计算 Python 中一系列关键事件之间的时间

Calculate the time between a range of key events in Python

作为我的自学项目的一部分,我决定在 Python 中创建一个应用程序,用于侦听某些 keyevents 并找出键入单词所需的时间。经过 12 多个小时的直接调试和 loop/logic 实验,我所能做的就是:

import Tkinter
import time

KeyWatch = Tkinter.Tk()

WordBoxLegend = Tkinter.Label(KeyWatch, text="Type the required word")
WordBoxLegend.pack(side=Tkinter.LEFT)

WordBox = Tkinter.Entry(KeyWatch)
WordBox.pack(side=Tkinter.LEFT)

TextBoxLegend = Tkinter.Label(KeyWatch, text="Type the same to calculate the time")
TextBoxLegend.pack(side=Tkinter.LEFT)

TextBox = Tkinter.Entry(KeyWatch) 
TextBox.pack(side=Tkinter.RIGHT)

WordBox.focus()

def pressed(keyevent):
    WordRequire = WordBox.get()
    LetterList = tuple(WordRequire)
    start = time.time()
    LastLetter = str(LetterList[-1])
    print len(LetterList())
    print LetterList[len(LetterList)]
    if keyevent.char in LetterList:
        for x in range(0, len(LetterList)):
            if LetterList[x] != LastLetter:
                print LetterList[x]
                TextBox.unbind(str(LetterList[x]))
                TextBox.bind(str(LetterList[x+1]))
            elif str(LetterList[x]) == LastLetter and x == len(LetterList):
                stop = time.time()
                totaltime = stop - start
                print LetterList[x]
                print totaltime
                break
                TextBox.unbind(LetterList[x])
            else:
                TextBox.unbind(str(LetterList))     
    else:   
        print "Type only the letters from" +str(LetterList)

KeyWatch.mainloop()
TextBox.bind("<Key>", pressed)

Gist

绑定键并将其传递给 pressed 方法后,我不确定如何取消绑定该特定键并绑定元组中的下一个值。

预期结果

如果我输入单词,'Obliteration' 程序应该告诉我从 keyevent "O" 到 keyevent "n" 需要多少时间。

Tkinter 中事件 <Key> 的预期结果是什么?

这很有趣!

我根据 pep8 重写了您的代码。

调整了变量名以提高可读性。

您定义开始时间的方式存在一些逻辑错误。

考虑到每次按下一个键时,都会调用 pressed()。如何检测按键也存在一个隐藏的问题。您的代码有:

word_repeat_box.bind("<Key>", pressed)

"<KEY>" 捕捉到按键的按下,你想要按键的RELEASE。 否则在释放键之前调用该函数,并将值输入重复框。我将其改为使用 <KeyRelease>。 (随意运行我调试的代码来比较"<KEY>""<KeyRelease>"

import Tkinter
import time

start = None
stop = None

key_watcher = Tkinter.Tk()

word_entry_label = Tkinter.Label(key_watcher, text="Type the required word")
word_entry_label.pack(side=Tkinter.LEFT)

word_entry_box = Tkinter.Entry(key_watcher)
word_entry_box.pack(side=Tkinter.LEFT)

word_repeat_label = Tkinter.Label(key_watcher,
                           text="Type the same to calculate the time")

word_repeat_label.pack(side=Tkinter.LEFT)

word_repeat_box = Tkinter.Entry(key_watcher)
word_repeat_box.pack(side=Tkinter.RIGHT)

word_entry_box.focus()


def pressed(keyevent):
  # print keyevent  #Left this in for debugging, you can see the key press.
  # start = time.time()  #start time will be reset every time this is called.
  first_word = word_entry_box.get()
  first_word_size = len(first_word)
  first_letter_list = tuple(first_word)
  first_word_first_letter = str(first_letter_list[0])
  first_word_last_letter = str(first_letter_list[-1])

  repeat_word = word_repeat_box.get()
  repeat_word_size = len(repeat_word)

  if repeat_word_size > 0:
      repeat_letter_list = tuple(repeat_word)
      repeat_word_first_letter = str(repeat_letter_list[0])
      repeat_word_last_letter = str(repeat_letter_list[-1])

  # Lets see whats happening on each press:
  print "Pressed:", keyevent.char
  print "First word:", first_word, "size", first_word_size
  print "repeat word:", repeat_word, "size", repeat_word_size

  # now that all the things we want to look at are defined, lets logic:

  if repeat_word_size == 1:
      global start
      start = time.time()
      print "time started!", start

  if first_word_size == repeat_word_size:
      print "words are same size"

      if first_word_last_letter == repeat_word_last_letter:
          global end
          stop = time.time()
          totaltime = stop - start
          print "Total time taken", totaltime

word_repeat_box.bind("<KeyRelease>", pressed)
key_watcher.mainloop()  # mainloop() should be called last, it starts your loop

带调试的输出:

Pressed: t
First word: test size 4
repeat word: t size 1
time started! 1434046015.13
Pressed: e
First word: test size 4
repeat word: te size 2
Pressed: s
First word: test size 4
repeat word: tes size 3
Pressed: t
First word: test size 4
repeat word: test size 4
words are same size
Total time taken 1.69620299339

这或许可以在没有全局变量的情况下完成,也许可以通过递归调用来完成,但我会把它留给你来解决。