python - 如何删除 for 循环中的标签

Question

那么，我如何删除所有 Tkinter.Label是在我的 for 中生成的循环并生成新的，同时保留我的生成方法？因为如果我想降低数字，它不会删除前一个数字的标签(例如，输入“6”，你得到 [3,4,5] ，下一个数字“14”生成 [5,12,13] 。如何我可以删除 [5,12,13] 如果我的 input < 14 吗？)，如果有人有更好的方法来生成这些标签的输出，如果您愿意尝试教育我一点，我将不胜感激。代码如下:

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

在这里，我认为这就是您想要的:(如果 python 2，将 import tkinter 替换为 import Tkinter as 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()

与最初发布的代码相比，更改内容如下:

• 将庞大函数中的每个连贯步骤重构为单独的和大部分是独立的功能，具有更 Eloquent 名称。变量的名称也已更改为更具可读性。
• 因此，结果的计算与显示格式无关。
• 结果汇总在一个列表中
• 列表被传递以创建 tkinter.Labels 准备显示；这些标签聚合在一个列表中。
• 显示先前计算结果的(旧)​​标签随后将从显示中删除并销毁。
• 随后会显示保存新计算结果的新标签。

原始代码试图在一个函数中完成所有这些工作；它未能丢弃旧的结果；结果是(1)显示是旧结果和新结果的奇怪混合(不准确和不精确)，以及(2)被丢弃但从未被破坏的小部件使应用程序空间变得困惑。

在 OSX 上，小部件如下所示: