我想从另一列中使用单元格值减去一列中的单元格值,并将总和写入excel文件中的新列。然后我希望将总和(如果不等于0)添加到列表中供以后使用。我的excel文件中的数据结构如下:
Name | Number | Name1 | Number1
Name2 | Number2 | Name3 | Number3
....
Namex | Numberx | Namey |Numbery
我想从彼此中减去数字,然后将总和添加到新列中,如下所示:
Name| Number | Name1 | Number1 | Sum of (Number - Number1)
我曾尝试使用openpyxl来做这件事,但我真的很困惑,因为文档与早期版本的Python有所不同。我在Python 3.4中工作。我很高兴得到您建议我使用哪个模块的建议。 到目前为止我的代码给了我错误,因为我将excelfile称为生成器,而不是可订阅者。我不确定如何搜索和阅读excelfile,同时使其可订阅,以便可以写入它。有人可以帮帮我吗?
这是我的代码:
from openpyxl import Workbook, load_workbook
def analyzexlsx(filepath):
numbers = []
excel_input = load_workbook(filepath)
filepath = [pth for pth in Path.cwd().iterdir()
if pth.suffix == '.xlsx'] #Want to iterate through several excel files in a folder.
ws = excel_input.active
cols = tuple(ws.columns)
col_b = cols[1]
col_e = cols[4]
for j, k in zip(col_e, col_b):
if None:
print('None')
equally = (int(j.value) - int(k.value)) #line 13, error. Trying to subtract column cell values.
if equally != 0: #If the columns sum is not equal to 0, it is to be added to the numbers list.
numbers.append(j.row)
else:
pass
col1 = []
col2 = []
col4 = []
col5 = []
col7 = []
col8 = []
mainlist = []
try:
for row in numbers:
col1.append(str(ws.cell(row=row, column=1).value))
col2.append(str(ws.cell(row=row, column=2).value))
col4.append(ws.cell(row=row, column=4).value)
col5.append(ws.cell(row=row, column=5).value)
col7.append(ws.cell(row=row, column=7).value)
col8.append(ws.cell(row=row, column=8).value)
finally:
for i, j, k, l, m, n in zip(col1, col2, col4, col5, col7, col8):
mainlist.append(i + ", " + j + ", " + k + ", " + l + ", " + m + ", " + n)
return mainlist
Traceback (most recent call last):
Line 13, in analyzexlsx
equally = (int(j.value) - int(k.value))
TypeError: int() argument must be a string or a number, not 'NoneType
我会很高兴得到答案,因为我已经研究了很长一段时间,现在我被卡住了。我对Python很新。
答案 0 :(得分:3)
首先通过read_excel
从excel创建DataFrame
。
然后需要使用2.
列减去4
:
df = pd.read_excel('file.xlsx')
#select by column name
df['E'] = df['B'] - df['D']
#select by positions, but python count from 0 so for 2. column need 1
df['E'] = df.iloc[:, 1] - df.iloc[:, 3]
也许还有助于检查documentation。