python - ValueError:无效的文件路径或缓冲区对象类型:

Question

这是我拥有的一些代码的简化版本。在第一帧中，用户使用“tk.filedialog”选择了一个 csv 文件，它被绘制在 Canvas 上的同一帧上。

还有第二个框架能够绘制图表，以防跨不同的框架更容易绘制。

运行此版本的代码会导致错误:“ValueError:无效的文件路径或缓冲区对象类型:”。我不确定如何在不发生此问题的情况下使此代码正常工作，以便用户选择的文件绘制在包含“a”和“b”列的空图形上。

import csv
import pandas as pd
import tkinter as tk
from tkinter import filedialog
from tkinter import ttk
from tkinter import messagebox
import matplotlib

matplotlib.use("TkAgg")

from matplotlib.backends.backend_tkagg import FigureCanvasTkAgg, NavigationToolbar2TkAgg

from matplotlib.figure import Figure


fig = Figure(figsize=(5,4), dpi=100)
ax= fig.add_subplot(111)

LARGE_FONT= ("Verdana", 12)

class GUI(tk.Tk):

    def __init__(self, *args, **kwargs):

        tk.Tk.__init__(self, *args, **kwargs)
        tk.Tk.wm_title(self, "GUI")

        container = tk.Frame(self)
        container.pack(side="top", fill="both", expand = True)
        container.grid_rowconfigure(0, weight=1)
        container.grid_columnconfigure(0, weight=1)

        self.frames = {}

        for F in (Home, Graph):

            frame = F(container, self)

            self.frames[F] = frame

            frame.grid(row=0, column=0, sticky="nsew")

        self.show_frame(Home)




    def show_frame(self, cont):

        frame = self.frames[cont]
        frame.tkraise()


class Home(tk.Frame):

    def __init__(self, parent, controller):
        self.controller = controller
        tk.Frame.__init__(self,parent)
        label = tk.Label(self, text="Start Page", font=LARGE_FONT)
        label.pack(pady=10, padx=10)



        ftypes = [
                ('CSV files','*.csv')
        ]

        def browsefunc2():
            filename = tk.filedialog.askopenfilename(filetypes=ftypes)
            pathlabel2.config(text=filename)

            filename = filename.get()
            return filename



        #this line is just used to check that hard-coding in a filename works, which it does providing 'filename = tk.StringVar()' is removed
        #filename = '...'


        filename = tk.StringVar()

        df = pd.read_csv(filename, encoding='latin-1')

        browsebutton = tk.Button(self, borderwidth=0, text="Browse", command=browsefunc2, height=1, width=10)
        browsebutton.pack()

        pathlabel2 = tk.Label(self, borderwidth=0)
        pathlabel2.pack()

        canvas = FigureCanvasTkAgg(fig, self)


        df.plot.scatter('a', 'b', ax=ax)

        canvas.draw()
        canvas.get_tk_widget().pack(side=tk.BOTTOM, fill=tk.BOTH, expand=True)


        button2 = ttk.Button(self, text="Graph",
                             command=lambda: controller.show_frame(Graph))
        button2.pack()

class Graph(tk.Frame):

    def __init__(self, parent, controller):
        self.controller = controller
        tk.Frame.__init__(self,parent)
        label = tk.Label(self, text="Graph", font=LARGE_FONT)
        label.pack(pady=10,padx=10)

        canvas = FigureCanvasTkAgg(fig, self)

       #this line causes a problem as the dataframe is not recognised across frames
        df.plot.scatter('a', 'b', ax=ax)

        canvas.draw()
        canvas.get_tk_widget().pack(side=tk.BOTTOM, fill=tk.BOTH, expand=True)

        button3 = ttk.Button(self, text="Back",
                            command=lambda: controller.show_frame(Home))
        button3.pack()

app = GUI()
app.mainloop()

据我所知，无法将 .csv 文件上传到 StackOverflow，因此我重新创建了一个示例，但文件类型必须是 .csv。

a,b
1,10
2,32
3,23
4,5
5,4
6,66
7,7
8,19
9,31
10,44

Answer 1

我没有运行您的“简化”版本的代码，因为它绝不 Minimal, Complete, and Verifiable example .

该错误告诉您当它是 StringVar 时您假设它是路径或缓冲区。我相信错误在线:

df = pd.read_csv(filename, encoding='latin-1')

这要求 filename 是一个路径或缓冲区对象，而在上面的那一行 filename 确实是一个 StringVar 对象:

filename = tk.StringVar()

df = pd.read_csv(filename, encoding='latin-1')

为了获取 StringVar 或任何 Variable 子类类型的值，需要使用 get 方法。

filename.get()

但是，这会产生一个空字符串 ''，这会引发另一个错误。