问题：如何针对一个值测试多个变量？
我正在尝试制作一个将多个变量与一个整数进行比较并输出三个字母的字符串的函数。我想知道是否有一种方法可以将其转换为Python。所以说：
x = 0
y = 1
z = 3
mylist = []
if x or y or z == 0 :
mylist.append("c")
if x or y or z == 1 :
mylist.append("d")
if x or y or z == 2 :
mylist.append("e")
if x or y or z == 3 :
mylist.append("f")
这将返回以下列表：
["c", "d", "f"]
这样的事情可能吗？
I’m trying to make a function that will compare multiple variables to an integer and output a string of three letters. I was wondering if there was a way to translate this into Python. So say:
x = 0
y = 1
z = 3
mylist = []
if x or y or z == 0 :
mylist.append("c")
if x or y or z == 1 :
mylist.append("d")
if x or y or z == 2 :
mylist.append("e")
if x or y or z == 3 :
mylist.append("f")
which would return a list of:
["c", "d", "f"]
Is something like this possible?
回答 0
您误解了布尔表达式是如何工作的。它们不像英文句子那样工作，并且猜测您在这里对所有名称都使用相同的比较。您正在寻找：
if x == 1 or y == 1 or z == 1:
x并y以其他方式自行评估（False如果为0，则为True）。
您可以使用针对元组的容纳测试来缩短该时间：
if 1 in (x, y, z):
还是更好：
if 1 in {x, y, z}:
使用aset来利用固定成本的成员资格测试（in无论左侧操作数是多少，都花费固定的时间）。
使用时or，python会将运算符的每一面视为单独的表达式。该表达式x or y == 1首先被视为的布尔测试x，然后如果为False，y == 1则测试该表达式。
这是由于运算符的优先级。的or操作者具有较低的优先级比所述==测试，所以后者被评估第一。
但是，即使不是这种情况，并且x or y or z == 1实际上该表达式被解释为(x or y or z) == 1，该表达式仍不会执行您期望的操作。
x or y or z会求值为第一个“真实的”参数，例如，不是False，数字0或为空（有关布尔值在Python上下文中认为Python为假的详细信息，请参见布尔值表达式）。
因此，对于values x = 2; y = 1; z = 0，x or y or z将解析为2，因为那是参数中的第一个真值。然后2 == 1是False，即使y == 1会True。
反之亦然；针对单个变量测试多个值；x == 1 or 2 or 3会因为相同的原因而失败。使用x == 1 or x == 2 or x == 3或x in {1, 2, 3}。
You misunderstand how boolean expressions work; they don’t work like an English sentence and guess that you are talking about the same comparison for all names here. You are looking for:
if x == 1 or y == 1 or z == 1:
x and y are otherwise evaluated on their own (False if 0, True otherwise).
You can shorten that using a containment test against a tuple:
if 1 in (x, y, z):
or better still:
if 1 in {x, y, z}:
using a set to take advantage of the constant-cost membership test (in takes a fixed amount of time whatever the left-hand operand is).
When you use or, python sees each side of the operator as separate expressions. The expression x or y == 1 is treated as first a boolean test for x, then if that is False, the expression y == 1 is tested.
This is due to operator precedence. The or operator has a lower precedence than the == test, so the latter is evaluated first.
However, even if this were not the case, and the expression x or y or z == 1 was actually interpreted as (x or y or z) == 1 instead, this would still not do what you expect it to do.
x or y or z would evaluate to the first argument that is ‘truthy’, e.g. not False, numeric 0 or empty (see boolean expressions for details on what Python considers false in a boolean context).
So for the values x = 2; y = 1; z = 0, x or y or z would resolve to 2, because that is the first true-like value in the arguments. Then 2 == 1 would be False, even though y == 1 would be True.
The same would apply to the inverse; testing multiple values against a single variable; x == 1 or 2 or 3 would fail for the same reasons. Use x == 1 or x == 2 or x == 3 or x in {1, 2, 3}.
回答 1
使用以下字典结构可以更轻松地解决您的问题：
x = 0
y = 1
z = 3
d = {0: 'c', 1:'d', 2:'e', 3:'f'}
mylist = [d[k] for k in [x, y, z]]
Your problem is more easily addressed with a dictionary structure like:
x = 0
y = 1
z = 3
d = {0: 'c', 1:'d', 2:'e', 3:'f'}
mylist = [d[k] for k in [x, y, z]]
回答 2
正如Martijn Pieters所说，正确且最快的格式是：
if 1 in {x, y, z}:
根据他的建议，您现在将具有单独的if语句，以便Python可以读取每个语句，无论前者是True还是False。如：
if 0 in {x, y, z}:
mylist.append("c")
if 1 in {x, y, z}:
mylist.append("d")
if 2 in {x, y, z}:
mylist.append("e")
...
这将起作用，但是如果您习惯使用字典（请参阅我在那做的事情），则可以通过制作一个初始字典来将数字映射到所需的字母，然后使用for循环来进行清理：
num_to_letters = {0: "c", 1: "d", 2: "e", 3: "f"}
for number in num_to_letters:
if number in {x, y, z}:
mylist.append(num_to_letters[number])
As stated by Martijn Pieters, the correct, and fastest, format is:
if 1 in {x, y, z}:
Using his advice you would now have separate if-statements so that Python will read each statement whether the former were True or False. Such as:
if 0 in {x, y, z}:
mylist.append("c")
if 1 in {x, y, z}:
mylist.append("d")
if 2 in {x, y, z}:
mylist.append("e")
...
This will work, but if you are comfortable using dictionaries (see what I did there), you can clean this up by making an initial dictionary mapping the numbers to the letters you want, then just using a for-loop:
num_to_letters = {0: "c", 1: "d", 2: "e", 3: "f"}
for number in num_to_letters:
if number in {x, y, z}:
mylist.append(num_to_letters[number])
回答 3
直接的写法x or y or z == 0是
if any(map((lambda value: value == 0), (x,y,z))):
pass # write your logic.
但我不认为，您喜欢它。:)这种方式很难看。
另一种方法（更好）是：
0 in (x, y, z)
BTW很多ifs可以写成这样的东西
my_cases = {
0: Mylist.append("c"),
1: Mylist.append("d")
# ..
}
for key in my_cases:
if key in (x,y,z):
my_cases[key]()
break
The direct way to write x or y or z == 0 is
if any(map((lambda value: value == 0), (x,y,z))):
pass # write your logic.
But I dont think, you like it. :)
And this way is ugly.
The other way (a better) is:
0 in (x, y, z)
BTW lots of ifs could be written as something like this
my_cases = {
0: Mylist.append("c"),
1: Mylist.append("d")
# ..
}
for key in my_cases:
if key in (x,y,z):
my_cases[key]()
break
回答 4
如果您非常懒惰，可以将值放在数组中。如
list = []
list.append(x)
list.append(y)
list.append(z)
nums = [add numbers here]
letters = [add corresponding letters here]
for index in range(len(nums)):
for obj in list:
if obj == num[index]:
MyList.append(letters[index])
break
您也可以将数字和字母放入字典中并执行此操作，但这可能比if语句简单得多。那就是你变得更加懒惰的原因:)
还有一件事，你的
if x or y or z == 0:
会编译，但不会以您希望的方式编译。当您简单地将变量放在if语句中时（示例）
if b
程序将检查变量是否不为null。编写以上语句的另一种方法（更有意义）是
if bool(b)
Bool是python中的一个内置函数，它基本上执行验证布尔语句的命令（如果您不知道这是什么，那么它就是您现在要在if语句中创建的内容：）
我发现的另一种懒惰方式是：
if any([x==0, y==0, z==0])
If you ARE very very lazy, you can put the values inside an array. Such as
list = []
list.append(x)
list.append(y)
list.append(z)
nums = [add numbers here]
letters = [add corresponding letters here]
for index in range(len(nums)):
for obj in list:
if obj == num[index]:
MyList.append(letters[index])
break
You can also put the numbers and letters in a dictionary and do it, but this is probably a LOT more complicated than simply if statements. That’s what you get for trying to be extra lazy :)
One more thing, your
if x or y or z == 0:
will compile, but not in the way you want it to. When you simply put a variable in an if statement (example)
if b
the program will check if the variable is not null. Another way to write the above statement (which makes more sense) is
if bool(b)
Bool is an inbuilt function in python which basically does the command of verifying a boolean statement (If you don’t know what that is, it is what you are trying to make in your if statement right now :))
Another lazy way I found is :
if any([x==0, y==0, z==0])
回答 5
要检查一组变量中是否包含值，可以使用内置模块 itertools和operator。
例如：
进口：
from itertools import repeat
from operator import contains
声明变量：
x = 0
y = 1
z = 3
创建值的映射（以您要检查的顺序）：
check_values = (0, 1, 3)
使用itertools允许的变量重复：
check_vars = repeat((x, y, z))
最后，使用该map函数创建一个迭代器：
checker = map(contains, check_vars, check_values)
然后，在检查值时（按原始顺序），请使用next()：
if next(checker) # Checks for 0
# Do something
pass
elif next(checker) # Checks for 1
# Do something
pass
等等…
这是一个优势，lambda x: x in (variables)因为operator它是内置模块，并且比必须使用lambda它来创建自定义就地功能的模块更快，更高效。
检查列表中是否存在非零（或False）值的另一个选项：
not (x and y and z)
当量：
not all((x, y, z))
To check if a value is contained within a set of variables you can use the inbuilt modules itertools and operator.
For example:
Imports:
from itertools import repeat
from operator import contains
Declare variables:
x = 0
y = 1
z = 3
Create mapping of values (in the order you want to check):
check_values = (0, 1, 3)
Use itertools to allow repetition of the variables:
check_vars = repeat((x, y, z))
Finally, use the map function to create an iterator:
checker = map(contains, check_vars, check_values)
Then, when checking for the values (in the original order), use next():
if next(checker) # Checks for 0
# Do something
pass
elif next(checker) # Checks for 1
# Do something
pass
etc…
This has an advantage over the lambda x: x in (variables) because operator is an inbuilt module and is faster and more efficient than using lambda which has to create a custom in-place function.
Another option for checking if there is a non-zero (or False) value in a list:
not (x and y and z)
Equivalent:
not all((x, y, z))
回答 6
设置是这里的好方法，因为它对变量进行排序，这似乎是您的目标。{z,y,x}是{0,1,3}参数的任何命令。
>>> ["cdef"[i] for i in {z,x,y}]
['c', 'd', 'f']
这样，整个解决方案就是O（n）。
Set is the good approach here, because it orders the variables, what seems to be your goal here. {z,y,x} is {0,1,3} whatever the order of the parameters.
>>> ["cdef"[i] for i in {z,x,y}]
['c', 'd', 'f']
This way, the whole solution is O(n).
回答 7
这里提供的所有出色答案都集中在原始海报的特定要求上，并集中在if 1 in {x,y,z}Martijn Pieters提出的解决方案上。
他们忽略了这个问题的更广泛含义：
如何针对多个值测试一个变量？
如果使用例如字符串，则提供的解决方案不适用于部分匹配：
测试字符串“ Wild”是否为多个值
>>> x = "Wild things"
>>> y = "throttle it back"
>>> z = "in the beginning"
>>> if "Wild" in {x, y, z}: print (True)
...
要么
>>> x = "Wild things"
>>> y = "throttle it back"
>>> z = "in the beginning"
>>> if "Wild" in [x, y, z]: print (True)
...
在这种情况下，最容易转换为字符串
>>> [x, y, z]
['Wild things', 'throttle it back', 'in the beginning']
>>> {x, y, z}
{'in the beginning', 'throttle it back', 'Wild things'}
>>>
>>> if "Wild" in str([x, y, z]): print (True)
...
True
>>> if "Wild" in str({x, y, z}): print (True)
...
True
但是，应注意，如所述@codeforester，使用此方法会丢失单词边界，例如：
>>> x=['Wild things', 'throttle it back', 'in the beginning']
>>> if "rot" in str(x): print(True)
...
True
这3个字母rot确实存在于列表中，但不是单个单词。测试“腐烂”将失败，但是如果列表项之一“在地狱腐烂”，那也将失败。
结果是，如果使用此方法，请注意您的搜索条件，并注意它确实有此限制。
All of the excellent answers provided here concentrate on the specific requirement of the original poster and concentrate on the if 1 in {x,y,z} solution put forward by Martijn Pieters.
What they ignore is the broader implication of the question:
How do I test one variable against multiple values?
The solution provided will not work for partial hits if using strings for example:
Test if the string “Wild” is in multiple values
>>> x = "Wild things"
>>> y = "throttle it back"
>>> z = "in the beginning"
>>> if "Wild" in {x, y, z}: print (True)
...
or
>>> x = "Wild things"
>>> y = "throttle it back"
>>> z = "in the beginning"
>>> if "Wild" in [x, y, z]: print (True)
...
for this scenario it’s easiest to convert to a string
>>> [x, y, z]
['Wild things', 'throttle it back', 'in the beginning']
>>> {x, y, z}
{'in the beginning', 'throttle it back', 'Wild things'}
>>>
>>> if "Wild" in str([x, y, z]): print (True)
...
True
>>> if "Wild" in str({x, y, z}): print (True)
...
True
It should be noted however, as mentioned by @codeforester, that word boundries are lost with this method, as in:
>>> x=['Wild things', 'throttle it back', 'in the beginning']
>>> if "rot" in str(x): print(True)
...
True
the 3 letters rot do exist in combination in the list but not as an individual word. Testing for ” rot ” would fail but if one of the list items were “rot in hell”, that would fail as well.
The upshot being, be careful with your search criteria if using this method and be aware that it does have this limitation.
回答 8
我认为这样会更好地处理它：
my_dict = {0: "c", 1: "d", 2: "e", 3: "f"}
def validate(x, y, z):
for ele in [x, y, z]:
if ele in my_dict.keys():
return my_dict[ele]
输出：
print validate(0, 8, 9)
c
print validate(9, 8, 9)
None
print validate(9, 8, 2)
e
I think this will handle it better:
my_dict = {0: "c", 1: "d", 2: "e", 3: "f"}
def validate(x, y, z):
for ele in [x, y, z]:
if ele in my_dict.keys():
return my_dict[ele]
Output:
print validate(0, 8, 9)
c
print validate(9, 8, 9)
None
print validate(9, 8, 2)
e
回答 9
如果要使用if，则以下else语句是另一种解决方案：
myList = []
aList = [0, 1, 3]
for l in aList:
if l==0: myList.append('c')
elif l==1: myList.append('d')
elif l==2: myList.append('e')
elif l==3: myList.append('f')
print(myList)
If you want to use if, else statements following is another solution:
myList = []
aList = [0, 1, 3]
for l in aList:
if l==0: myList.append('c')
elif l==1: myList.append('d')
elif l==2: myList.append('e')
elif l==3: myList.append('f')
print(myList)
回答 10
d = {0:'c', 1:'d', 2:'e', 3: 'f'}
x, y, z = (0, 1, 3)
print [v for (k,v) in d.items() if x==k or y==k or z==k]
d = {0:'c', 1:'d', 2:'e', 3: 'f'}
x, y, z = (0, 1, 3)
print [v for (k,v) in d.items() if x==k or y==k or z==k]
回答 11
此代码可能会有所帮助
L ={x, y, z}
T= ((0,"c"),(1,"d"),(2,"e"),(3,"f"),)
List2=[]
for t in T :
if t[0] in L :
List2.append(t[1])
break;
This code may be helpful
L ={x, y, z}
T= ((0,"c"),(1,"d"),(2,"e"),(3,"f"),)
List2=[]
for t in T :
if t[0] in L :
List2.append(t[1])
break;
回答 12
您可以尝试以下显示的方法。在这种方法中，您可以自由指定/输入要输入的变量数。
mydict = {0:"c", 1:"d", 2:"e", 3:"f"}
mylist= []
num_var = int(raw_input("How many variables? ")) #Enter 3 when asked for input.
for i in range(num_var):
''' Enter 0 as first input, 1 as second input and 3 as third input.'''
globals()['var'+str('i').zfill(3)] = int(raw_input("Enter an integer between 0 and 3 "))
mylist += mydict[globals()['var'+str('i').zfill(3)]]
print mylist
>>> ['c', 'd', 'f']
You can try the method shown below. In this method, you will have the freedom to specify/input the number of variables that you wish to enter.
mydict = {0:"c", 1:"d", 2:"e", 3:"f"}
mylist= []
num_var = int(raw_input("How many variables? ")) #Enter 3 when asked for input.
for i in range(num_var):
''' Enter 0 as first input, 1 as second input and 3 as third input.'''
globals()['var'+str('i').zfill(3)] = int(raw_input("Enter an integer between 0 and 3 "))
mylist += mydict[globals()['var'+str('i').zfill(3)]]
print mylist
>>> ['c', 'd', 'f']
回答 13
一线解决方案：
mylist = [{0: 'c', 1: 'd', 2: 'e', 3: 'f'}[i] for i in [0, 1, 2, 3] if i in (x, y, z)]
要么：
mylist = ['cdef'[i] for i in range(4) if i in (x, y, z)]
One line solution:
mylist = [{0: 'c', 1: 'd', 2: 'e', 3: 'f'}[i] for i in [0, 1, 2, 3] if i in (x, y, z)]
Or:
mylist = ['cdef'[i] for i in range(4) if i in (x, y, z)]
回答 14
也许您需要直接的公式来设置输出位。
x=0 or y=0 or z=0 is equivalent to x*y*z = 0
x=1 or y=1 or z=1 is equivalent to (x-1)*(y-1)*(z-1)=0
x=2 or y=2 or z=2 is equivalent to (x-2)*(y-2)*(z-2)=0
让我们映射到位： 'c':1 'd':0xb10 'e':0xb100 'f':0xb1000
isc（是’c’）的关系：
if xyz=0 then isc=1 else isc=0
如果公式https://youtu.be/KAdKCgBGK0k?list=PLnI9xbPdZUAmUL8htSl6vToPQRRN3hhFp&t=315使用数学
[C]： (xyz=0 and isc=1) or (((xyz=0 and isc=1) or (isc=0)) and (isc=0))
[d]： ((x-1)(y-1)(z-1)=0 and isc=2) or (((xyz=0 and isd=2) or (isc=0)) and (isc=0))
…
通过以下逻辑连接这些公式：
你会有一个总和表示总和，你总和公式
那么sum＆1是c，sum＆2是d，sum＆4是e，sum＆5是f
之后，您可以形成预定义的数组，其中字符串元素的索引将对应于就绪字符串。
array[sum] 给你字符串。
Maybe you need direct formula for output bits set.
x=0 or y=0 or z=0 is equivalent to x*y*z = 0
x=1 or y=1 or z=1 is equivalent to (x-1)*(y-1)*(z-1)=0
x=2 or y=2 or z=2 is equivalent to (x-2)*(y-2)*(z-2)=0
Let’s map to bits: 'c':1 'd':0xb10 'e':0xb100 'f':0xb1000
Relation of isc (is ‘c’):
if xyz=0 then isc=1 else isc=0
Use math if formula https://youtu.be/KAdKCgBGK0k?list=PLnI9xbPdZUAmUL8htSl6vToPQRRN3hhFp&t=315
[c]: (xyz=0 and isc=1) or (((xyz=0 and isc=1) or (isc=0)) and (isc=0))
[d]: ((x-1)(y-1)(z-1)=0 and isc=2) or (((xyz=0 and isd=2) or (isc=0)) and (isc=0))
…
Connect these formulas by following logic:
- logic
and is the sum of squares of equations
- logic
or is the product of equations
and you’ll have a total equation
express sum and you have total formula of sum
then sum&1 is c, sum&2 is d, sum&4 is e, sum&5 is f
After this you may form predefined array where index of string elements would correspond to ready string.
array[sum] gives you the string.
回答 15
它可以很容易地完成
for value in [var1,var2,var3]:
li.append("targetValue")
It can be done easily as
for value in [var1,var2,var3]:
li.append("targetValue")
回答 16
用Python表示伪代码的最简便的方法是：
x = 0
y = 1
z = 3
mylist = []
if any(v == 0 for v in (x, y, z)):
mylist.append("c")
if any(v == 1 for v in (x, y, z)):
mylist.append("d")
if any(v == 2 for v in (x, y, z)):
mylist.append("e")
if any(v == 3 for v in (x, y, z)):
mylist.append("f")
The most mnemonic way of representing your pseudo-code in Python would be:
x = 0
y = 1
z = 3
mylist = []
if any(v == 0 for v in (x, y, z)):
mylist.append("c")
if any(v == 1 for v in (x, y, z)):
mylist.append("d")
if any(v == 2 for v in (x, y, z)):
mylist.append("e")
if any(v == 3 for v in (x, y, z)):
mylist.append("f")
回答 17
要使用一个值测试多个变量： if 1 in {a,b,c}:
要使用一个变量测试多个值： if a in {1, 2, 3}:
To test multiple variables with one single value: if 1 in {a,b,c}:
To test multiple values with one variable: if a in {1, 2, 3}:
回答 18
看起来您正在构建某种凯撒密码。
更为通用的方法是：
input_values = (0, 1, 3)
origo = ord('c')
[chr(val + origo) for val in inputs]
输出
['c', 'd', 'f']
不确定这是否是代码的理想副作用，但是输出的顺序将始终排序。
如果这是您想要的，可以将最后一行更改为：
sorted([chr(val + origo) for val in inputs])
Looks like you’re building some kind of Caesar cipher.
A much more generalized approach is this:
input_values = (0, 1, 3)
origo = ord('c')
[chr(val + origo) for val in inputs]
outputs
['c', 'd', 'f']
Not sure if it’s a desired side effect of your code, but the order of your output will always be sorted.
If this is what you want, the final line can be changed to:
sorted([chr(val + origo) for val in inputs])
回答 19
您可以使用字典：
x = 0
y = 1
z = 3
list=[]
dict = {0: 'c', 1: 'd', 2: 'e', 3: 'f'}
if x in dict:
list.append(dict[x])
else:
pass
if y in dict:
list.append(dict[y])
else:
pass
if z in dict:
list.append(dict[z])
else:
pass
print list
You can use dictionary :
x = 0
y = 1
z = 3
list=[]
dict = {0: 'c', 1: 'd', 2: 'e', 3: 'f'}
if x in dict:
list.append(dict[x])
else:
pass
if y in dict:
list.append(dict[y])
else:
pass
if z in dict:
list.append(dict[z])
else:
pass
print list
回答 20
如果没有字典，请尝试以下解决方案：
x, y, z = 0, 1, 3
offset = ord('c')
[chr(i + offset) for i in (x,y,z)]
并给出：
['c', 'd', 'f']
Without dict, try this solution:
x, y, z = 0, 1, 3
offset = ord('c')
[chr(i + offset) for i in (x,y,z)]
and gives:
['c', 'd', 'f']
回答 21
这将为您提供帮助。
def test_fun(val):
x = 0
y = 1
z = 2
myList = []
if val in (x, y, z) and val == 0:
myList.append("C")
if val in (x, y, z) and val == 1:
myList.append("D")
if val in (x, y, z) and val == 2:
myList.append("E")
test_fun(2);
This will help you.
def test_fun(val):
x = 0
y = 1
z = 2
myList = []
if val in (x, y, z) and val == 0:
myList.append("C")
if val in (x, y, z) and val == 1:
myList.append("D")
if val in (x, y, z) and val == 2:
myList.append("E")
test_fun(2);
回答 22
你可以团结起来
x = 0
y = 1
z = 3
在一个变量中。
In [1]: xyz = (0,1,3,)
In [2]: mylist = []
将我们的条件更改为：
In [3]: if 0 in xyz:
...: mylist.append("c")
...: if 1 in xyz:
...: mylist.append("d")
...: if 2 in xyz:
...: mylist.append("e")
...: if 3 in xyz:
...: mylist.append("f")
输出：
In [21]: mylist
Out[21]: ['c', 'd', 'f']
You can unite this
x = 0
y = 1
z = 3
in one variable.
In [1]: xyz = (0,1,3,)
In [2]: mylist = []
Change our conditions as:
In [3]: if 0 in xyz:
...: mylist.append("c")
...: if 1 in xyz:
...: mylist.append("d")
...: if 2 in xyz:
...: mylist.append("e")
...: if 3 in xyz:
...: mylist.append("f")
Output:
In [21]: mylist
Out[21]: ['c', 'd', 'f']
回答 23
问题
同时测试多个值的模式
>>> 2 in {1, 2, 3}
True
>>> 5 in {1, 2, 3}
False
具有很高的可读性，并且可以在许多情况下工作，但有一个陷阱：
>>> 0 in {True, False}
True
但是我们想要
>>> (0 is True) or (0 is False)
False
解
先前表达式的一种概括是基于ytpillai的答案：
>>> any([0 is True, 0 is False])
False
可以写成
>>> any(0 is item for item in (True, False))
False
虽然此表达式返回正确的结果，但它不如第一个表达式可读：
Problem
While the pattern for testing multiple values
>>> 2 in {1, 2, 3}
True
>>> 5 in {1, 2, 3}
False
is very readable and is working in many situation, there is one pitfall:
>>> 0 in {True, False}
True
But we want to have
>>> (0 is True) or (0 is False)
False
Solution
One generalization of the previous expression is based on the answer from ytpillai:
>>> any([0 is True, 0 is False])
False
which can be written as
>>> any(0 is item for item in (True, False))
False
While this expression returns the right result it is not as readable as the first expression :-(
