问题:如何在Python中将’false’转换为0并将’true’转换为1

有没有一种方法可以将true类型转换unicode为1并将false类型转换unicode为0(在Python中)?

例如: x == 'true' and type(x) == unicode

我想要 x = 1

PS:我不想使用ifelse

Is there a way to convert true of type unicode to 1 and false of type unicode to 0 (in Python)?

For example: x == 'true' and type(x) == unicode

I want x = 1

PS: I don’t want to use ifelse.


回答 0

使用int()一个布尔测试:

x = int(x == 'true')

int()将布尔值转换为10。请注意,任何等于的值'true'都将导致0返回。

Use int() on a boolean test:

x = int(x == 'true')

int() turns the boolean into 1 or 0. Note that any value not equal to 'true' will result in 0 being returned.


回答 1

如果B是布尔数组,则写

B = B*1

(一些代码golfy。)

If B is a Boolean array, write

B = B*1

(A bit code golfy.)


回答 2

您可以使用x.astype('uint8')where x是布尔数组。

You can use x.astype('uint8') where x is your Boolean array.


回答 3

这是您的问题的另一种解决方案:

def to_bool(s):
    return 1 - sum(map(ord, s)) % 2
    # return 1 - sum(s.encode('ascii')) % 2  # Alternative for Python 3

它的工作原理因为ASCII码的总和'true'就是448,这是偶数,而的ASCII码的总和'false'就是523这是奇怪的。


关于此解决方案的有趣之处在于,如果输入不是'true' or 之一,则其结果是非常随机的'false'。一半的时间会回来0,另一半1encode如果输入不是ASCII ,变体using 将引发编码错误(从而增加行为的不确定性)。


认真地说,我认为最易读,更快捷的解决方案是使用if

def to_bool(s):
    return 1 if s == 'true' else 0

查看一些微基准测试:

In [14]: def most_readable(s):
    ...:     return 1 if s == 'true' else 0

In [15]: def int_cast(s):
    ...:     return int(s == 'true')

In [16]: def str2bool(s):
    ...:     try:
    ...:         return ['false', 'true'].index(s)
    ...:     except (ValueError, AttributeError):
    ...:         raise ValueError()

In [17]: def str2bool2(s):
    ...:     try:
    ...:         return ('false', 'true').index(s)
    ...:     except (ValueError, AttributeError):
    ...:         raise ValueError()

In [18]: def to_bool(s):
    ...:     return 1 - sum(s.encode('ascii')) % 2

In [19]: %timeit most_readable('true')
10000000 loops, best of 3: 112 ns per loop

In [20]: %timeit most_readable('false')
10000000 loops, best of 3: 109 ns per loop

In [21]: %timeit int_cast('true')
1000000 loops, best of 3: 259 ns per loop

In [22]: %timeit int_cast('false')
1000000 loops, best of 3: 262 ns per loop

In [23]: %timeit str2bool('true')
1000000 loops, best of 3: 343 ns per loop

In [24]: %timeit str2bool('false')
1000000 loops, best of 3: 325 ns per loop

In [25]: %timeit str2bool2('true')
1000000 loops, best of 3: 295 ns per loop

In [26]: %timeit str2bool2('false')
1000000 loops, best of 3: 277 ns per loop

In [27]: %timeit to_bool('true')
1000000 loops, best of 3: 607 ns per loop

In [28]: %timeit to_bool('false')
1000000 loops, best of 3: 612 ns per loop

请注意该怎么if解决办法是至少 2.5倍速度所有其他解决方案。避免使用s 是没有意义的,if除非这是某种家庭作业(在这种情况下,您本来不应该首先问这个问题)。

Here’s a yet another solution to your problem:

def to_bool(s):
    return 1 - sum(map(ord, s)) % 2
    # return 1 - sum(s.encode('ascii')) % 2  # Alternative for Python 3

It works because the sum of the ASCII codes of 'true' is 448, which is even, while the sum of the ASCII codes of 'false' is 523 which is odd.


The funny thing about this solution is that its result is pretty random if the input is not one of 'true' or 'false'. Half of the time it will return 0, and the other half 1. The variant using encode will raise an encoding error if the input is not ASCII (thus increasing the undefined-ness of the behaviour).


Seriously, I believe the most readable, and faster, solution is to use an if:

def to_bool(s):
    return 1 if s == 'true' else 0

See some microbenchmarks:

In [14]: def most_readable(s):
    ...:     return 1 if s == 'true' else 0

In [15]: def int_cast(s):
    ...:     return int(s == 'true')

In [16]: def str2bool(s):
    ...:     try:
    ...:         return ['false', 'true'].index(s)
    ...:     except (ValueError, AttributeError):
    ...:         raise ValueError()

In [17]: def str2bool2(s):
    ...:     try:
    ...:         return ('false', 'true').index(s)
    ...:     except (ValueError, AttributeError):
    ...:         raise ValueError()

In [18]: def to_bool(s):
    ...:     return 1 - sum(s.encode('ascii')) % 2

In [19]: %timeit most_readable('true')
10000000 loops, best of 3: 112 ns per loop

In [20]: %timeit most_readable('false')
10000000 loops, best of 3: 109 ns per loop

In [21]: %timeit int_cast('true')
1000000 loops, best of 3: 259 ns per loop

In [22]: %timeit int_cast('false')
1000000 loops, best of 3: 262 ns per loop

In [23]: %timeit str2bool('true')
1000000 loops, best of 3: 343 ns per loop

In [24]: %timeit str2bool('false')
1000000 loops, best of 3: 325 ns per loop

In [25]: %timeit str2bool2('true')
1000000 loops, best of 3: 295 ns per loop

In [26]: %timeit str2bool2('false')
1000000 loops, best of 3: 277 ns per loop

In [27]: %timeit to_bool('true')
1000000 loops, best of 3: 607 ns per loop

In [28]: %timeit to_bool('false')
1000000 loops, best of 3: 612 ns per loop

Notice how the if solution is at least 2.5x times faster than all the other solutions. It does not make sense to put as a requirement to avoid using ifs except if this is some kind of homework (in which case you shouldn’t have asked this in the first place).


回答 4

如果您需要从本身不是布尔值的字符串进行通用转换,则最好编写类似于以下所示的例程。秉承鸭子打字的精神,我没有默默地传递错误,而是将其转换为适合当前情况的错误。

>>> def str2bool(st):
try:
    return ['false', 'true'].index(st.lower())
except (ValueError, AttributeError):
    raise ValueError('no Valid Conversion Possible')


>>> str2bool('garbaze')

Traceback (most recent call last):
  File "<pyshell#106>", line 1, in <module>
    str2bool('garbaze')
  File "<pyshell#105>", line 5, in str2bool
    raise TypeError('no Valid COnversion Possible')
TypeError: no Valid Conversion Possible
>>> str2bool('false')
0
>>> str2bool('True')
1

If you need a general purpose conversion from a string which per se is not a bool, you should better write a routine similar to the one depicted below. In keeping with the spirit of duck typing, I have not silently passed the error but converted it as appropriate for the current scenario.

>>> def str2bool(st):
try:
    return ['false', 'true'].index(st.lower())
except (ValueError, AttributeError):
    raise ValueError('no Valid Conversion Possible')


>>> str2bool('garbaze')

Traceback (most recent call last):
  File "<pyshell#106>", line 1, in <module>
    str2bool('garbaze')
  File "<pyshell#105>", line 5, in str2bool
    raise TypeError('no Valid COnversion Possible')
TypeError: no Valid Conversion Possible
>>> str2bool('false')
0
>>> str2bool('True')
1

回答 5

布尔到整数: x = (x == 'true') + 0

现在x包含1,x == 'true'否则为0。

注意:x == 'true'将返回bool,然后将其与0一起转换为具有值(如果bool值为True则为1,否则为0)的int类型。

bool to int: x = (x == 'true') + 0

Now the x contains 1 if x == 'true' else 0.

Note: x == 'true' will return bool which then will be typecasted to int having value (1 if bool value is True else 0) when added with 0.


回答 6

仅与此:

const a = true; const b = false;

console.log(+ a); // 1 console.log(+ b); // 0

only with this:

const a = true; const b = false;

console.log(+a);//1 console.log(+b);//0


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