python - 用于在python中连接数据excel数据的循环文件名？

Question

我有一些用于连接 Excel 文件的代码，如何编辑我的代码以便不必每次都键入完整的文件名？

谢谢

代码如下:

import pandas as pd

excel_names = ["Market_Information_System_Control_daily_trading_day_170701.xlsx",
               "Market_Information_System_Control_daily_trading_day_170702.xlsx",
               "Market_Information_System_Control_daily_trading_day_170703.xlsx",
               "Market_Information_System_Control_daily_trading_day_170704.xlsx",
               "Market_Information_System_Control_daily_trading_day_170731.xlsx"]

excels = [pd.ExcelFile(name) for name in excel_names]
frames = [x.parse(x.sheet_names[1], header=None,index_col=None) for x in excels]

frames[1:] = [df[1:] for df in frames[1:]]
combined = pd.concat(frames)
combined.to_excel("c.xlsx", header=False, index=False)

编辑1:

我只是想要时间流动，而不是有标题中断。

我认为这与此有关:frames[1:] = [df[1:] for df inframes[1:]]

Answer 1

这是您要找的吗？这将打印出所有文件名，您只需循环遍历文件名然后组合

from os import listdir
from os.path import isfile, join

folder_path = './'

onlyfiles = [f for f in listdir(folder_path) if isfile(join(folder_path, f))]
print(onlyfiles);

附注这是来自How do I list all files of a directory?

编辑，如果您确实想要进行某种文件扩展名过滤，我使用 json 文件作为示例，请将其更改为您的需要

from os import listdir
from os.path import isfile, join

folder_path = './'

onlyfiles = [f for f in listdir(folder_path) if isfile(join(folder_path, f))]
filter_extension = []
my_ext = 'json'

for f in listdir(folder_path):
    extension = f.split('.')[-1]
    if (extension == my_ext):
        filter_extension.extend([f])

print(filter_extension)