Question

我试图在python中找到列表的前k个元素，而不使用heapq或对列表进行排序。

这就是我试过的，

list = [20,4,67,22,445,1,34]
k = 3
newList=[]

for i in range(0,k):
    newList.append(list[i])
for i in list:
    mini = min(newList)
    if i <= mini:
        continue
    else:
        newList.remove(mini)
        newList.append(i)
print newList

但我得到67,67,445。我在这里做错了什么？

Answer 1

添加一些跟踪时，问题很明显：

>>> list = [20,4,67,22,445,1,34]
>>> k = 3
>>> newList=[]
>>> 
>>> for i in range(0,k):
...     newList.append(list[i])
... 
>>> for i in list:
...     mini = min(newList)
...     if i <= mini:
...         continue
...     else:
...         print newList
...         print "Replacing {0} with {1}".format(mini, i)
...         newList.remove(mini)
...         newList.append(i)
...         print newList
...         print '-' * 20
... 
[20, 4, 67]
Replacing 4 with 20
[20, 67, 20]
--------------------
[20, 67, 20]
Replacing 20 with 67
[67, 20, 67]
--------------------
[67, 20, 67]
Replacing 20 with 22
[67, 67, 22]
--------------------
[67, 67, 22]
Replacing 22 with 445
[67, 67, 445]

当您对其进行迭代并再次添加时，您已经在列表中有67个。

我会把它重写为：

>>> numbers = [20,4,67,22,445,1,34]
>>> k = 3
>>> newList = numbers[:k]
>>> 
>>> for i in numbers[k:]:
...     mini = min(newList)
...     if i > mini:
...         print "Replacing {0} with {1}".format(mini, i)
...         newList.remove(mini)
...         newList.append(i)
... 
Replacing 4 with 22
Replacing 20 with 445
Replacing 22 with 34
>>> print newList
[67, 445, 34]

但我不会将您的列表命名为list。

Answer 2

你可以简单地这样做：

a = [20,4,67,22,445,1,34]
k = 3
newList=[]

for i in range(k):
    pos = a.index(max(a))
    newList.append(a[pos])
    a.pop(pos)

>>> print newList
[67, 445, 34]

Answer 3

你的新名单中有67个，而67个从不弹出

Answer 4

hughdbrown的解决方案有一个我注意到的错误。如果列表具有类似的条目，则结果将仅显示这些条目之一。例如，如果列表为[1, 2, 3, 4, 5, 5]，则结果将显示[3, 4, 5]而不是[4, 5, 5]

在没有heapq或排序的情况下在数组python中查找前k个元素

4 个答案: