在Python中使用Lambda进行递延评估 [英] Deferred evaluation with lambda in Python

问题描述

在一个循环中，我试图将比较两个节点的两个value（）推迟到以后。

In a loop, I am trying to defer the comparison the two value()s of two Nodes to a later time.

class Node():
    def __init__(self, v):
        self.v = v
    def value(self):
        return self.v

nodes = [Node(0), Node(1), Node(2), Node(3), Node(4), Node(2)]
results = []
for i in [0, 1, 2]:
    j = i + 3
    results.append(lambda: nodes[i].value() == nodes[j].value())

for result in results:
    print result

结果均为True （因为对于所有的λ，i，j == 2,5）。如何将lambda的执行推迟到实际被调用之前，但是具有正确的变量绑定？而且lambda中的表达式不一定都相等……还有更多其他涉及的表达式。

The results are all True (because i,j==2,5 for all the lambdas). How can I defer the execution of the lambda until it is actually called, but with the correct variable bindings? And the expressions in the lambda are not all necessarily equality... there are a bunch of other more involved expressions.

感谢您的帮助！

推荐答案

绑定 i 和 j 而不是让函数在外部范围中显示，可以使用闭包或默认参数值。最简单的方法是在lambda中使用默认参数值：

To bind the current values of i and j to the function instead of having it look in the outer scope, you can use either a closure or default argument values. The easiest way to do this is to use default argument values in your lambda:

for i in [0, 1, 2]:
    j = i + 3
    results.append(lambda i=i, j=j: nodes[i].value() == nodes[j].value())

这里是闭包的样子：

def make_comp_func(i, j):
    return lambda: nodes[i].value() == nodes[j].value()

for i in [0, 1, 2]:
    j = i + 3
    results.append(make_comp_func(i, j))

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