Question

我正在尝试迭代Excel中的列并检查字符串是否存在。如果字符串存在，我想将列表重置为[]并重复该过程。花了太多时间在这上面，我似乎无法弄清楚我做错了什么。

示例数据：

Open Ended Schemes(Balanced)    

Aditya Birla Sun Life Mutual Fund   

120518  Aditya Birla Sun Life Equity Hybrid'95 Fund - Direct Plan-Dividend
120517  Aditya Birla Sun Life Equity Hybrid'95 Fund - Direct Plan-Growth


Open Ended Schemes(Debt Scheme - Banking and PSU Fund)  

Axis Mutual Fund    

128953  Axis Banking & PSU Debt Fund - Bonus Option
117447  Axis Banking & PSU Debt Fund - Daily Dividend Option

代码：

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

keys = ['1', '2']
m_dict = {}
scheme_codes = []
for g in groups[0:2]:
    for row in ws.iter_rows('A{}:A{}'.format(ws.min_row +1, ws.max_row)):
        # scheme_codes = []
        for cell in row:
            if cell.value != None:
                if 'Schemes' in cell.value:
                    print('Found Schemes' + str(cell.value))
                    scheme_codes=[]
                    break
                else:
                    scheme = cell.value

                    scheme_codes.append(scheme)
            m_dict[g] = scheme_codes

我每个方案只获得1个项目，我尝试了各种方法，并且它只是一直在rows。该文件有18000行。

预期输出

{1：[首次重复＆＃39;计划之前的所有项目＆＃39;在＆＃39; A＆＃39;专栏]，2：[第二次重复之前的所有项目＆＃39;计划＆＃39;在＆＃39; A＆＃39;柱]

现在当我运行代码时，我得到一个len（scheme_codes）= 8069，这是我看不到的错误。第一个列表应该接近80个项目。

Answer 1

这不是你要求的，它实际上提供了一些额外的信息......

它给你一个dicts的字典，它包含一组scheme_code和scheme_names元组，如：

{scheme: {sub_scheme : {(code, name), (code, name), ...}}}

如果您真的只需要顶级方案及其代码，您应该能够简化它。只需删除一级defaultdict，然后使用scheme_codes[scheme].add(cell.value)代替......

from openpyxl import load_workbook
import os

from collections import defaultdict

wb = load_workbook("mfcodes.xlsx")
ws = wb.active

scheme_codes = defaultdict(lambda: defaultdict(set))

scheme = 'N/A'
sub_scheme = 'N/A'
for row in ws[f'A{ws.min_row}:B{ws.max_row}']:
    cell = row[0]
    if not cell.value:
        continue

    if 'Schemes' in cell.value:
        scheme = cell.value
    else:
        if not cell.value.isdigit():
            sub_scheme = cell.value
        else:
            scheme_codes[scheme][sub_scheme].add((cell.value, row[1].value))

print(repr(next(iter(scheme_codes.items()))))

输出：

{'Open Ended Schemes(Balanced)' :
    {'Aditya Birla Sun Life Mutual Fund': {('120518', "Aditya Birla Sun Life Equity Hybrid'95 Fund - Direct Plan-Dividend"),
                                           ('120517', "Aditya Birla Sun Life Equity Hybrid'95 Fund - Direct Plan-Growth"),
                                           ('103154', "Aditya Birla Sun Life Equity Hybrid'95 Fund - Regular Plan-Dividend"),
                                           ('103155', "Aditya Birla Sun Life Equity Hybrid'95 Fund - Regular Plan-Growth"),
                                           ('131671', 'Aditya Birla Sun Life Balanced Advantage Fund - Direct Plan - Dividend Option'),
                                           ('131670', 'Aditya Birla Sun Life Balanced Advantage Fund - Direct Plan - Growth Option'),
                                           ('131665', 'Aditya Birla Sun Life Balanced Advantage Fund - Regular Plan - Dividend Option'),
                                           ('131666', 'Aditya Birla Sun Life Balanced Advantage Fund - Regular Plan - Growth Option')},
    'Baroda Pioneer Mutual Fund': {('125112', 'Baroda Pioneer Balance Fund - Plan A - Bonus Option'),
                                   ('101913', 'BARODA PIONEER BALANCE FUND - Plan A - Dividend Option'),
                                   ('101912', 'BARODA PIONEER BALANCE FUND - Plan A - Growth Option'),
                                   ('119325', 'BARODA PIONEER BALANCE FUND - Plan B (Direct) - Dividend Option'),
                                   ('119326', 'BARODA PIONEER BALANCE FUND - Plan B (Direct) - Growth Option')},
    # et cetera ...
    }
}

顺便说一下：第一个方案有67个代码......

如果字符串不存在于单元格值中，则添加到列表，如果存在则中断并启动新列表？

1 个答案: