需要了解包含lambda的函数
我发现了一个函数,它获取列表列表并返回字符串。 然而,我很难理解它到底在做什么?请对以下代码进行注释:需要了解包含lambda的函数,lambda,python-2.7,Lambda,Python 2.7,我发现了一个函数,它获取列表列表并返回字符串。 然而,我很难理解它到底在做什么?请对以下代码进行注释: def getAsTable(self, arrays): """ This method takes an array of arrays and returns string (which is really a table) :param arrays: An array of arrays :returns: string (this is really a
def getAsTable(self, arrays):
""" This method takes an array of arrays and returns string (which is really a table)
:param arrays: An array of arrays
:returns: string (this is really a table of the input)
>>> [[a, b, b], [1, 2, 3], [4, 5, 6]]
>>> a b c
>>> 1 2 3
>>> 4 5 6
"""
def areAllEqual(lst):
return not lst or [lst[0]] * len(lst) == lst
if not areAllEqual(map(len, arrays)):
return "Cannot print a table with unequal array lengths"
verticalMaxLengths = [max(value) for value in map(lambda * x:x, *[map(len, a) for a in arrays])]
spacedLines = []
for array in arrays:
spacedLine = ''
for i, field in enumerate(array):
diff = verticalMaxLengths[i] - len(field)
spacedLine += field + ' ' * diff + '\t'
spacedLines.append(spacedLine)
return '\n '.join(spacedLines)
一个简短的解释让我不必在下面的代码中添加注释:
map
函数将其第一个参数(通常是函数,但也可以是类或任何其他可调用的对象)应用于第二个参数中的每个值,并返回结果列表。将其视为使用给定函数变换每个元素。与两个参数一起使用时,其工作原理如下:
def map(fct, iterable): return [fct(x) for x in iterable]
与三个或三个以上参数一起使用,map
假定第一个参数之后的所有参数都是iterable,并并行迭代,将每个iterable的第n个元素传递给第n次传递的函数:
def p(a,b,c): print "a: %s, b:%s, c:%s"
map(p, "abc", "123", "456") #-> prints "a 1 4", then "b 2 5", then "c 3 6"
代码的注释版本:
def getAsTable(self, arrays):
#helper function checking that all values contained in lst are equal
def areAllEqual(lst):
#return true for the empty list, or if a list of len times the first
#element equals the original list
return not lst or [lst[0]] * len(lst) == lst
#check that the length of all lists contained in arrays is equal
if not areAllEqual(map(len, arrays)):
#return an error message if this is not the case
#this should probably throw an exception instead...
return "Cannot print a table with unequal array lengths"
verticalMaxLengths = [max(value) for value in map(lambda * x:x, *[map(len, a) for a in arrays])]
让我们把这条线分成几部分:
(1) [map(len, a) for a in arrays]
这会将len映射到数组中的每个列表,这意味着您将得到一个
元素的长度列表。例如,对于输入[[“a”、“b”、“c”]、[“1”、“11”、“111”]、[“n”、“n^2”、“n^10”]
,结果将是[[1,1,1]、[1,2,3]、[1,2,4]
(2) map(lambda *x:x, *(1))
*
打开(1)中获得的列表,这意味着每个元素都是
作为单独的参数传递给map。如上所述,具有
如果有多个参数,映射会将一个参数传递给函数。兰姆达酒店
这里定义的只是将其所有参数作为元组返回。
继续上面的示例,对于输入[[1,1,1],[1,2,3],[1,2,4]
结果将是[(1,1,1)、(1,2,2)、(1,3,4)]
这基本上导致了输入的矩阵转置
(3) [max(value) for value in (2)]
这将调用(2)中返回的列表中所有元素的max(请记住这些元素是元组)。对于输入[(1,1,1)、(1,2,2)、(1,3,4)]
结果将是[1,2,4]
因此,在这里的上下文中,整行接受输入数组并计算每列中元素的最大长度
代码的其余部分:
#initialize an empty list for the result
spacedLines = []
#iterate over all lists
for array in arrays:
#initialize the line as an empty string
spacedLine = ''
#iterate over the array - enumerate returns position (i) and value
for i, field in enumerate(array):
#calculate the difference of the values length to the max length
#of all elements in the column
diff = verticalMaxLengths[i] - len(field)
#append the value, padded with diff times space to reach the
#max length, and a tab afterwards
spacedLine += field + ' ' * diff + '\t'
#append the line to the list of lines
spacedLines.append(spacedLine)
#join the list of lines with a newline and return
return '\n '.join(spacedLines)