python中lambda函数 list comprehension 和 zip函数使用指南
lambda 函数
Python 支持一种有趣的语法,它允许你快速定义单行的最小函数。这些叫做 lambda 的函数,是从 Lisp 借用来的,可以用在任何需要函数的地方。
def f(x): return x*2,用lambda函数来替换可以写成:g = lambda x: x*2`g(3)结果是6.(lambda x: x*2)(3)`也是同样的效果。
这是一个 lambda 函数,完成同上面普通函数相同的事情。注意这里的简短的语法:在参数列表周围没有括号,而且忽略了 return 关键字 (隐含存在,因为整个函数只有一行)。而且,该函数没有函数名称,但是可以将它赋值给一个变量进行调用
使用 lambda 函数时甚至不需要将它赋值给一个变量。这可能不是世上最有用的东西,它只是展示了 lambda 函数只是一个内联函数。
总的来说,lambda 函数可以接收任意多个参数 (包括可选参数) 并且返回单个表达式的值。lambda 函数不能包含命令,包含的表达式不能超过一个。不要试图向 lambda 函数中塞入太多的东西;如果你需要更复杂的东西,应该定义一个普通函数,然后想让它多长就多长。 我将它们用在需要封装特殊的、非重用代码上,避免令我的代码充斥着大量单行函数。
列表推导式(list comprehension)
看一段简单代码
testList = [1,2,3,4]
def mul2(x):
print x*2
[mul2(i) for i in testList]
[mul2(i) for i in testList if i%2==0]
多维数组初始化
multilist = [[0 for col in range(5)] for row in range(3)]
zip 函数
>>> a = [1,2,3]
>>> b = [4,5,6]
>>> c = [4,5,6,7,8]
>>> zipped = zip(a,b)
[(1, 4), (2, 5), (3, 6)]
>>> zip(a,c)
[(1, 4), (2, 5), (3, 6)]
>>> zip(*zipped)
[(1, 2, 3), (4, 5, 6)]
学习资源
学以致用
一
m = [[-1.0, 2.0/c-1, -2.0/c+1, 1.0],
[2.0, -3.0/c+1, 3.0/c-2, -1.0],
[-1.0, 0.0, 1.0, 0.0],
[0.0, 1.0/c, 0.0, 0.0]]
multiply = lambda x: x*c
m = [[multiply(m[col][row]) for col in range(4)] for row in range(4)]
print [[m[col][row] for col in range(4)] for row in range(4)]
它所作的工作:m是一个包含参数c的矩阵,他计算了c*m的结果
想了一下,最后一句改成
print [[multiply(each) for each in row] for row in m]更加pythonic
二 矩阵相乘
学习资源
def matrixMul(A, B):
res = [[0] * len(B[0]) for i in range(len(A))] for i in range(len(A)):
for j in range(len(B[0])):
for k in range(len(B)):
res[i][j] += A[i][k] * B[k][j] return res
def matrixMul2(A, B):
return [[sum(a * b for a, b in zip(a, b)) for b in zip(*B)] for a in A]
a = [[1,2], [3,4], [5,6], [7,8]]
b = [[1,2,3,4], [5,6,7,8]]
print matrixMul(a,b) print matrixMul(b,a) print "-"*90
print matrixMul2(a,b) print matrixMul2(b,a) print "-"*90

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