Quick and easy file dialog in Python?

Try with wxPython:

import wx

def get_path(wildcard):
    app = wx.App(None)
    style = wx.FD_OPEN | wx.FD_FILE_MUST_EXIST
    dialog = wx.FileDialog(None, 'Open', wildcard=wildcard, style=style)
    if dialog.ShowModal() == wx.ID_OK:
        path = dialog.GetPath()
    else:
        path = None
    dialog.Destroy()
    return path

print get_path('*.txt')

You can use easygui:

import easygui

path = easygui.fileopenbox()

To install easygui, you can use pip:

pip3 install easygui

It is a single pure Python module (easygui.py) that uses tkinter.

Tkinter is the easiest way if you don't want to have any other dependencies. To show only the dialog without any other GUI elements, you have to hide the root window using the withdraw method:

import tkinter as tk
from tkinter import filedialog

root = tk.Tk()
root.withdraw()

file_path = filedialog.askopenfilename()

Python 2 variant:

import Tkinter, tkFileDialog

root = Tkinter.Tk()
root.withdraw()

file_path = tkFileDialog.askopenfilename()