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Passing a function as an argument in Python I wanted to pass a lambda expression when writing a function, but in Python2 I couldn't put it in a lambda expression because print is a statement rather than an expression. did. Therefore, import the \ _ \ _ future__ module and change the print statement to a print expression as in Python3.

For the difference between expressions and statements, see StackOverFlow Post Is easy to understand. In summary, an expression can evaluate its value by itself, but a statement cannot be evaluated without completing the defined procedure.

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#Cannot be defined in Python2(Because print is a statement)
# a = lambda: print "hoge"

#Can be defined in Python 3(Because print is an expression)
a = lambda: print("hoge")

#How to define in Python2
from __future__ import print_function
a = lambda: print("hoge")

However, the problem is that you have to change all __print "hogehoge" to print ("hogehoge") __. When using it in Python2, it is better to limit the use such as importing it in the method.

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For example, in C # you can write sentences in lambda expressions, so it's better to keep in mind that there are differences between languages.

Action<string> act = (string x) =>
{
    if(x == "hoge") Console.WriteLine(x);
    else Console.WriteLine("Except for hoge");
}

Also, I didn't know how to write a lambda expression with no arguments in Python.

a = lambda: 5 #Lambda expression that returns 5

It seems that you can write with.

\ _ \ _ Future__ Modules are summarized in Contents of Salinger's blog. It may be helpful if you want to gradually move from Python2 to Python3.

Use print in a Python2 lambda expression

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