If not, what is the best way to do this?
如果没有,最好的方法是什么?
Right now I'm doing (for a django project):
现在我正在做(一个django项目):
if not 'thing_for_purpose' in request.session:
request.session['thing_for_purpose'] = 5
but its pretty awkward. In Ruby it would be:
但是它非常尴尬。在Ruby中是:
request.session['thing_for_purpose'] ||= 5
which is much nicer.
这是好得多。
5 个解决方案
#1
137
The accepted answer is good for dicts, but the title seeks a general equivalent to Ruby's ||= operator. A common way to do something like ||= in Python is
公认的答案对dicts来说是好的,但是标题寻求的是与Ruby的||=操作符相对应的通用答案。在Python中执行||=之类操作的常见方法是
x = x or new_value
#2
14
dict
has setdefault()
.
dict setdefault()。
So if request.session
is a dict
:
如果请求。会话是一个东西:
request.session.setdefault('thing_for_purpose', 5)
#3
9
Setting a default makes sense if you're doing it in a middleware or something, but if you need a default value in the context of one request:
如果您是在中间件或其他方面进行设置,那么设置默认值是有意义的,但是如果您在一个请求的上下文中需要一个默认值:
request.session.get('thing_for_purpose', 5) # gets a default
bonus: here's how to really do an ||=
in Python.
好处:这里是如何在Python中真正做||=。
def test_function(self, d=None):
'a simple test function'
d = d or {}
# ... do things with d and return ...
#4
2
Precise answer: No. Python does not have a single built-in operator op
that can translate x = x or y
into x op y
.
精确的答案:没有。Python没有一个内置的操作符op,可以将x = x或y转换为x op y。
But, it almost does. The bitwise or-equals operator (|=
) will function as described above if both operands are being treated as booleans, with a caveat. (What's the caveat? Answer is below of course.)
但是,它几乎一样。如果两个操作数都被当作布尔值来处理,那么位运算符(|=)将像上面描述的那样发挥作用。(有什么警告?答案当然在下面。
First, the basic demonstration of functionality:
首先,功能的基本演示:
x = True
x
Out[141]: True
x |= True
x
Out[142]: True
x |= False
x
Out[143]: True
x &= False
x
Out[144]: False
x &= True
x
Out[145]: False
x |= False
x
Out[146]: False
x |= True
x
Out[147]: True
The caveat is due python not being strictly-typed, and thus even if the values are being treated as booleans in an expression they will not be short-circuited if given to a bitwise operator. For example, suppose we had a boolean function which clears a list and returns True
iff there were elements deleted:
需要注意的是,python并不是严格类型的,因此即使将值作为布尔值处理,也不会在给定一个位运算符时发生短路。例如,假设我们有一个布尔函数,它清除一个列表并返回True iff,有一些元素被删除:
def my_clear_list(lst):
if not lst:
return False
else:
del lst[:]
return True
Now we can see the short-circuited behavior as so:
现在我们可以看到短路的行为是这样的:
x = True
lst = [1, 2, 3]
x = x or my_clear_list(lst)
print(x, lst)
Output: True [1, 2, 3]
However, switching the or
to a bitwise or (|
) removes the short-circuit, so the function my_clear_list
executes.
但是,将or切换到位或(|)将删除短路,因此函数my_clear_list将执行。
x = True
lst = [1, 2, 3]
x = x | my_clear_list(lst)
print(x, lst)
Output: True []
Above, x = x | my_clear_list(lst)
is equivalent to x |= my_clear_list(lst)
.
上面,x = x | my_clear_list(lst)相当于x |= my_clear_list(lst)。
#5
0
In general, you can use dict[key] = dict.get(key, 0) + val
.
一般来说,您可以使用dict[key] = dict.get(key, 0) + val。
#1
137
The accepted answer is good for dicts, but the title seeks a general equivalent to Ruby's ||= operator. A common way to do something like ||= in Python is
公认的答案对dicts来说是好的,但是标题寻求的是与Ruby的||=操作符相对应的通用答案。在Python中执行||=之类操作的常见方法是
x = x or new_value
#2
14
dict
has setdefault()
.
dict setdefault()。
So if request.session
is a dict
:
如果请求。会话是一个东西:
request.session.setdefault('thing_for_purpose', 5)
#3
9
Setting a default makes sense if you're doing it in a middleware or something, but if you need a default value in the context of one request:
如果您是在中间件或其他方面进行设置,那么设置默认值是有意义的,但是如果您在一个请求的上下文中需要一个默认值:
request.session.get('thing_for_purpose', 5) # gets a default
bonus: here's how to really do an ||=
in Python.
好处:这里是如何在Python中真正做||=。
def test_function(self, d=None):
'a simple test function'
d = d or {}
# ... do things with d and return ...
#4
2
Precise answer: No. Python does not have a single built-in operator op
that can translate x = x or y
into x op y
.
精确的答案:没有。Python没有一个内置的操作符op,可以将x = x或y转换为x op y。
But, it almost does. The bitwise or-equals operator (|=
) will function as described above if both operands are being treated as booleans, with a caveat. (What's the caveat? Answer is below of course.)
但是,它几乎一样。如果两个操作数都被当作布尔值来处理,那么位运算符(|=)将像上面描述的那样发挥作用。(有什么警告?答案当然在下面。
First, the basic demonstration of functionality:
首先,功能的基本演示:
x = True
x
Out[141]: True
x |= True
x
Out[142]: True
x |= False
x
Out[143]: True
x &= False
x
Out[144]: False
x &= True
x
Out[145]: False
x |= False
x
Out[146]: False
x |= True
x
Out[147]: True
The caveat is due python not being strictly-typed, and thus even if the values are being treated as booleans in an expression they will not be short-circuited if given to a bitwise operator. For example, suppose we had a boolean function which clears a list and returns True
iff there were elements deleted:
需要注意的是,python并不是严格类型的,因此即使将值作为布尔值处理,也不会在给定一个位运算符时发生短路。例如,假设我们有一个布尔函数,它清除一个列表并返回True iff,有一些元素被删除:
def my_clear_list(lst):
if not lst:
return False
else:
del lst[:]
return True
Now we can see the short-circuited behavior as so:
现在我们可以看到短路的行为是这样的:
x = True
lst = [1, 2, 3]
x = x or my_clear_list(lst)
print(x, lst)
Output: True [1, 2, 3]
However, switching the or
to a bitwise or (|
) removes the short-circuit, so the function my_clear_list
executes.
但是,将or切换到位或(|)将删除短路,因此函数my_clear_list将执行。
x = True
lst = [1, 2, 3]
x = x | my_clear_list(lst)
print(x, lst)
Output: True []
Above, x = x | my_clear_list(lst)
is equivalent to x |= my_clear_list(lst)
.
上面,x = x | my_clear_list(lst)相当于x |= my_clear_list(lst)。
#5
0
In general, you can use dict[key] = dict.get(key, 0) + val
.
一般来说,您可以使用dict[key] = dict.get(key, 0) + val。