python - 小改动使脚本挂起 - 不明白为什么

Question

我做了一个单词搜索游戏。它会打印出一个包含一些单词的网格。代码如下:

import random
DirectionList = []

# Create a grid filled with "." representing a blank
def createGrid():
    grid=[]
    for row in range(15):
        grid.append([])
        for column in range(50):
            grid[row].append(".")
    return grid

# Print the grid to the screen
def printGrid(grid):
    for row in range(len(grid)):
        for column in range(len(grid[row])):
            print(grid[row][column],end="")
        print()

# Try to place the word. Return True if successful
# False if it failed and we need to try again.
def tryToPlaceWord(grid,word):
    # Figure out the direction of the work.
    # Change the 8 to a 7 if you don't want backwards
    # words.
    status_check = []
    direction=random.randrange(0,8)
    DirectionList.append(direction)
    if( direction == 0 ):
        x_change=-1
        y_change=-1
    if( direction == 1 ):
        x_change=0
        y_change=1
    if( direction == 2 ):
        x_change=1
        y_change=-1
    if( direction == 3 ):
        x_change=1
        y_change=0
    if( direction == 4 ):
        x_change=1
        y_change=1
    if( direction == 5 ):
        x_change=0
        y_change=1
    if( direction == 6 ):
        x_change=-1
        y_change=1
    if( direction == 7 ):
        x_change=-1
        y_change=0

    # Find the length and height of the grid
    height=len(grid)
    width=len(grid[0])

    # Create a random start point
    column=random.randrange(width)
    row=random.randrange(height)

    # Check to make sure  the word won't run off the edge of the grid.
    # If it does, return False. We failed.
    if( x_change < 0 and column < len(word) ):
        status_check.append(False)
        status_check.append("None")
        return status_check
    if( x_change > 0 and column > width-len(word) ):
        status_check.append(False)
        status_check.append("None")
        return status_check
    if( y_change < 0 and row < len(word) ):
        status_check.append(False)
        status_check.append("None")
        return status_check
    if( y_change > 0 and row > height-len(word) ):
        status_check.append(False)
        status_check.append("None")
        return status_check

    # Now check to make sure there isn't another letter in our way
    current_column=column
    current_row=row
    for letter in word:
        # Make sure it is blank, or already the correct letter.
        if grid[current_row][current_column]==letter or grid[current_row][current_column]=='.':
            current_row += y_change
            current_column += x_change
        else:
            # Oh! A different letter is already here. Fail.
            status_check.append(False)
            status_check.append("None")
            return status_check

    # Everything is good so far, actually place the letters.
    current_column=column
    current_row=row
    for letter in word:
        grid[current_row][current_column]=letter
        current_row += y_change
        current_column += x_change
    if 7 in DirectionList:
        status_check.append(True)
        status_check.append("True")
        return status_check
    else:
        status_check.append(True)
        status_check.append("None")
        return status_check

# This just calls tryToPlaceWord until we succeed. It could
# repeat forever if there is no possible place to put the word.
def placeWord(grid,words):
    for word in words:
        succeed=False

        while not(succeed):
            status_check=tryToPlaceWord(grid,word)
            succeed=status_check[0]
    backward = status_check[1]
    return backward

# Create an empty grid
grid = createGrid()

# A list of words
words = ["pandabear","fish","snake","porcupine","dog","cat","tiger","bird","alligator","ant","camel","dolphin"]

# Place some words
placeWord(grid,words)

backward = placeWord(grid,words)
print("Current status:\n"
      "\tGenerating a word searching diagram...\n"
      "\tWords :",len(words),"\n"
      "\tBackword :",backward)

# Print it out
printGrid(grid)

然后，我想让代码打印出随机字母而不是点(.)，所以我做了一些小改动。

import random
import string
DirectionList = []

# Create a grid filled random letters representing a blank
def createGrid():
    grid=[]
    for row in range(15):
        grid.append([])
        for column in range(50):
            grid[row].append(random.choice(string.ascii_uppercase))
    return grid

但是 python 停止运行，很长时间没有打印任何内容。我想也许python需要很多时间来生成15 x 50个随机字母，所以我将代码更改为grid[row].append("a")，但是python仍然无法运行。

出了什么问题？

编辑:

将代码更改为

for letter in word:
        # Make sure it is blank, or already the correct letter.
        if grid[current_row][current_column]==letter or grid[current_row][current_column]==' ':
            current_row += y_change
            current_column += x_change

但是代码仍然挂起......

Answer 1

您使用“.”还要检查网格中的单元格是否空闲(第 87 行)，因此如果您用随机字母填充单元格，它们将不再空闲。您可以在输入单词后用随机字母填充空格...