Choosing a file in Python with simple Dialog

Question

How about using tkinter?

from Tkinter import Tk     # from tkinter import Tk for Python 3.x
from tkinter.filedialog import askopenfilename

Tk().withdraw() # we don't want a full GUI, so keep the root window from appearing
filename = askopenfilename() # show an "Open" dialog box and return the path to the selected file
print(filename)

Done!

Python 3.x version of Etaoin's answer for completeness:

from tkinter.filedialog import askopenfilename
filename = askopenfilename()

With EasyGui:

import easygui
print(easygui.fileopenbox())

To install:

pip install easygui

Demo:

import easygui
easygui.egdemo()

In Python 2 use the tkFileDialog module.

import tkFileDialog

tkFileDialog.askopenfilename()

In Python 3 use the tkinter.filedialog module.

import tkinter.filedialog

tkinter.filedialog.askopenfilename()

Another option to consider is Zenity: http://freecode.com/projects/zenity.

I had a situation where I was developing a Python server application (no GUI component) and hence didn't want to introduce a dependency on any python GUI toolkits, but I wanted some of my debug scripts to be parameterized by input files and wanted to visually prompt the user for a file if they didn't specify one on the command line. Zenity was a perfect fit. To achieve this, invoke "zenity --file-selection" using the subprocess module and capture the stdout. Of course this solution isn't Python-specific.

Zenity supports multiple platforms and happened to already be installed on our dev servers so it facilitated our debugging/development without introducing an unwanted dependency.