根据差异对列表中最近的元素进行分组

Question

我的清单如下：

tst = [1,3,4,6,8,22,24,25,26,67,68,70,72]

我想根据列表中连续元素之间的差异（相差1或2），将上面列表中的元素分组到单独的组/列表中。你知道吗

如果连续元素之间的差异大于4，则这些元素应形成一个单独的列表。你知道吗

以上输入的预期输出是：

[[1, 3, 4, 6, 8], [22, 24, 25, 26], [67, 68, 70, 72]]

我尝试了下面的代码，我认为这不是完美的方法。你知道吗

def lsp(litt):
    lia = []
    for i in range(len(litt)-1):
        if len(litt)>=2:
            if litt[i+1]-litt[i] >= 4:
                lia.append(litt[i])

    litti = []
    for i in lia:
        if i in litt:
            litti.append(litt.index(i))
    litti.insert(0,0)

    littil = []
    for i in range(len(litti)-1):
        littil.append([litti[i],litti[i+1]])

    t1 = []
    for i,j in enumerate(littil):
        t2 = []
        if i==0:
            t2.append([j[0], j[1]])
        if i!=0:
            t2.append([j[0]+1,j[1]])
        t1.append(t2)
    t1 = [i for j in t1 for i in j]

    fl = []
    for i,j in t1:
        fl.append(litt[i:j+1])
    fl.append(litt[t1[-1][1]+1:])
    return fl

我想通过使用itertools.groupby组，但不知道怎么做。你知道吗

Answer 1

使用简单for循环的另一个解决方案：

tst = [1,3,4,6,8,22,24,25,26,67,68,70,72]    # considering this as already sorted. else use tst.sort()

il = []
ol = []
for k, v in enumerate(tst):                  # enumerate is used give index to list element
    if k > 0:                                # to avoid tst[-1] which will get the last element of the list
        if abs(tst[k] - tst[k-1]) < 3:       # check if differnce is less than 3
            if tst[k-1] not in il:           # insert to inner list "il" only if it doesn't already exist
                il.append(tst[k-1])
            if tst[k] not in il:             # insert to inner list "il" only if it doesn't already exist
                il.append(tst[k])
        else:
            ol.append(list(il))              # if difference is greater than 2 then append it to outer list "ol"
            il = []                          # clear the inner list "il"
ol.append(list(il))                          # finaly append the last "il" to "ol" which didnt went in else for our case "[67, 68, 70, 72]"
print (ol)

#Result: [[1, 3, 4, 6, 8], [22, 24, 25, 26], [67, 68, 70, 72]]

Answer 2

我喜欢这样，定义切片方法并传递lambda谓词：

def slice_when(predicate, iterable):
  i, x, size = 0, 0, len(iterable)
  while i < size-1:
    if predicate(iterable[i], iterable[i+1]):
      yield iterable[x:i+1]
      x = i + 1
    i += 1
  yield iterable[x:size]

tst = [1,3,4,6,8,22,24,25,26,67,68,70,72]
slices = slice_when(lambda x,y: y - x > 2, tst)
print(list(slices))
#=> [[1, 3, 4, 6, 8], [22, 24, 25, 26], [67, 68, 70, 72]]

在许多情况下有用。你知道吗