Python 遍历数组
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Iterating through array
提问by Swift
I have an array of bools and now I want to swap those entries for numbers.
我有一组布尔值，现在我想将这些条目交换为数字。
False => 0
True => 1
I have written two different pieces of code and I would like to know, which one is better and why. This is not so much about actually solving the problem, as about learning.
我写了两段不同的代码，我想知道哪一段更好，为什么。这与其说是关于实际解决问题，不如说是关于学习。
arr = [[True,False],[False,True],[True,True]]
for i,row in enumerate(arr):
for j,entry in enumerate(row):
if entry:
arr[i][j] = 1
else:
arr[i][j] = 0
print(arr)
And the second approach:
第二种方法：
arr = [[True,False],[False,True],[True,True]]
for i in range(len(arr)):
for j in range(len(arr[i])):
if arr[i][j]:
arr[i][j] = 1
else:
arr[i][j] = 0
print(arr)
I read that there are ways to do this with importing itertoolsor similar. I am really not a fan of importing things if it can be done with “on-board tools”, but should I rather be using them for this problem?
我读到有一些方法可以通过导入itertools或类似方法来做到这一点。如果可以使用“板载工具”完成，我真的不喜欢导入东西，但我应该使用它们来解决这个问题吗？
采纳答案by John1024
Let's define your array:
让我们定义你的数组：
>>> arr = [[True,False],[False,True],[True,True]]
Now, let's convert the booleans to integer:
现在，让我们将布尔值转换为整数：
>>> [[int(i) for i in row] for row in arr]
[[1, 0], [0, 1], [1, 1]]
Alternatively, if we want to be more flexible about what gets substituted in, we can use a ternary statement:
或者，如果我们想要更灵活地替换什么，我们可以使用三元语句：
>>> [[1 if i else 0 for i in row] for row in arr]
[[1, 0], [0, 1], [1, 1]]
回答by thebjorn
If you want to stay with a for-loop (e.g. because you want to mutate the existing array instead of creating a new one), you should simplify the code.
如果您想继续使用 for 循环（例如，因为您想改变现有数组而不是创建新数组），您应该简化代码。
I would first simplify the outer loop by removing the indexing (there is no need for it since it's even easier to modify a row than a nested array):
我将首先通过删除索引来简化外部循环（不需要它，因为修改行比修改嵌套数组更容易）：
for row in arr:
for j, entry in enumerate(row):
if entry:
row[j] = 1
else:
row[j] = 0
these kinds of simple if statement can often be simplified by using an if expression:
这些简单的 if 语句通常可以通过使用 if 表达式来简化：
row[j] = 1 if entry else 0
but in this case we can do even better. boolis a subclass of int(ie. all bool's are int's), and Trueand Falseare defined to be exactly 1and 0respectively -- if you scroll down to the specification section of PEP 285 (https://www.python.org/dev/peps/pep-0285/) you'll see that that equivalence is not accidental but very much by design.
但在这种情况下，我们可以做得更好。bool是int（即所有bool的都是int）的子类，并且True和False被定义为完全1和0分别——如果你向下滚动到 PEP 285 的规范部分（https://www.python.org/dev/ peps/pep-0285/）你会看到这种等价性不是偶然的，而是设计的。
We can therefore use the intconstructor to grab the underlying integer values[*], since int(True) == 1and int(False) == 0, the if-expression can be simplified to:
因此，我们可以使用int构造函数来获取底层整数值 [*]，因为int(True) == 1和int(False) == 0，if 表达式可以简化为：
row[j] = int(entry)
[*] technically this is an explicit upcast to a base class, and not a conversion constructor..
[*] 从技术上讲，这是对基类的显式向上转换，而不是转换构造函数..
The simplified code:
简化代码：
for row in arr:
for j, entry in enumerate(row):
row[j] = int(entry)
