如何检查循环查询中的所有项目并仅在全部匹配所需结果时继续? [英] How to check all items in loop query and continue only if all match the desired result?
问题描述
faces_all = list(range(1,25)) # Create list numbered 1-24
for x in itertools.combinations((faces_all),3): # check combinations in numbers 2-23
if faces_all[0] + sum(x) == 50: # if 1 + combination = 50 continue
for pair in itertools.combinations([faces_all[0],x],2): # HELP
if pair != 25:
这是我的代码.
我想要做的只是继续如果 faces_all[0] 和 x 的每个组合(应该是 4 个数字，2 个数字的 6 个组合)不等于 25.
What I want to be able to do is ONLY continue if every combination of faces_all[0] and x (should be 4 numbers, 6 combinations of 2 numbers) is NOT equal to 25.
如果即使是 2 个数字的一个组合 = 25，也应该迭代 x 的下一个组合...
If even one combination of 2 numbers = 25, the next combination for x should be iterated...
如果有道理.
我是 Python 新手，我以前唯一的经验是非结构化 BASIC，所以请保持温和.
I'm a python novice and my only previous experience is unstructured BASIC, so please be gentle.
已编辑
非常感谢@wovano.很明显，我有很多关于流量控制的知识.这是我根据@wovano 的回答编写的新代码...
Many thanks to @wovano. It is clear I have much to learn about flow control. Here is my new code based on @wovano's answer...
n = 0
faces_all = list(range(1,25))
for x in itertools.combinations((faces_all[2:24]),3):
if faces_all[0] + sum(x) == 50:
side1 = (faces_all[0], x[0], x[1], x[2])
for pair in combinations(side1, 2):
if sum(pair) == 25:
break
else:
n += 1
print (n,":",side1,":",sum(side1))
推荐答案
[faces_all[0],x] 部分创建了一个包含两个元素的列表.第一个元素是 faces_all[0](在这种情况下总是 1)，第二个元素是 x，它是 3 个数字本身的元组.在这种情况下，您想要的是一个包含 4 个数字的列表.您可以像这样创建它:[faces_all[0], x[0], x[1], x[2]].
The part [faces_all[0],x] creates a list of two elements. The first element is faces_all[0] (in this case always 1) and the second element is x, which is a tuple of 3 numbers itself. What you want in this case is a list of 4 numbers. You can create it like this: [faces_all[0], x[0], x[1], x[2]].
提示:如果在最后一个 if 语句之前添加 print(repr(pair))，您可以看到第二次调用 itertools.combinations() 的结果.
Tip: if you add print(repr(pair)) before the last if-statement you can see the results of the second call to itertools.combinations().
之前:
(1, (2, 23, 24))
(1, (3, 22, 24))
(1, (4, 21, 24))
...
之后:
(1, 2)
(1, 23)
(1, 24)
(2, 23)
(2, 24)
(23, 24)
(1, 3)
(1, 22)
(1, 24)
(3, 22)
(3, 24)
(22, 24)
(1, 4)
(1, 21)
(1, 24)
(4, 21)
(4, 24)
(21, 24)
...
更新
现在，一旦您有了正确的数字对，您就可以检查数字之和是否为 25，如果是这种情况，则中断 for 循环.(请注意，itertools.combinations() 返回一个生成器，这意味着如果您中断循环，则不会计算"下一对.)
Now, once you have the correct number pairs, you could check if the sum of the numbers is 25 and break the for-loop if this is the case. (Note that itertools.combinations() returns a generator, meaning that the next pairs are not 'calculated' if you break from the loop.)
完整的代码可能是这样的.请注意 for-else 语句，它在大多数语言中都不存在，在这种情况下很方便.当 for 循环正常"完成(即没有中断)时，会到达 else 语句.
The complete code could be something like this. Note the for-else statement, which does not exist in most languages, and is convenient in this case. The else statement is reached when the for-loop is completed 'normally' (i.e. without break).
import itertools
faces_all = list(range(1, 25)) # Create list numbered 1-24
for x in itertools.combinations((faces_all), 3): # check combinations in numbers 2-23
if faces_all[0] + sum(x) == 50: # if 1 + combination = 50 continue
four_numbers = (faces_all[0], x[0], x[1], x[2])
for pair in itertools.combinations(four_numbers, 2):
if sum(pair) == 25:
break
else:
print(repr(four_numbers))
以上代码的输出为:
(1, 4, 22, 23)
(1, 5, 21, 23)
(1, 6, 20, 23)
(1, 6, 21, 22)
(1, 7, 19, 23)
(1, 7, 20, 22)
(1, 8, 18, 23)
(1, 8, 19, 22)
(1, 8, 20, 21)
(1, 9, 17, 23)
(1, 9, 18, 22)
(1, 9, 19, 21)
(1, 10, 16, 23)
(1, 10, 17, 22)
(1, 10, 18, 21)
(1, 10, 19, 20)
(1, 11, 15, 23)
(1, 11, 16, 22)
(1, 11, 17, 21)
(1, 11, 18, 20)
(1, 12, 14, 23)
(1, 12, 15, 22)
(1, 12, 16, 21)
(1, 12, 17, 20)
(1, 12, 18, 19)
(1, 13, 14, 22)
(1, 13, 15, 21)
(1, 13, 16, 20)
(1, 13, 17, 19)
(1, 14, 15, 20)
(1, 14, 16, 19)
(1, 14, 17, 18)
(1, 15, 16, 18)
这是您要寻找的结果吗?
Is this the result you're looking for?
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