如何解决python 3 lambda的参数绑定似乎已损坏
我在CentOS 7环境中使用Python 3.6.10。我试图根据结构化规范创建要执行的命令列表。将其视为lambda列表似乎是自然而然的。我通过遍历规范来构建lambda列表。令我惊讶的是，当我执行结果时，我发现每个lambda都是相同的，因为在创建lambda时它不会捕获其参数。我认为那是一个错误。
以下是说明行为的示例代码：
specification = {
'labelOne': ['labelOne.one','labelOne.two','labelOne.three','labelOne.four','labelOne.five'],'labelTwo': ['labelTwo.one','labelTwo.two','labelTwo.three','labelTwo.four','labelTwo.five'],'labelThree': ['labelThree.one','labelThree.two','labelThree.three','labelThree.four','labelThree.five'],'labelFour': ['labelFour.one','labelFour.two','labelFour.three','labelFour.four','labelFour.five'],'labelFive': ['labelFive.one','labelFive.two','labelFive.three','labelFive.four','labelFive.five'],}
lambdas = []
for label,labelStrings in specification.items():
for labelString in labelStrings:
lambdaString = f"""Label: \"{label}\" with labelString: \"{labelString}\""""
oneArgLambda = lambda someArg: print(someArg,lambdaString)
lambdas.append(oneArgLambda)
for each in lambdas:
each('Show: ')
我希望看到这个：
Show: Label: "labelOne" with labelString: "labelOne.one"
Show: Label: "labelOne" with labelString: "labelOne.two"
Show: Label: "labelOne" with labelString: "labelOne.three"
Show: Label: "labelOne" with labelString: "labelOne.four"
Show: Label: "labelOne" with labelString: "labelOne.five"
Show: Label: "labelTwo" with labelString: "labelTwo.one"
Show: Label: "labelTwo" with labelString: "labelTwo.two"
Show: Label: "labelTwo" with labelString: "labelTwo.three"
Show: Label: "labelTwo" with labelString: "labelTwo.four"
Show: Label: "labelTwo" with labelString: "labelTwo.five"
Show: Label: "labelThree" with labelString: "labelThree.one"
Show: Label: "labelThree" with labelString: "labelThree.two"
Show: Label: "labelThree" with labelString: "labelThree.three"
Show: Label: "labelThree" with labelString: "labelThree.four"
Show: Label: "labelThree" with labelString: "labelThree.five"
Show: Label: "labelFour" with labelString: "labelFour.one"
Show: Label: "labelFour" with labelString: "labelFour.two"
Show: Label: "labelFour" with labelString: "labelFour.three"
Show: Label: "labelFour" with labelString: "labelFour.four"
Show: Label: "labelFour" with labelString: "labelFour.five"
Show: Label: "labelFive" with labelString: "labelFive.one"
Show: Label: "labelFive" with labelString: "labelFive.two"
Show: Label: "labelFive" with labelString: "labelFive.three"
Show: Label: "labelFive" with labelString: "labelFive.four"
Show: Label: "labelFive" with labelString: "labelFive.five"
相反，我看到了：
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
Show: Label: "labelFive" with labelString: "labelFive.five"
lambda的参数绑定发生在执行lambda时，而不是在创建lambda时。至少这是出乎意料的，我认为可以说是错误的。
我认为lambda的局限性在于它应该创建一个CLOSURE，它的整个目的是在其创建时捕获其自变量的状态，以便稍后在lambda时使用它们。被评估。这就是为什么将其称为“关闭”的原因，因为它在创建时会关闭其参数的值。
我误会什么？
解决方法
这是另一种选择
lambdas = []
for label,labelStrings in specification.items():
for labelString in labelStrings:
lambdaString = f"""Label: \"{label}\" with labelString: \"{labelString}\""""
oneArgLambda = lambda someArg,lambdaString: print(someArg,lambdaString)
lambdas.append((oneArgLambda,lambdaString))
for f,lambdaString in lambdas:
f('Show: ',lambdaString)
正如您所说的，它创建CLOSURE，并且闭包在 lambdaString 的上限范围内使用声明的变量，但诀窍是所有lambda都使用相同的ref到 lambdaString >并且由于每次您都记住它的最后一个时都会对其进行更改。例如：
c = ['one','two']
res = []
for i in range(2):
for y in c:
def la(x):
print(x,y)
res.append(la)
for la in res:
la('Show: ')
# Show: two
# Show: two
# Show: two
# Show: two
您只需要另一个关闭即可防止这种情况
c = ['one','two']
res = []
for i in range(2):
for y in c:
def closure_y(y):
def la(x):
print(x,y)
return la
res.append(closure_y(y))
for la in res:
la('Show: ')
# Show: one
# Show: two
# Show: one
# Show: two
完整代码可能是
specification = {
'labelOne': ['labelOne.one','labelOne.two','labelOne.three','labelOne.four','labelOne.five'],'labelTwo': ['labelTwo.one','labelTwo.two','labelTwo.three','labelTwo.four','labelTwo.five'],'labelThree': ['labelThree.one','labelThree.two','labelThree.three','labelThree.four','labelThree.five'],'labelFour': ['labelFour.one','labelFour.two','labelFour.three','labelFour.four','labelFour.five'],'labelFive': ['labelFive.one','labelFive.two','labelFive.three','labelFive.four','labelFive.five'],}
lambdas = []
for label,labelStrings in specification.items():
for labelString in labelStrings:
lambdaString = f"""Label: \"{label}\" with labelString: \"{labelString}\""""
def clousure(lambdaString):
oneArgLambda = lambda someArg: print(someArg,lambdaString)
return oneArgLambda
lambdas.append(clousure(lambdaString))
for each in lambdas:
each('Show: ')
您还有其他评论和答案来说明正在发生的事情，并且您的 问题是“有趣”是指它迫使读者感到困惑 代码和各种绑定问题。
但是，如果我在审核过程中从同事那里看到了您的代码，我会要求您 重写–不是因为我会立即知道存在错误，而是因为它 对于周围的约束是否需要太多的思考 （并不断变化）作用域将完全符合预期。
相反，请在程序中坚持更严格的纪律，从而减轻 读者的认知负担（大部分时间是您）。具体来说，移动 将函数创建到真正隔离的范围，并传递所有变化 该函数创建者的输入。这种方法是可靠的，因为它将 初次尝试或完全失败（如果您忽略了所有 函数创建者所需的参数。
一种方法：
# A function to create another function,with non-surprising argument binding.
# We expect nothing from the surrounding scope. All business can be done locally.
def get_func(label,x):
return lambda prefix: print(f'''{prefix} => {label}: {x}''')
# Some input data.
specification = {
label : [label + str(n) for n in range(3)]
for label in ('A','B','C')
}
# Use that data to create some functions.
funcs = [
get_func(label,x)
for label,xs in specification.items()
for x in xs
]
# Run 'em.
for f in funcs:
f('Show')
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