减少python执行时间

编程问答 2022-09-05

问题描述

我正在尝试创建按钮以获取文件路径，然后将内容复制到主文件。我的代码如下：

import tkinter as tk
from tkinter import *
from tkinter import ttk
from tkinter import filedialog

def get_loan():
    global file_loan
    file_loan = filedialog.askopenfilename(title = "Select A File",filetypes = (("xlsx","*.xlsx"),("xls","*.xls"),("xlsm","*.xlsm"),("All Files","*.*")))
    l1 = Label(window,text = file_loan).pack()

window = tk.Tk()
window.title("place() method")
window.geometry('550x300')
b1 = tk.Button(window,text = "Loan File",command = get_loan,bg='#6eb5ff',fg='white').pack()
b2 = tk.Button(window,text = "Run",command = window.destroy,bg='#97c1a9',fg='white').pack()
window.mainloop()
print(file_loan)

import openpyxl as xl
path1 = ("r'" and file_loan)
path2 = ('Input_Data_Merge.xlsx')
wb1 = xl.load_workbook(filename=path1)
ws1 = wb1["BASEL_LN_CTR"]
wb2 = xl.load_workbook(filename=path2)
ws2 = wb2["Export Worksheet"]

for row in ws1:
    for cell in row:
        ws2[cell.coordinate].value = cell.value

wb2.save(path2)

它运行良好，但如果添加更多文件需要很长时间才能执行。我想问一下有没有其他方法可以更快地打开文件并将内容复制到主文件中？

解决方法

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