在python中循环一行
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for loop one line in python
提问by z-one ng
When I tried to use for statement in one line like below,
当我尝试在如下一行中使用 for 语句时，
def get_cubes(x):
ls=[]
ls.append(pow(item*3, 3)) for item in range(int((x-x%3)/3)+1)
return ls
there always an error:
总是有一个错误：
File "<ipython-input-47-8c391c3d568a>", line 3
ls.append(pow(item*3, 3)) for item in range(int((x-x%3)/3)+1)
^
SyntaxError: invalid syntax
But when I write it in regular for loop method, it works well. Can anyone tell me why?
但是当我用常规的 for 循环方法编写它时，它运行良好。谁能告诉我为什么？
回答by PabloTrinidad
?Solution
？解决方案
If you strictly want a one-liner, then this is the solution:
如果您严格想要单线，那么这就是解决方案：
get_cubes = lambda x: [pow(i, 3) for i in range(0, x+1, 3)]
But since clarity and readability should always be a priority in Python, it's better to write it as a function just like @daniel did:
但由于清晰性和可读性在 Python 中始终是优先考虑的事项，因此最好像 @daniel 那样将其编写为函数：
def get_cubes(x):
return [pow(i, 3) for i in range(0, x+1, 3)]
?Output
？输出
Using a for loop to test the function you get the following results:
使用 for 循环来测试该函数，您会得到以下结果：
for i in range(20):
print(i, ':', get_cubes(i))
0 : [0]
1 : [0]
2 : [0]
3 : [0, 27]
4 : [0, 27]
5 : [0, 27]
6 : [0, 27, 216]
7 : [0, 27, 216]
8 : [0, 27, 216]
9 : [0, 27, 216, 729]
10 : [0, 27, 216, 729]
11 : [0, 27, 216, 729]
12 : [0, 27, 216, 729, 1728]
13 : [0, 27, 216, 729, 1728]
14 : [0, 27, 216, 729, 1728]
15 : [0, 27, 216, 729, 1728, 3375]
16 : [0, 27, 216, 729, 1728, 3375]
17 : [0, 27, 216, 729, 1728, 3375]
18 : [0, 27, 216, 729, 1728, 3375, 5832]
19 : [0, 27, 216, 729, 1728, 3375, 5832]
Explanation
解释
For those who asked why @daniel's code work:
对于那些问为什么@daniel 的代码有效的人：
The original code does the following:
原始代码执行以下操作：
- Given an
x, iterate from0to the number of times3 divides xplus1. - For each x, multiply it by 3 and raise it to the power of 3
- Return the list containing the results of step 2
- 给定
x，从0到迭代次数3 divides xplus1。 - 对于每个 x，乘以 3 并将其提高到 3 的幂
- 返回包含第 2 步结果的列表
Step 1was originally written as
步骤 1最初写为
range(int((x - x%3) / 3) + 1)
which subtracts the reminder of a number when divided by 3and then divides that result by 3before adding 1. But the same result can be achieved by just getting the integer part of dividing xby 3and then adding 1which will look something like this:
它在被3除时减去一个数字的提醒，然后3在添加之前将该结果除以1。但同样的结果可以通过刚开划分的整数部分来实现x的3，然后加入1，这将是这个样子：
int(x / 3) + 1
Step 2multiplies each iteration by 3before raising it to the power of 3but that "multiplication"can be achieved as well by using ranges (just like @daniel did) using:
第 2 步将每次迭代乘以 ，3然后再将其提高到 的幂，3但也可以通过使用范围（就像@daniel 所做的那样）使用以下方法实现“乘法”：
range(0, x+1, 3)
which iterate from 0to x+1on steps of 3and yields the same results. I.e xbeing 10:
从0到x+1的步骤进行迭代3并产生相同的结果。即x是10：
range(0, 10 + 1, 3) | i*3 for i in range(int(x / 3) + 1)
============================================================
0 | 0*3 = 0
3 | 1*3 = 3
6 | 2*3 = 6
9 | 3*3 = 9
For the Step 3we just need to apply pow(x, 3)(or x ** 3) and we can fit everything in a single line using list comprehension, then we can fit that in a lambdafunction or inside the returnstatement of a function.
对于第 3 步，我们只需要应用pow(x, 3)(或x ** 3) 并且我们可以使用列表推导将所有内容放入一行中，然后我们可以将其放入lambda函数或函数的return语句中。
[pow(i, 3) for i in range(0, x+1, 3)]
回答by Daniel
The correct syntax for for-loops is
for 循环的正确语法是
def get_cubes(x):
ls = []
for item in range(int((x-x%3)/3)+1):
ls.append(pow(item*3, 3))
return ls
Perhaps, you are looking for list comprehension, which is a way to generate lists, when the loop body only appends to a list:
也许，您正在寻找列表理解，这是一种生成列表的方法，当循环体仅附加到列表时：
def get_cubes(x):
ls = [(item*3) ** 3 for item in range(int((x-x%3)/3)+1)]
return ls
or in short:
或简而言之：
def get_cubes(x):
return [i ** 3 for i in range(0, x+1, 3)]
回答by Sheldore
My answer follows (modifies) your approach (code) instead of using the list comprehension but in a rectified manner. Comparing it to your code, you will figure out where the error was. I added comments #to explain the changes
我的回答遵循（修改）您的方法（代码），而不是使用列表理解，而是以更正的方式。将它与您的代码进行比较，您将找出错误所在。我添加了注释#来解释更改
def get_cubes(x):
ls=[] # create an empty list outside the for loop once
for item in range(int((x-x%3)/3)+1): # for loop for iterating through elements
ls.append(pow(item*3, 3)) # append one item at a time
return ls
print (get_cubes(10))
> [0, 27, 216, 729]
