[英]Can someone explain the behaviour of `<<-` assignament operator in R?
After some hours I found a bug in my code because an unexpected behaviour of the <<-
assinament operator in r. 几个小时后,我在代码中发现了一个错误,因为r中的
<<-
assinament运算符的意外行为。 I have read the documentation and browsed on internet but I still do not understand the behaviour of the operator. 我已经阅读了文档并浏览了Internet,但是我仍然不了解操作员的行为。
Look these two functions: 查看以下两个功能:
# Define a function a_counter
a_counter <- function(){
i <<- i + 1
print(i)
}
> i <- 0
> a_counter()
[1] 1
> print(i)
[1] 1
> a_counter()
[1] 2
> print(i)
[1] 2
# Define a function not_a_counter
not_a_counter <- function(){
i <- 0
i <<- i + 1
print(i)
}
> i <- 0
> not_a_counter()
[1] 0
> print(i)
[1] 1
> not_a_counter()
[1] 0
> print(i)
[1] 1
The first chunk of code runs as I expected, the variable i
in both environments (function env. and global env.) are increased in each function call. 第一部分代码按我的预期运行,在每个环境调用(函数环境和全局环境)中的变量
i
都增加了。
The second chunk of code is absolutely unexpected for me. 第二段代码对我来说绝对是意外的。 The
i <<- i + 1
does not assign the value to the i
located in the function environment, but it does so in the i
located in the global environment. 在
i <<- i + 1
不分配值到i
所在的功能环境,但在这样做, i
位于全球环境。 I expected both enviroments to be updated. 我希望两个环境都可以更新。
In your a_counter
, there is only one value of i
. 在您的
a_counter
,只有i
一个值。 Not two. 不是两个。 When a "free variable" is found in a function, it's looked up in the environment in which the function was defined.
当在函数中找到“自由变量”时,将在定义该函数的环境中对其进行查找。 So when you call
i
in that function, it goes up to the global environment to find the value. 因此,当您在该函数中调用
i
时,它将取决于全局环境以查找值。 Then when you do <<-
, assignment doesn't happen in the function environment at all. 然后,当您执行
<<-
,函数环境中根本不会发生分配。 <<-
always starts looking in the parent environment. <<-
总是开始在父环境中查找。 If you look at 如果你看
counter_vars <- function() {
a <- 4
i <<- i + 1
ls()
}
counter_vars()
# [1] "a"
you'll see that the only variable inside the function environment/closure is the a
variable. 您会看到函数环境/闭包内部的唯一变量是
a
变量。 The i
variable doesn't exist there. i
变量在那里不存在。 So with the original function all the i
s come from global scope 因此,与原来的功能全部
i
š来自全球范围内
a_counter <- function(){
i <<- i + 1 # global i = global i + 1
print(i) # global i still (no local variable has been created)
}
So the behavior you see in not_a_counter
should be expected because <<-
will not change values in current environment. 所以应该在
not_a_counter
看到您的行为,因为<<-
不会在当前环境中更改值。 It always start looking one environment "up". 它总是开始寻找一个“向上”的环境。 When you have
当你有
not_a_counter <- function(){
i <- 0 # local i (always 0)
i <<- i + 1 # global i == local i + 1 (always 0+1)
print(i) # local i (always still 0, local value not changed)
}
The i
variable is no longer "free" once you define it in the function. 在函数中定义后,
i
变量不再“空闲”。 So i <- 0
creates a local variable, in the i <<- i + 1
part, the right hand side i+1
uses that local variable and assigns to the i
in the parent environment. 所以
i <- 0
创建一个局部变量,在i <<- i + 1
一部分,右边i+1
使用该本地变量并分配给i
在父环境。
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