问题：为什么在lambda中无法打印？
为什么不起作用？
lambda: print "x"
这不是一个单一的陈述，还是其他？该文档对于lambda允许的内容似乎有点稀疏…
Why doesn’t this work?
lambda: print "x"
Is this not a single statement, or is it something else?
The documentation seems a little sparse on what is allowed in a lambda…
回答 0
一个lambda人的身体必须是一个单一的表情。在Python 2.x中，print是一条语句。但是，在Python 3中，print是函数（而函数应用程序是表达式，因此它将在lambda中工作）。如果您使用的是最新的Python 2.x，则可以（并且应该，为了向前兼容：）使用向后打印功能：
In [1324]: from __future__ import print_function
In [1325]: f = lambda x: print(x)
In [1326]: f("HI")
HI
A lambda‘s body has to be a single expression. In Python 2.x, print is a statement. However, in Python 3, print is a function (and a function application is an expression, so it will work in a lambda). You can (and should, for forward compatibility :) use the back-ported print function if you are using the latest Python 2.x:
In [1324]: from __future__ import print_function
In [1325]: f = lambda x: print(x)
In [1326]: f("HI")
HI
回答 1
在我将其用于简单存根的情况下，请使用以下方法：
fn = lambda x: sys.stdout.write(str(x) + "\n")
完美地运作。
In cases where I am using this for simple stubbing out I use this:
fn = lambda x: sys.stdout.write(str(x) + "\n")
which works perfectly.
回答 2
你写的等同于
def anon():
return print "x"
这也会导致SyntaxError，python不允许您分配要在2.xx中打印的值；在python3中，你可以说
lambda: print('hi')
这样做是可行的，因为他们将print更改为函数而不是语句。
what you’ve written is equivalent to
def anon():
return print "x"
which also results in a SyntaxError, python doesn’t let you assign a value to print in 2.xx; in python3 you could say
lambda: print('hi')
and it would work because they’ve changed print to be a function instead of a statement.
回答 3
Lambda的主体必须是一个返回值的表达式。 print作为语句，不会返回任何东西，甚至也不返回None。同样，您不能将的结果分配给print变量：
>>> x = print "hello"
File "<stdin>", line 1
x = print "hello"
^
SyntaxError: invalid syntax
您也不能将变量赋值放在lambda中，因为赋值是语句：
>>> lambda y: (x = y)
File "<stdin>", line 1
lambda y: (x = y)
^
SyntaxError: invalid syntax
The body of a lambda has to be an expression that returns a value. print, being a statement, doesn’t return anything, not even None. Similarly, you can’t assign the result of print to a variable:
>>> x = print "hello"
File "<stdin>", line 1
x = print "hello"
^
SyntaxError: invalid syntax
You also can’t put a variable assignment in a lambda, since assignments are statements:
>>> lambda y: (x = y)
File "<stdin>", line 1
lambda y: (x = y)
^
SyntaxError: invalid syntax
回答 4
你可以做这样的事情。
创建一个函数以将打印语句转换为函数：
def printf(text):
print text
并打印：
lambda: printf("Testing")
You can do something like this.
Create a function to transform print statement into a function:
def printf(text):
print text
And print it:
lambda: printf("Testing")
回答 5
使用Python 3.x，打印可以在lambda中工作，而无需更改lambda的语义。
以特殊的方式使用，这对于调试非常方便。我发布此“最新答案”，因为这是我经常使用的实用技巧。
假设您的“非工具化” lambda为：
lambda: 4
然后，您的“工具化” lambda为：
lambda: (print (3), 4) [1]
With Python 3.x, print CAN work in a lambda, without changing the semantics of the lambda.
Used in a special way this is very handy for debugging.
I post this ‘late answer’, because it’s a practical trick that I often use.
Suppose your ‘uninstrumented’ lambda is:
lambda: 4
Then your ‘instrumented’ lambda is:
lambda: (print (3), 4) [1]
回答 6
Lambda的主体必须是单个表达式。print是一个声明，很遗憾，它已经退出了。
The body of a lambda has to be a single expression. print is a statement, so it’s out, unfortunately.
回答 7
在这里，您会看到问题的答案。 print它说不是在Python中表达。
Here, you see an answer for your question. print is not expression in Python, it says.
