Python: two dimensional array

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

# 假设puzzle是一个包含多个字符串的列表，每个字符串都是同一长度的，代表字母网格的一行  puzzle = [      "JGJGDDAOYD",      "IDGFHSPOSA",      "FGDIOSAFSC",      "INTERNETSO",      "FJKCOSAFSM",      "DJSGAPAHDP",      "HAUSTRFBFU",      "KDGFUCNSKT",      "WSJDYCFXDE",      "ODVFKXJVCR"  ]      # 获取网格的宽和高（即每一行的长度和行数）  grid_width = len(puzzle[0])  grid_height = len(puzzle)      # 定义要搜索的单词  word = input("Enter a word to search for: ")      # 搜索函数  def solve(puzzle, word):      found = False      for i in range(grid_height - len(word) + 1):  # 确保从第一行到能放下整个单词的行          for j in range(grid_width - len(word) + 1):  # 确保从第一列到能放下整个单词的列              # 检查在i, j位置开始的子网格是否包含目标单词              for k in range(len(word)):                  if puzzle[i + k][j + k] != word[k]:                      break  # 如果字符不匹配，则跳出内层循环              else:  # 只有当内层循环正常结束时（即没有break），才说明找到了单词                  found = True                  print(f"Word '{word}' found at ({i}, {j})")                  break  # 找到后，可以退出外层循环，因为不需要继续搜索      return found      # 调用搜索函数  found = solve(puzzle, word)  if not found:      print(f"Word '{word}' not found in the puzzle.")

　　

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65

def find_diagonal_word(letter_array, word):     """     查找字列表中的字符串     :param letter_array: 字符串列表     :param word: 所需要查找字符     :return:     """     rows = len(letter_array)     cols = len(letter_array[0]) if rows > 0 else 0       # Iterate over all possible starting positions for the diagonal     for i in range(rows):         for j in range(cols):             # Check if the current position is within the bounds of the array             # and if the word can fit in the diagonal             if i + len(word) <= rows and j + len(word) <= cols:                 # Check if the word matches the diagonal                 match = True                 for k in range(len(word)):                     if letter_array[i + k][j + k] != word[k]:                         match = False                         break                 if match:                     return True       return False     def find_column_word(letter_array, word):     """       :param letter_array:     :param word:     :return:     """     cols = len(letter_array[0]) if letter_array else 0       # Iterate over all columns     for j in range(cols):         # Check if the word matches the current column         match = True         for i in range(len(letter_array)):             # Check if the current row is within the bounds of the array             # and if the character matches the word             if i < len(letter_array) and letter_array[i][j] != word[i]:                 match = False                 break                   # If the word matches the column, return True         if match and i == len(word) - 1:             return True       return False   def suc(score):     s6grade=''     if score >= 80:         s6grade = 'A'     if score >= 65 and score <= 79:         s6grade = 'B'     if score >= 50 and score < 65:         s6grade = 'C'     if score < 50:         s6grade = 'F'     return s6grade

 

调用：

　　

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47

if __name__ == '__main__':       letter_array = [         "JGJGDDAOYD",         "IDGFHSPOSA",         "FGDIOSAFSC",         "INTERNETSO",         "FJKCOSAFSM",         "DJSGAPAHDP",         "HAUSTRFBFU",         "KDGFUCNSKT",         "WSJDYCFXDE",         "ODVFKXJVCR"     ]       word = "coMPUTER"     word = input("please word:")     # Convert the word to uppercase since the letter array seems to be in uppercase     word = word.upper()       if find_column_word(letter_array, word):         print(f"The word '{word}' exists in the letter array as a column.")     else:         print(f"The word '{word}' does not exist in the letter array as a column.")           letter_array = [         "JGJGDDAOYD",         "IDGFHSPOSA",         "FGDIOSAFSC",         "INT ERNETSO",         "FJKCOSAFSM",         "DJSGAPAHDP9",         "HAUSTRFBFU",         "KDGFUCNSKT",         "WSJDYCFXDE",         "ODVFKXJVCR"     ]       word = "CAR"     word=input("please word:")       if find_diagonal_word(letter_array, word):         print(f"The word '{word}' exists in the letter array as a diagonal.")     else:         print(f"The word '{word}' does not exist in the letter array as a diagonal.")

　　

 

 

 

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

score=[     [58,80,74,90,45,82],     [71,70,64,85,50,86],     [87,63,65,84,62,83],     [91,66,67,92,65,90],     [83,74,81,82,57,82]] k=0 while k<5:     subavg=0     a=0     while a<6:         subavg=subavg+score[k][a]         a=a+1     subavg =subavg/6     print(subavg)     k=k+1 print("***************8") rows, cols = (5, 6) s6grade=[['0'] * cols] * rows  # 赋值为零 #[[]] # '''''' for i in range(0,5):     for j in range(0,6):         s6grade[i][j]=suc(score[i][j])         print(s6grade[i][j], end=" ")     print() print('**********') for i in range(0,5):     col=[]     for j in range(0,6):         col.append(suc(score[i][j]))         #s6grade[i][j] =suc(score[i][j])  # 出问题     s6grade.insert(i,col) #print(s6grade) #print(suc(score[0][4])) # 0,4 45

　　

 

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

score=[     [58,80,74,90,45,82],     [71,70,64,85,50,86],     [87,63,65,84,62,83],     [91,66,67,92,65,90],     [83,74,81,82,57,82]] k=0 while k<5:     subavg=0     a=0     while a<6:         subavg=subavg+score[k][a]         a=a+1     subavg =subavg/6     print(subavg)     k=k+1 print("***************8") rows, cols = (5, 6) s6grade=[[]] #[['0'] * cols] * rows  # 赋值为零 ''' for i in range(0,5):     for j in range(0,6):         s6grade[i][j]='1'         print(s6grade[i][j], end=" ")     print() ''' for ii in range(0,5):     col=[]     for ji in range(0,6):         d=suc(score[ii][ji])         col.append(d)         #print(d)         #s6grade[ii][ji] = d     print(col)     s6grade.insert(ii,col) print(s6grade) #print(suc(score[0][4])) # 0,4 45   for i in range(0,5):     for j in range(0,6):         print(s6grade[i][j],end=" ")     print()

　　

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

while True:     char=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']     morsecode=['.-','-...','-.-.','-..','.','..-.','--.','....','..','.---','-.-','.-..','--',                '-.','---','.--.','--.-','.-.','...','-','..-','...-','.--','-..-','-.--','--..']     message=input('please word')     l=len(message)     code=""     mos=""     for k in range(0,l):         for index in range(0,len(char)):             if(message[k]==char[index]):                 code=code+" "+str(index+1)                 mos=mos+" "+morsecode[index]       print(code)     print(mos)     yn=input('y/N')     if yn=='Y' or yn=='y':         continue     else:         break

　　

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

def find_column_word(letter_array, word):      cols = len(letter_array[0]) if letter_array else 0          # Iterate over all columns      for j in range(cols):          # Check if the word matches the current column          match = True          for i in range(len(letter_array)):              # Check if the current row is within the bounds of the array              # and if the character matches the word              if i < len(letter_array) and letter_array[i][j] != word[i]:                  match = False                  break              # If the word matches the column, return True          if match and i == len(word) - 1:              return True          return False      # Example usage  letter_array = [      "JGJGDDAOYD",      "IDGFHSPOSA",      "FGDIOSAFSC",      "INTERNETSO",      "FJKCOSAFSM",      "DJSGAPAHDP",      "HAUSTRFBFU",      "KDGFUCNSKT",      "WSJDYCFXDE",      "ODVFKXJVCR"  ]      word = "coMPUTER"      # Convert the word to uppercase since the letter array seems to be in uppercase  word = word.upper()      if find_column_word(letter_array, word):      print(f"The word '{word}' exists in the letter array as a column.")  else:      print(f"The word '{word}' does not exist in the letter array as a column.")

　

 

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

def getMoseCode():     """       :return:     """     char = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u',             'v', 'w', 'x', 'y', 'z']     morsecode = ['.-', '-...', '-.-.', '-..', '.', '..-.', '--.', '....', '..', '.---', '-.-', '.-..', '--',                  '-.', '---', '.--.', '--.-', '.-.', '...', '-', '..-', '...-', '.--', '-..-', '-.--', '--..']     message = input('please word')     l = len(message)     code = ""     mos = ""     for k in range(0, l):         for index in range(0, len(char)):             if (message[k] == char[index]):                  code = code + " " + str(index + 1)                 mos = mos + " " + morsecode[index]                 break# 加这行效率更快Geovin Du     print(code)     print(mos)   def getMoseCodeWhile():     """     这个快1秒     :return:     """     char = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u',             'v', 'w', 'x', 'y', 'z']     morsecode = ['.-', '-...', '-.-.', '-..', '.', '..-.', '--.', '....', '..', '.---', '-.-', '.-..', '--',                  '-.', '---', '.--.', '--.-', '.-.', '...', '-', '..-', '...-', '.--', '-..-', '-.--', '--..']     message = input('please word')     l = len(message)     code = ""     mos = ""     j=0     while j<l:         find = False         index=0         while index<len(char) and find==False: # 加这行效率更快             if (message[j] == char[index]):                 code = code + " " + str(index + 1)                 mos = mos + " " + morsecode[index]                 find=True             index=index+1         j=j+1     print(code)     print(mos)

　　

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

# Press the green button in the gutter to run the script. if __name__ == '__main__':     #print_hi('PyCharm,python language ，Geovin Du')       while True:         time_start = time.time()  # 记录开始时间         #getMoseCode()         getMoseCodeWhile()         time_end = time.time()  # 记录结束时间         time_sum = time_end - time_start  # 计算的时间差为程序的执行时间，单位为秒/s         print(time_sum)         yn = input('y/N')         if yn == 'Y' or yn == 'y':             continue         else:             break

　　

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98

def find_column_wordPart(letter_array, word):     """     在给定的字符串数组列表中按列查找指定的单词的部分位置列中。       :param letter_array: 字符串数组列表     :param word: 所需要查找的单词     :return: 如果找到单词的部分则返回True，否则返回False     """     if not letter_array or not word:         return False       cols = len(letter_array[0])     rows = len(letter_array)     word_length = len(word)       # 遍历所有列     for j in range(cols):         # 在每一列中，从顶部到底部滑动单词长度的窗口         for start_row in range(rows - word_length + 1):             # 检查当前窗口的字符串是否与单词匹配             match = True             for i in range(word_length):                 if letter_array[start_row + i][j] != word[i]:                     match = False                     break             if match:                 return True  # 如果在某一列的窗口中找到单词，则返回True       return False  # 如果检查完所有列都没有找到匹配的单词，则返回False     def find_column_word(letter_array, word):     """     在给定的字符串数组列表中按一列查找指定的单词。       :param letter_array: 字符串数组列表     :param word: 所需要查找的单词     :return: 如果找到单词则返回True，否则返回False     """     if not letter_array or not word:         return False       cols = len(letter_array[0])     rows = len(letter_array)     word_length = len(word)       # 如果单词长度大于列数，则无法找到该单词     if word_length > cols:         return False       # 遍历所有列     for j in range(cols):         # 检查单词是否与当前列匹配         match = True         for i in range(word_length):             # 检查当前列的字符是否与单词的对应字符匹配             if letter_array[i][j] != word[i]:                 match = False                 break         if match:             return True  # 如果单词与某一列完全匹配，则返回True       return False  # 如果检查完所有列都没有找到匹配的单词，则返回False     # Press the green button in the gutter to run the script. C:\Users\geovindu\PycharmProjects\EssentialAlgorithms\ if __name__ == '__main__':       print_hi('PyCharm,python,geovindu,Geovin Du,涂聚文，你好！')     letter_array = [         "JGJGDDAOYD",         "IDGFHSPOSA",         "FGDIOSAFSC",         "INTERNETSO",         "FJKCOSAFSM",         "DJSGAPAHDP",         "HAUSTRFBFU",         "KDGFUCNSKT",         "WSJDYCFXDE",         "ODVFKXJVCR"     ]       word = "dacoMPUTER" #整个一列中     # 转换为大写以便与letter_array中的字符进行比较     word = word.upper()       if find_column_word(letter_array, word):         print(f"The word '{word}' exists in the letter array as a column.")     else:         print(f"The word '{word}' does not exist in the letter array as a column.")       word = "MPUTER"  # 在一列部分的位置     # 转换为大写以便与letter_array中的字符进行比较     word = word.upper()     if find_column_wordPart(letter_array, word):         print(f"The word Part '{word}' exists in the letter array as a column.")     else:         print(f"The word Part '{word}' does not exist in the letter array as a column.")