Question

我的清单是:

[ 02 ,  03 ,  03 ,  16 ,  17 ,  17 ,  28 ,  29 ,  29 ]

我要知道,我如何能够从名单上删除这些重复之处。

Would it be also possible when I add an item to the list to check if the item is already in the list (to avoid duplicates?)

Answer 1

提出

let list=[ 02 ,  03 ,  03 ,  16 ,  17 ,  17 ,  28 ,  29 ,  29 ]
let unduplist=filter(copy(list),  index(list, v:val, v:key+1)==-1 )

。关于第二个问题,见:h Index()。

如果是,

all list items are strings;
there is no empty strings possible;
you don t care about order of list items

那么,你或许应当使用假肢:对于大量探照相机,速度较快(实际上不需要)。

Answer 2

这种排位外法(即,在<代码>后生效:)的取代将消除复制件(条件是这些复制件是连续的):

s/( [0-9]+ ),s+1/1/g

Answer 3

您还可以将名单转换为一位理论家的钥匙:

let list=[ 02 ,  03 ,  03 ,  16 ,  17 ,  17 ,  28 ,  29 ,  29 ]
let dict = {}
for l in list
   let dict[l] =   
endfor
let uniqueList = keys(dict)

这既符合分类清单,也符合未经编辑的名单。

Answer 4

Vim 7.4.218 (25 Mar 2014) and newer

规定

uniq({list} [, {func} [, {dict}]])          *uniq()* *E882*
        Remove second and succeeding copies of repeated adjacent
        {list} items in-place.  Returns {list}.  If you want a list
        to remain unmodified make a copy first: >
            :let newlist = uniq(copy(mylist))
<       The default compare function uses the string representation of
        each item.  For the use of {func} and {dict} see |sort()|.

Vim 7.4.218 (25 Mar 2014) and newer

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