关于python：如何将数字列表拆分并插入到另一个列表中

How to get a split up a list of numbers and insert into another list

目前我有一个文件，有6行数字，每行包含9个数字。关键是测试文件中每一行数字，如果它完成了一个魔方。例如，假设文件中的一行数字是4 3 8 9 5 1 2 7 6。前三个数字必须是矩阵中的第一行。接下来的三个数字需要是第二行，第三行也一样。因此，您需要得到一个矩阵：['4'、'3'、'8']、['9'、'5'、'1']、['2'、'7'、'6']

我需要测试这个矩阵，看看它是否是一个有效的魔方(行加15，列加15，对角线加15)。

我的代码当前为：

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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)

每当我运行代码来测试它时，我都会得到：

1	[['4', '3', '8'], ['4', '3', '8'], ['4', '3', '8']]

所以你可以看到，我只是把前3个数放入我的矩阵中。最后，我的问题是，我将如何继续我的矩阵与以下6个数字行的数字？我不确定这是否是我在循环中可以做的事情，或者我的线路分裂错误，或者我完全走错了轨道？

谢谢。

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。我用了一种不同的方法来填充矩阵。也许更难理解zip(*[iter(seq)] * size)是如何工作的，但这是一个非常有用的模式。如果你需要解释的话，请告诉我。

我的代码为矩阵使用了一个元组列表，而不是一个列表列表，但是无论如何，元组在这里更适合，因为矩阵中的数据不需要修改。另外，我将str的输入数据转换为int，因为您需要对数字进行算术运算以测试matrix是否是一个魔方。

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#! /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()

这里有一个修改过的make_square()版本，它返回列表而不是元组列表，但是请记住，如果不需要列表提供的可变性，元组列表实际上比列表列表更好。

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def make_square(seq, size):
square = zip(*[iter(seq)] * size)
return [list(t) for t in square]

我想我应该提到，实际上只有一个可能的3x 3魔方使用了从1到9的所有数字，不计算旋转和反射。但我想，对这一事实进行一次蛮力的演示是没有坏处的。：)

另外，我还有几年前(当我第一次学习Python时)编写的Python代码，它生成了大小为n x n的魔法方块(奇数n>=5)。如果你想看，请告诉我。

zip和迭代器对象

下面是一些代码，简要说明了zip()和iter()函数的作用。

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''' 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)

下面是在交互式解释器中运行的相同代码，因此您可以轻松地看到每个语句的输出。

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>>> 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')]
>>>