Question

我希望按子列表

中的唯一项目筛选列表

示例：

#       ---------
#       v       v
lst = [[1, 0], [1, 1], [2, 2]]

# result
[[1, 0], [2, 2]]

我正在使用，想要：

def setx0(arr):
   r = []
   filter(lambda x : r.append(x) if x[0] not in set(map(lambda x: x[0], r)) else False, arr)
   return r

更好的方式？

Answer 1

假设列表已经排序，您可以使用itertools.groupby：

>>> from itertools import groupby 
>>> from operator import itemgetter
>>> lst = [[1, 0], [1, 1], [2, 2]]
>>> [next(g) for k, g in groupby(lst, itemgetter(0))] 
[[1, 0], [2, 2]]

Answer 2

我会直接前进：

lst = [[1, 0], [1, 1], [2, 2]]

seen = set()
new_list = []

for ele in lst:
    if ele[0] not in seen:
        new_list.append(ele)
        seen.add(ele[0])

print new_list

结果：

[[1, 0], [2, 2]]

Answer 3

不是非常有效，但如果你喜欢单行：

try:
    reduce
except NameError:  # Python 3
    from functools import reduce

lst = [[1, 0], [1, 1], [2, 2]]

filtered = reduce(lambda a, b: a+[b] if b[0] not in set(x[0] for x in a) else a,
                  lst, [])
print(filtered)  # -> [[1, 0], [2, 2]]

Answer 4

With sorted list, the easiest would be

>>> lst = [[1, 0], [1, 1], [2, 2]]
>>> dict(reversed(lst)).items()
[(1, 0), (2, 2)]

Answer 5

试试这个

const lst = [[1, 0], [1, 1], [2, 2]],
      foundItems = [],
      result = []

for (let i in lst) {
  const lstItem = lst[i]
  if (foundItems.indexOf(lstItem[0]) === -1) {
    foundItems.push(lstItem[0])
    result.push(lstItem)
  }
}

console.log(JSON.stringify(result)) // print [[1, 0], [2, 2]]

按列表中的唯一项筛选数组？

5 个答案: