Python: two dimensional array
# 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.")
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：
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.")
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
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()
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
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.")
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)
# 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
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.")
哲学管理(学)人生, 文学艺术生活, 自动(计算机学)物理(学)工作, 生物(学)化学逆境, 历史(学)测绘(学)时间, 经济(学)数学金钱(理财), 心理(学)医学情绪, 诗词美容情感, 美学建筑(学)家园, 解构建构(分析)整合学习, 智商情商(IQ、EQ)运筹(学)生存.---Geovin Du(涂聚文）
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