我將如何刪除我的for循環中的標籤

Question

那麼，我將如何刪除我的for循環內生成的所有Tkinter.Label，並在保留我的生成方法的同時生成新標籤？因爲如果我想降低數字，它不會刪除前一個數字的標籤（例如，輸入'6'，你會得到[3,4,5]，下一個數字'14'會生成[5,12,13]。刪除[5,12,13]如果我的input < 14？），如果有人有更好的方法來生成這些標籤的輸出，將不勝感激，如果你會嘗試教育我一點。 下面的代碼：

import Tkinter 
import sys 
from fractions import gcd 


def func(event): 
    x = int(e1.get()) # get max number 
    row = 0 
    column = 0 
    count = 0 
    for a in range(1, x): # loops to get each value in range of x 
     for b in range(a, x): 
      for c in range(b, x): 
       if a**2 + b**2 == c**2 and gcd(a, b) == 1: # if it is a primitive pyth triple, print 
        row += 1 
        l = Tkinter.Label(root, text=('[',a,',',b,',',c,']')) 
        assert isinstance(l, object) 
        l.grid(row=row, column=column, ipadx=5, ipady=5, sticky='W''E') # display each group of triples to root 
        root.title('Primitive Triples') 
        if count > 1: 
         l.destroy() 
        if row == 7: 
         column += 1 
         row -= 8 


def close(): # close program 
    Tkinter.sys.exit(0) 
    sys.exit(0) 


root = Tkinter.Tk() # establish main gui 
root.title('Generator') 
e1 = Tkinter.Entry(root) 
assert isinstance(e1, object) # only method I've found to allow for Entry().grid() 
e1.grid(ipadx=5, ipady=5, sticky='W''E') 
root.bind('<Return>', func) # bind to Enter, cleaner and quicker than a button 
root.mainloop() 

Answer 1

在這裏，我想這是你想要的東西：（如果蟒蛇2，用進口的Tkinter作爲Tkinter的替代進口的Tkinter）

import tkinter 
import sys 
from fractions import gcd 

CURRENT_LABELS = [] 

def pythagorean_primitive(a, b, c): 
    """returns True if a,b,c are pythagorean primitives, False otherwise""" 
    return a**2 + b**2 == c**2 and gcd(a, b) == 1 

def generate_results(n): 
    """lists each triplet of distinct integers <n that is a pythagorean primitive""" 
    results = [] 
    for a in range(1, n): 
     for b in range(a, n): 
      for c in range(b, n): 
       if pythagorean_primitive(a, b, c): 
        results.append([a, b, c]) 
    return results 

def generate_labels(sequence): 
    """returns a list of tkinter labels from the sequence provided""" 
    labels = [] 
    for elt in sequence: 
     a, b, c = elt[0], elt[1], elt[2] 
     labels.append(tkinter.Label(root, text='[' + str(a) + ', '+ str(b) + ", " + str(c) + "]")) 
    return labels 

def destroy_old(): 
    """purges the current tkinter labels from root, and destroys them""" 
    global CURRENT_LABELS 
    for elt in CURRENT_LABELS: 
     elt.grid_forget() 
     elt.destroy() 

def show_new_labels(sequence): 
    """assembles a new display of tkinter labels from the sequence provided""" 
    r, c = 1, 0 
    for label in sequence: 
     label.grid(row=r, column=c, ipadx=5, ipady=5, sticky='W''E') 
     r += 1 
     if not r % 10: 
      r = 1 
      c += 1 

def event_handler(event): 
    """deals with the input of a number in the Entry field""" 
    global CURRENT_LABELS 
    x = int(e1.get()) # get max number 
    results = generate_results(x) 
    try: 
     destroy_old() 
    except IndexError: 
     pass 
    CURRENT_LABELS = generate_labels(results) 
    show_new_labels(CURRENT_LABELS) 


def close(): # close program 
    tkinter.sys.exit(0) 
    sys.exit(0) 


root = tkinter.Tk() # establish main gui 
root.title('Generator') 
e1 = tkinter.Entry(root) 
assert isinstance(e1, object) # only method I've found to allow for Entry().grid() 
e1.grid(ipadx=5, ipady=5, sticky='W''E') 
root.bind('<Return>', event_handler) # bind to Enter, cleaner and quicker than a button 
root.mainloop() 

什麼已經從碼本來改變貼如下：

• 重構每個相干步驟出來的膨鬆函數的成單獨和 大多獨立的功能具有更多口才名稱。變量的名稱也更改爲更易讀。
• 因此，結果的計算獨立於其顯示格式。
• 結果彙總在列表中
• 該列表被傳遞以創建tkinter.Labels以供顯示;這些標籤彙總在一個列表中。
• 顯示先前計算結果的（舊）​​標籤將從顯示屏上消失，並被銷燬。
• 然後顯示保存新計算結果的新標籤。

原代碼試圖在一個函數中做所有的事情;它沒有放棄舊的結果;其結果是（1）顯示是新舊結果的奇怪組合（不準確和不精確），（2）丟棄，但從未銷燬的小部件混亂了應用程序空間。

在OSX，小部件看起來是這樣的：