目前我有一个包含6行数字的文件，每行包含9个数字。关键是如果它完成一个魔术方格，则测试文件中的每一行数字。例如，假设文件中的一行数字是4 3 8 9 5 1 2 7 6.前三个数字需要是矩阵中的第一行。接下来的三个数字需要是第二行，第三行相同。 因此，您需要最终得到以下矩阵： [[＆＃39; 4＆＃39;＆＃39 3＆＃39;＆＃39 8＆＃39;]，[＆＃39; 9＆＃39;＆＃39 5＆＃39;，＆ ＃39; 1＆＃39;]，[＆＃39; 2＆＃39;＆＃39 7＆＃39;＆＃39; 6＆＃39;]]
我需要测试矩阵以查看它是否是一个有效的魔术方（行数加15，列加15，对角线加15）。
我的代码目前是：
def readfile(fname):
"""Return a list of lines from the file"""
f = open(fname, 'r')
lines = f.read()
lines = lines.split()
f.close()
return lines
def assignValues(lines):
magicSquare = []
rows = 3
columns = 3
for row in range(rows):
magicSquare.append([0] * columns)
for row in range(len(magicSquare)):
for column in range(len(magicSquare[row])):
magicSquare[row][column] = lines[column]
return magicSquare
def main():
lines = readfile(input_fname)
matrix = assignValues(lines)
print(matrix)
每当我运行我的代码进行测试时，我都会：
[['4', '3', '8'], ['4', '3', '8'], ['4', '3', '8']]
所以你可以看到我只将前3个数字输入矩阵 最后，我的问题是如何通过以下6个数字行继续我的矩阵？我不确定这是否是我在循环中可以做的事情，或者我是否将错误分开，或者我完全走错了路线？
感谢。
答案 0 :(得分：0)
它只获得前3列，因为
magicSquare[row][column] = lines[column]
从而
def assignValues(lines):
magicSquare = []
rows = 3
columns = 3
for row in range(rows):
magicSquare.append([0] * columns)
for line in range((sizeof(lines)/9)) #since the input is already split this means that the size of 'lines' divided by 9 is equal to the number of rows of numbers
for row in range(len(magicSquare)):
for column in range(len(magicSquare[row])):
magicSquare[row][column] = lines[(9*line)+(3*row)+column]
return magicSquare
请注意，（3 * row）+列将在每次迭代时向右移动3列 并且（9 *行）+（3 *行）+列将每次迭代向右移动9列（整行）
一旦你得到了这个，你现在已经准备好了解魔术广场
def testMagicSquare(matrix):
rows = 3
columns = 3
for a in len(matrix)
test1 = 0
test2 = 0
test3 = 0
for b in range(3)
if(sum(matrix[a][b])==15) test1=1 #flag true if whole row is 15 but turns false if a row is not 15
else test1=0
if((matrix[a][0][b]+matrix[a][1][b]+matrix[a][2][b])==15) test2=1 #flag true if column is 15 but turns false if a column is not 15
else test2=0
if(((matrix[a][0][0]+matrix[a][1][1]+matrix[a][2][2])==15) and
((matrix[a][0][2]+matrix[a][1][1]+matrix[a][2][0])==15)) test3=1 #flag true if diagonal is 15 but turns false if diagonal is not 15
else test3=0
if(test1>0 and test2>0 and test3>0) println('line ' + a + ' is a magic square')
else println('line ' + a + ' is not a magic square')
答案 1 :(得分：0)
要测试输入文件中的每一行是否包含幻方数据，您需要稍微重新组织代码。我用弗朗西斯的不同技术来填充矩阵。理解zip(*[iter(seq)] * size)如何工作可能有点困难，但它是非常有用的模式。如果您需要解释，请告诉我。
我的代码使用矩阵的元组列表，而不是列表列表，但无论如何，元组更适合，因为矩阵中的数据不需要修改。此外，我将输入数据从str转换为int，因为您需要对数字进行算术运算，以测试matrix是否为幻方。
#! /usr/bin/env python
def make_square(seq, size):
return zip(*[iter(seq)] * size)
def main():
fname = 'mydata'
size = 3
with open(fname, 'r') as f:
for line in f:
nums = [int(s) for s in line.split()]
matrix = make_square(nums, size)
print matrix
#Now call the function to test if the data in matrix
#really is a magic square.
#test_square(matrix)
if __name__ == '__main__':
main()
这是make_square()的修改版本，它返回列表而不是元组列表，但请记住，如果您不需要，元组列表实际上比列表列表更好列表给你的可变性。
def make_square(seq, size):
square = zip(*[iter(seq)] * size)
return [list(t) for t in square]
我想我应该提到的是，实际上只有一个可能的3 x 3魔方使用了1到9之间的所有数字，不计算旋转和反射。但我认为对这一事实进行蛮力证明是没有害处的。 ：）
另外，我有几年前写的Python代码（当我第一次学习Python的时候），它生成大小为nxn的奇数n＆gt; = 5的魔术方块。如果你想看到它，请告诉我。< / p>
这里有一些代码，简要说明了zip（）和iter（）函数的作用。
''' Fun with zip '''
numbers = [1, 2, 3, 4, 5, 6]
letters = ['a', 'b', 'c', 'd', 'e', 'f']
#Using zip to create a list of tuples containing pairs of elements of numbers & letters
print zip(numbers, letters)
#zip works on other iterable objects, including strings
print zip(range(1, 7), 'abcdef')
#zip can handle more than 2 iterables
print zip('abc', 'def', 'ghi', 'jkl')
#zip can be used in a for loop to process two (or more) iterables simultaneously
for n, l in zip(numbers, letters):
print n, l
#Using zip in a list comprehension to make a list of lists
print [[l, n] for n, l in zip(numbers, letters)]
#zip stops if one of the iterables runs out of elements
print [[n, l] for n, l in zip((1, 2), letters)]
print [(n, l) for n, l in zip((3, 4), letters)]
#Turning an iterable into an iterator object using the iter function
iletters = iter(letters)
#When we take some elements from an iterator object it remembers where it's up to
#so when we take more elements from it, it continues from where it left off.
print [[n, l] for n, l in zip((1, 2, 3), iletters)]
print [(n, l) for n, l in zip((4, 5), iletters)]
#This list will just contain a single tuple because there's only 1 element left in iletters
print [(n, l) for n, l in zip((6, 7), iletters)]
#Rebuild the iletters iterator object
iletters = iter('abcdefghijkl')
#See what happens when we zip multiple copies of the same iterator object.
print zip(iletters, iletters, iletters)
#It can be convenient to put multiple copies of an iterator object into a list
iletters = iter('abcdefghijkl')
gang = [iletters] * 3
#The gang consists of 3 references to the same iterator object
print gang
#We can pass each iterator in the gang to zip as a separate argument
#by using the "splat" syntax
print zip(*gang)
#A more compact way of doing the same thing:
print zip(* [iter('abcdefghijkl')]*3)
这是在交互式解释器中运行的相同代码，因此您可以轻松查看每个语句的输出。
>>> numbers = [1, 2, 3, 4, 5, 6]
>>> letters = ['a', 'b', 'c', 'd', 'e', 'f']
>>>
>>> #Using zip to create a list of tuples containing pairs of elements of numbers & letters
... print zip(numbers, letters)
[(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd'), (5, 'e'), (6, 'f')]
>>>
>>> #zip works on other iterable objects, including strings
... print zip(range(1, 7), 'abcdef')
[(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd'), (5, 'e'), (6, 'f')]
>>>
>>> #zip can handle more than 2 iterables
... print zip('abc', 'def', 'ghi', 'jkl')
[('a', 'd', 'g', 'j'), ('b', 'e', 'h', 'k'), ('c', 'f', 'i', 'l')]
>>>
>>> #zip can be used in a for loop to process two (or more) iterables simultaneously
... for n, l in zip(numbers, letters):
... print n, l
...
1 a
2 b
3 c
4 d
5 e
6 f
>>> #Using zip in a list comprehension to make a list of lists
... print [[l, n] for n, l in zip(numbers, letters)]
[['a', 1], ['b', 2], ['c', 3], ['d', 4], ['e', 5], ['f', 6]]
>>>
>>> #zip stops if one of the iterables runs out of elements
... print [[n, l] for n, l in zip((1, 2), letters)]
[[1, 'a'], [2, 'b']]
>>> print [(n, l) for n, l in zip((3, 4), letters)]
[(3, 'a'), (4, 'b')]
>>>
>>> #Turning an iterable into an iterator object using using the iter function
... iletters = iter(letters)
>>>
>>> #When we take some elements from an iterator object it remembers where it's up to
... #so when we take more elements from it, it continues from where it left off.
... print [[n, l] for n, l in zip((1, 2, 3), iletters)]
[[1, 'a'], [2, 'b'], [3, 'c']]
>>> print [(n, l) for n, l in zip((4, 5), iletters)]
[(4, 'd'), (5, 'e')]
>>>
>>> #This list will just contain a single tuple because there's only 1 element left in iletters
... print [(n, l) for n, l in zip((6, 7), iletters)]
[(6, 'f')]
>>>
>>> #Rebuild the iletters iterator object
... iletters = iter('abcdefghijkl')
>>>
>>> #See what happens when we zip multiple copies of the same iterator object.
... print zip(iletters, iletters, iletters)
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'l')]
>>>
>>> #It can be convenient to put multiple copies of an iterator object into a list
... iletters = iter('abcdefghijkl')
>>> gang = [iletters] * 3
>>>
>>> #The gang consists of 3 references to the same iterator object
... print gang
[<iterator object at 0xb737eb8c>, <iterator object at 0xb737eb8c>, <iterator object at 0xb737eb8c>]
>>>
>>> #We can pass each iterator in the gang to zip as a separate argument
... #by using the "splat" syntax
... print zip(*gang)
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'l')]
>>>
>>> #A more compact way of doing the same thing:
... print zip(* [iter('abcdefghijkl')]*3)
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'l')]
>>>