为什么排序列表检测在这种情况下不起作用？ [英] Why sorted list detection does not work in this situation?

问题描述

我的目标是让python代码允许检测列表是否已排序。

My goal is to have python code allowing to detect if a list is sorted or not.

我想了解为什么以下代码返回 True 而不是我的预期猜测 False

I would like to understand why the following code return True instead of my expected guess False

l = [1, 2, 3, 4, 1, 6, 7, 8, 7]
all(l[i] <= l[i+1] for i in xrange(len(l)-1)) # return "True"

注意：

• 我在 iPython 0.10中使用python 2.6.4


• 我使用非常大的列表所以我我宁愿避免
  类型的解决方案l == l.sort（）

• I'm using python 2.6.4 inside iPython 0.10
• I use very huge list so I'd prefer to avoid solution of type l == l.sort()

在理解这一点的方式中，我已经从以下两篇主要内容中读取（和测试）信息：

In the way to understand this I already read (and test) info from the main two following post:

• < a href =https://stackoverflow.com/a/3755251/4716013> https://stackoverflow.com/a/3755251/4716013


• https://stackoverflow.com/a/4710776/4716013

• https://stackoverflow.com/a/3755251/4716013
• https://stackoverflow.com/a/4710776/4716013

编辑：
显然问题显然只出现在iPython内部，而不是只使用python命令行！

OK apparently the problem appear inside iPython ONLY, not when using only python command line!

iPython 0.10

In [93]: l = [1, 2, 3, 4, 1, 6, 7, 8, 7]

In [94]: all(l[i] <= l[i+1] for i in xrange(len(l)-1))
Out[94]: True

python 2.6.4

>>> l = [1, 2, 3, 4, 1, 6, 7, 8, 7]
>>> all(l[i] <= l[i+1] for i in xrange(len(l)-1))
False

推荐答案

我能够在我的机器上重现类似的问题。然而，在挖掘之后，发生了所有函数不是内置函数，而是来自numpy（ all .__ module__ =='numpy。 core.fromnumeric'）。

I was able to reproduce a similar problem on my machine. However, after digging, it occurred that the all function was not the built-in function but came from numpy (all.__module__ == 'numpy.core.fromnumeric').

问题是你要创建一个生成器而不是列表。例如：

The problem is that you are creating a generator rather than a list. For example:

all(x>5 for x in xrange(3))
# <generator object <genexpr> at 0x1153bf7d0>

all([x>5 for x in xrange(3)])
# False

if all(x>5 for x in xrange(3)):
   print True
else:
   print False
# prints True

if all([x>5 for x in xrange(3)]):
   print True
else:
   print False
# prints False

只需在表达式中添加 [...] ：

all([l[i] <= l[i+1] for i in xrange(len(l)-1)])
# False

如果你需要创建一个列表，那么更有效的解决方案就是做一个简单的for循环：

If it is the case that you need to create a list, a more efficient solution would be to do a simple for loop:

for i in xrange(len(l)-1):
    if l[i] > l[i+1]:
        result = False
        break
else:
    result = True

如果像我一样，你覆盖了内置的所有功能，你可以做 del all 来恢复它。之后，你应该有 all .__ module__ =='__ builtin __'。如果仍然不是这样，只需要 all = __builtin __。所有

If like me, you overrode the built-in all function, you can do del all to recover it. After that, you should have all.__module__ == '__builtin__'. If that's still not the case, just do all = __builtin__.all

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