Python 在 Ruby 中是否有像 ||= 这样的“或等于”函数？
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Does Python have an "or equals" function like ||= in Ruby?
提问by Sean W.
If not, what is the best way to do this?
如果不是，那么最好的方法是什么？
Right now I'm doing (for a django project):
现在我正在做（对于 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:
但它很尴尬。在 Ruby 中，它将是：
request.session['thing_for_purpose'] ||= 5
which is much nicer.
这是更好的。
采纳答案by Joseph Sheedy
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
接受的答案适用于 dicts，但标题寻求与 Ruby 的 ||= 运算符的一般等效项。在 Python 中执行 ||= 之类的常用方法是
x = x or new_value
回答by Jon-Eric
dicthas setdefault().
dict有setdefault().
So if request.sessionis a dict:
所以如果request.session是dict：
request.session.setdefault('thing_for_purpose', 5)
回答by Brian Hicks
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:
如果您在中间件或其他东西中设置默认值是有意义的，但如果您需要在一个请求的上下文中使用默认值：
request.session.get('thing_for_purpose', 5) # gets a default
bonus: here's how to really do an ||=in Python.
奖励：这是||=在 Python 中真正执行操作的方法。
def test_function(self, d=None):
'a simple test function'
d = d or {}
# ... do things with d and return ...
回答by curtissv
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.
回答by Apollys supports Monica
Precise answer: No.Python does not have a single built-in operator opthat can translate x = x or yinto x op y.
准确答案：不可以。Python 没有一个内置运算符op可以转换x = x or y为x op y.
But, it almost does.The bitwise or-equals operator (|=) willfunction 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，几乎可以。如果两个操作数都被视为布尔值， 按位或等于运算符 ( |=)将按上述方式运行，但需要注意。（有什么警告？答案当然在下面。）
First, the basic demonstration of functionality:
一、功能基本演示：
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 Trueiff there were elements deleted:
警告是由于 python 不是严格类型的，因此即使在表达式中将值视为布尔值，如果将它们提供给按位运算符，它们也不会短路。例如，假设我们有一个布尔函数，它清除列表并True在删除元素时返回：
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行为如下：
x = True
lst = [1, 2, 3]
x = x or my_clear_list(lst)
print(x, lst)
Output: True [1, 2, 3]
However, switching the orto a bitwise or (|) removes the short-circuit, so the function my_clear_listexecutes.
但是，将 切换or为按位或 ( |) 会消除短路，因此函数会my_clear_list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).
以上，x = x | my_clear_list(lst)相当于x |= my_clear_list(lst)。
