这样的库来写出一个新的 excel 文件(在顶部插入新行)。Excel 不像您可以读取和追加的数据库那样工作。不幸的是,您只需要读入信息并在内存中进行操作,然后写出本质上是一个新文件的内容。

于 2013-06-25T19:02:33.250 回答