我正在使用txt文件作为“单词搜索板”,其中包含单词。我还在另一个txt文件中使用了一个单词列表,这个单词可能位于单词板中,也可能不位于单词板中。在所有方向,北,南,东,西,NE,NW,SE,SW和相反的方向发现词。通过电路板搜索的功能必须是递归的。在我当前的searchBoard()
中,我试图只搜索单词的第一个字母然后检查周围的下一个字母,依此类推。我的递归功能按照我想要的方式工作,任何提示都会受到赞赏。
董事会的示例文本文件:
G J T P B A V K U V L V
M N Q H S G M N T C E E
Y H I J S G Q E N Y C W
G S K M G H C B M U T H
R A T V M N V D G V U T
E P G U E A B P W Q R T
T J C I D D R Q T E E C
U P C I S E N G B U O B
P S J C I V N F O U N N
M P R O J E C T R R A M
O H Q T P P D S H A P G
C O W U K Q E G I J M S
包含单词列表的其他txt文件:
COMPUTER
SCRAM
COURSE
LECTURE
PROGRAMMING
PROJECT
SCIENCE
STUDENT
我的代码:
#Making the word file into a list
def getWords(wordList):
fname = open(wordList,'r')
lines = fname.read().split()
return(lines)
#Making the puzzle file into a 2d list.
def buildBoard(puzzleBoard):
fname = open(puzzleBoard,'r')
boardList = []
for line in fname:
number_strings = line.split()
letters = [n for n in number_strings]
boardList.append(letters)
fname.close()
print(boardList[1][0])#row ,then col. Both start at 0s
return(boardList)
#Recursively search board
def searchBoard(pos,puzzleBoard,wordList):
for word in wordList:
firstLetter = word[:1]
if puzzleBoard[pos[0]][pos[1]] == firstLetter:
newPos = pos[0][0]
searchBoard(newPos,puzzleBoard,wordList)
#Checking above
if puzzleBoard[pos[0]-1][pos[1]] == firstLetter:
newPos = [pos[0]-1,pos[1]]
searchBoard(newPos,puzzleBoard,wordList)
def main():
print("Welcome to the Word Search")
print("For this, you will import two files: 1. The puzzle board, and 2. The word list.")
puzzleBoard = input("What is the puzzle file you would like to import?: ")
wordList = input("What is the word list file you would like to import?: ")
puzzleBoard = buildBoard(puzzleBoard)
wordList = getWords(wordList)
pos = [puzzleBoard[0][0]]
searchBoard(pos,puzzleBoard,wordList)
main()
错误:
if puzzleBoard[pos[0]][pos[1]] == firstLetter:
TypeError: list indices must be integers or slices, not str
答案 0 :(得分:2)
这是一个很好的练习:)
我拿了你的输入拼图板并写了一个小函数,递归搜索拼图板。首先,它尝试找到单词text.find(word[0], pos)
的第一个字母,然后使用floor和modulo division更改每个方向的x,y坐标,并查找下一个字母。如果下一个字母不匹配,它会尝试下一个方向,直到找到整个单词,或者字符不匹配或者它离开拼图板的边界。
除了SCRAM之外的所有单词都可以在你的拼图中找到。
text = """
G J T P B A V K U V L V
M N Q H S G M N T C E E
Y H I J S G Q E N Y C W
G S K M G H C B M U T H
R A T V M N V D G V U T
E P G U E A B P W Q R T
T J C I D D R Q T E E C
U P C I S E N G B U O B
P S J C I V N F O U N N
M P R O J E C T R R A M
O H Q T P P D S H A P G
C O W U K Q E G I J M S"""
#cleans the input puzzleBoard
text = text.replace(' ', '').strip()
#gets the width and height of the puzzleboard
width = len(text.splitlines()[0])
text = text.replace('\n', '')
height = len(text) / width
#a dictionary storing the human readable directions
directions={0:'NW',1:'N',2:'NE',3:'W',4:'',5:'E',6:'SW',7:'S',8:'SE'}
#tries to find a word in a text
#returns x,y of the first character and the orientation of the word
def find_word(text, word):
pos = 0
while pos != -1:
pos = text.find(word[0], pos)
if pos > -1:
for ori in [0,1,2,3,5,6,7,8]:
found = True
i = 0
x = pos % width
y = pos // height
while found:
i += 1
if i == len(word) and found:
return (pos % width, pos // height, directions[ori])
#moves x,y in the selected direction
x += ori % 3 - 1
y += ori // 3 - 1
if x < width and y < height and x > -1 and y > -1:
found = text[width * y + x] == word[i]
else:
found = False
pos += 1
#nothing found
return(-1, -1, directions[4])
解决同样的问题,但通过递归调用相同的函数。该函数返回找到的单词,最后一个字母的x,y位置和方向,即需要向后查找整个单词。
def find_char(text, pos, word, ori):
x = int(pos % width)
y = int(pos // height)
x += ori % 3 - 1
y += ori // 3 - 1
if text[pos] != word[0]:
return None
if len(word) == 1:
return (x,y)
if x < width and y < height and x > -1 and y > -1:
pos = int(width * y + x)
if text[pos] == word[1]:
if len(word) > 1:
resp = find_char(text, pos, word[1:], ori)
if resp:
return resp
else:
return None
word_list = ['COMPUTER', 'SCRAM', 'COURSE', 'LECTURE', 'PROGRAMMING', 'PROJECT', 'SCIENCE', 'STUDENT']
for i in range(len(text)):
for ori in [0,1,2,3,5,6,7,8]:
for word in word_list:
resp = find_char(text, i, word, ori)
if resp:
print(word, resp, ori)
答案 1 :(得分:1)
如果在将pos
传递给searchBoard()
之前打印['G']
,python会将此打印出来:
pos = [puzzleBoard[0][0]]
您创建了一个包含pos = ['G']
的列表,该列表等于puzzleBoard[pos[0]][pos[1]]
因此,当您尝试访问searchBoard()
函数中的puzzleBoard['G'][<nonsense>]
时,您会尝试访问pos
。使用的两个索引都是错误的。
加号searchBoard()
不是位置,,但是包含双列表中的一个字符的列表,并且不包含更多信息(实际上甚至不是位置)。
我建议你将整数作为整数传递给searchBoard(0, 0, puzzleBoard,wordList) # xpos and ypos being the first parameters for instance
:
searchBoard((0, 0), puzzleBoard,wordList)
或者你甚至可以这样做:
pos
这里newPos
是一个元组,它包含两个索引,但只包含这些索引,没有其他信息
编辑:正如robyschek在评论中指出的那样,第一个解决方案会更清晰!
您必须从newPos = pos[0][0]
更改新的分配:
newPos = (pos[0],pos[0])
类似
{{1}}
我并不了解您计划如何管理索引,但重点是:使其成为元组