如何使用行和列创建矩阵。 当我打印矩阵时，输出应该是这样的：
O X X X X X X X X X X X X X X X
N X X X X X X X X X X X X X X X
M X X X X X X X X X X X X X X X
L X X X X X X X X X X X X X X X
K X X X X X X X X X X X X X X X
J X X X X X X X X X X X X X X X
I X X X X X X X X X X X X X X X
H X X X X X X X X X X X X X X X
G X X X X X X X X X X X X X X X
F X X X X X X X X X X X X X X X
E X X X X X X X X X X X X X X X
D X X X X X X X X X X X X X X X
C X X X X X X X X X X X X X X X
B X X X X X X X X X X X X X X X
A X X X X X X X X X X X X X X X
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
我认为我需要在字典中使用列表，并使用矩阵来编辑“ X”，以便以后进行编辑。
hall_dictionary = {}
hall_dictionary["merhaba"] = []
rows = 10
columns = 15
x = [[hall_dictionary["merhaba"] for i in range(columns)] for j in range(rows)]
答案 0 :(得分：1)
您可以将整个数据存储封装为一个类。它可以处理所有“簿记”，您只需使用A到...和1到...来更改X。
内部使用一个简单的一维列表：
class Field:
def __init__(self, rows, cols, init_piece="x"):
self.rows = rows
self.cols = cols
self.field = [init_piece] * rows * cols
def place_at(self, row, col, piece):
"""Changes one tile on the field. Does all the reverse-engineering to compute
1-dim place of A..?,1..? given tuple of coords."""
def validation():
"""Raises error when out of bounds."""
error = []
if not (isinstance(row,str) and len(row) == 1 and row.isalpha()):
error.append("Use rows between A and {}".format(chr(ord("A") +
self.rows - 1)))
if not (0 < col <= self.cols):
error.append("Use columns between 1 and {}".format(self.cols))
if error:
error = ["Invalid row/column: {}/{}".format(row,col)] + error
raise ValueError('\n- '.join(error))
validation()
row = ord(row.upper()[0]) - ord("A")
self.field[row * self.cols + col - 1] = piece
def print_field(self):
"""Prints the playing field."""
for c in range(self.rows - 1,-1,-1):
ch = chr(ord("A") + c)
print("{:<4} ".format(ch), end = "")
print(("{:>2} " * self.cols).format(*self.field[c * self.cols:
(c + 1) * self.cols], sep = " "))
print("{:<4} ".format(""), end = "")
print(("{:>2} " * self.cols).format(*range(1,self.cols + 1)))
然后您可以像这样使用它：
rows = 10
cols = 15
f = Field(rows,cols)
f.print_field()
# this uses A...? and 1...? to set things
for r,c in [(0,0),("A",1),("ZZ",99),("A",99),("J",15)]:
try:
f.place_at(r,c,"i") # set to 'i'
except ValueError as e:
print(e)
f.print_field()
输出（之前）：
J x x x x x x x x x x x x x x x
I x x x x x x x x x x x x x x x
H x x x x x x x x x x x x x x x
G x x x x x x x x x x x x x x x
F x x x x x x x x x x x x x x x
E x x x x x x x x x x x x x x x
D x x x x x x x x x x x x x x x
C x x x x x x x x x x x x x x x
B x x x x x x x x x x x x x x x
A x x x x x x x x x x x x x x x
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
输出（设置&&之后的内容）
Invalid row/column: 0/0
- Use rows between A and J
- Use columns between 1 and 15
Invalid row/column: ZZ/99
- Use rows between A and J
- Use columns between 1 and 15
Invalid row/column: A/99
- Use columns between 1 and 15
J x x x x x x x x x x x x x x i
I x x x x x x x x x x x x x x x
H x x x x x x x x x x x x x x x
G x x x x x x x x x x x x x x x
F x x x x x x x x x x x x x x x
E x x x x x x x x x x x x x x x
D x x x x x x x x x x x x x x x
C x x x x x x x x x x x x x x x
B x x x x x x x x x x x x x x x
A i x x x x x x x x x x x x x x
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
答案 1 :(得分：0)
听起来像二维数组（类似于回答How to define a two-dimensional array in Python），所以类似这样：
vertical = list(string.ascii_uppercase[0:15][::-1]) # ['O', 'N', ..., 'A']
columns = 15
hall_dictionary = {}
hall_dictionary["merhaba"] = [[x for x in range(columns)] for y in vertical]
for i in range(len(vertical)):
for j in range(columns):
hall_dictionary["merhaba"][i][j] = 'X'
然后您可以根据需要编制索引：
hall_dictionary["merhaba"][0][1] # Always prints 'X'
显示整个数组：
for row in hall_dictionary["merhaba"]:
print(row)
['X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X']
... 15 rows ...
并分配，更新新值：
hall_dictionary["merhaba"][0][2] = 'O'
for row in hall_dictionary["merhaba"]:
print(row)
['X', 'X', 'O', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X']
...
确认元素[0][2]已更新。
答案 2 :(得分：0)
如果您有兴趣，也可以使用pandas DataFrame。
import pandas as pd
rows = 10
columns = 15
def indexToLetter(index:int): # This function might be a bit too verbose
if index == 0: # but all this does is convert an
return 'A' # integer index [0, ∞) to an
# alphabetical index [A..Z, AA..ZZ, AAA...]
ret = ''
while index > 0:
length = len(ret)
letter = chr(ord('A') + index % 26 - [0, 1][length >= 1])
ret = letter + ret
index //= 26
return ret
# create the row labels
rLabels = [*map(indexToLetter, range(rows))]
# create the dataframe, note that we can simplify
# [['X' for i in range(columns)] for j in range(rows)]
# to [['X'] * columns] * rows
df = pd.DataFrame([['X'] * columns] * rows, index=rLabels)
print(df)
输出：
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
A X X X X X X X X X X X X X X X
B X X X X X X X X X X X X X X X
C X X X X X X X X X X X X X X X
D X X X X X X X X X X X X X X X
E X X X X X X X X X X X X X X X
F X X X X X X X X X X X X X X X
G X X X X X X X X X X X X X X X
H X X X X X X X X X X X X X X X
I X X X X X X X X X X X X X X X
J X X X X X X X X X X X X X X X
输出看起来很难看，可能不是您想要的。但是有了数据框，操作矩阵和数据表非常方便。
您可以先指定列，然后指定行（不同于其他解决方案）来访问它。
df[1][0] = 'O'
df[1][2] = 'O'
df[1][3] = 'O'
print(df)
输出：
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
A X O X X X X X X X X X X X X X
B X X X X X X X X X X X X X X X
C X O X X X X X X X X X X X X X
D X O X X X X X X X X X X X X X
E X X X X X X X X X X X X X X X
F X X X X X X X X X X X X X X X
G X X X X X X X X X X X X X X X
H X X X X X X X X X X X X X X X
I X X X X X X X X X X X X X X X
J X X X X X X X X X X X X X X X
说，有人想预订大厅的整行'E'。
if any(df.loc['E'] == 'O'): # check if any seats were taken
print('Error: some seats in Row <E> are taken.')
else:
df.loc['E'] = 'O'
print(df)
输出：
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
A X O X X X X X X X X X X X X X
B X X X X X X X X X X X X X X X
C X O X X X X X X X X X X X X X
D X O X X X X X X X X X X X X X
E O O O O O O O O O O O O O O O
F X X X X X X X X X X X X X X X
G X X X X X X X X X X X X X X X
H X X X X X X X X X X X X X X X
I X X X X X X X X X X X X X X X
J X X X X X X X X X X X X X X X
注意：您也可以执行df.iloc[4]访问E行。是否要访问B到E行？使用df.loc['B':'E']或df.iloc[1:5]。
您还可以通过访问df[<column_index>] = 'O'对列进行相同操作。