pyqt5 运行慢

Question

所以我的程序是一个数独，至此，我已经完成了生成棋盘的代码。我正在尝试将其实现到 pyqt5 中。我通过使按钮对应于数独板中的每个数字（因此有 81 个按钮）来完成此操作。我有一台能够以高帧率玩运行三 A 游戏的电脑（如果这很重要的话）。可能是什么问题？我把我的代码放在下面。

import random
import numpy as np

import sys
from PyQt5 import QtWidgets as qtw
from PyQt5 import QtCore as qtc
from PyQt5 import QtGui as qtg


# populates a row in random spaces
def populate():
    row = [0] * 9
    num_in_box = random.randint(3, 6)
    while True:
        spot = random.randint(0, 8)
        r = random.randint(1, 9)
        if r not in row:
            row[spot] = r
        if row.count(0) == (9 - num_in_box):
            break
    return row


# populates a grid in random spaces - has duplicates in column, row and box
mapped = list()
for x in range(9): mapped.append(populate())


# checks every number in column and row and returns numbers in list
def col_row_nums(array, row, col):
    check_col = [j for j in array[:, col] if j != 0]
    check_row = [i for i in array[row] if i != 0]
    if array[row][col] in check_col:
        check_col.remove(array[row][col])
    return check_col + check_row


# checks every number box and returns numbers in list
def box_nums(array, box_row):
    reshaped_box = np.reshape(array, (27, 3))
    list_boxes_in_rows = list()
    for a in range(3):
        for b in range(3):
            for c in range(3):
                p2 = list(np.reshape((reshaped_box[c::3]), (3, 9)))
                for d in range(3): list_boxes_in_rows.append(p2[a])
    array_boxes_in_rows = np.array(list_boxes_in_rows)
    return [k for k in array_boxes_in_rows[box_row] if k != 0]


# Basically goes through each number and removes any duplicates so each column, row and box all have only one set of numbers 1 - 9
def finalize_board():
    box_rows_num = -1
    for x in range(9):
        for y in range(9):
            box_rows_num += 1
            arr = np.array(mapped)
            possible_nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]
            col_row_duplicates = list()
            box_duplicates = list()
            used_nums = set(col_row_nums(arr, x, y)) | set(box_nums(arr, box_rows_num))

            for w in used_nums:
                col_row_count = col_row_nums(arr, x, y).count(w)
                box_count = box_nums(arr, box_rows_num).count(w)
                if col_row_count > 1: col_row_duplicates.append(w)
                if box_count > 1: box_duplicates.append(w)
            if mapped[x][y] in col_row_duplicates or mapped[x][y] in box_duplicates:
                remaining_nums = list(set(possible_nums) - set(used_nums))
                if len(remaining_nums) > 0: mapped[x][y] = random.choice(remaining_nums)
    return mapped


class MainWindow(qtw.QWidget):

    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)

        # -------------------------------------------------------------------------------------------
        all_buttons = list()
        layout = qtw.QGridLayout()

        for x in range(9):
            for y in range(9):
                if finalize_board()[x][y] == 0: finalize_board()[x][y] = ' '
                all_buttons.append(layout.addWidget(qtw.QPushButton(str(finalize_board()[x][y])), x, y))

        self.setLayout(layout)

        # -------------------------------------------------------------------------------------------
        self.show()


if __name__ == '__main__':
    app = qtw.QApplication(sys.argv)
    w = MainWindow()
    sys.exit(app.exec_())

Answer 1

问题是你正在为每个按钮执行 finalize_board 函数，所以如果函数需要 T 秒并且你有 N 个按钮（在你的例子中是 81），同样在每次迭代中你使用 K乘以该函数（在您的情况下为 3），则总时间为 T * N * K。

解决方案是只调用 finalize_board 一次。

class MainWindow(qtw.QWidget):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        all_buttons = list()
        layout = qtw.QGridLayout(self)

        board = finalize_board()

        for x, rows in enumerate(board):
            for y, cell in enumerate(rows):
                button = qtw.QPushButton(str(cell))
                layout.addWidget(button, x, y)
                all_buttons.append(button)
        self.show()

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