[英]Group consecutive integers together
有以下代碼:
import sys
ints = [1,2,3,4,5,6,8,9,10,11,14,34,14,35,16,18,39,10,29,30,14,26,64,27,48,65]
ints.sort()
ints = list(set(ints))
c = {}
for i,v in enumerate(ints):
if i+1 >= len(ints):
continue
if ints[i+1] == v + 1 or ints[i-1] == v - 1:
if len(c) == 0:
c[v] = [v]
c[v].append(ints[i+1])
else:
added=False
for x,e in c.items():
last = e[-1]
if v in e:
added=True
break
if v - last == 1:
c[x].append(v)
added=True
if added==False:
c[v] = [v]
else:
if v not in c:
c[v] = [v]
print('input ', ints)
print('output ', c))
目標:
給定一個整數列表,創建一個包含連續整數的字典,以減少列表的總長度。
這是我當前解決方案的輸出:
input [1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 14, 16, 18, 26, 27, 29, 30, 34, 35, 39, 48, 64, 65]
output {1: [1, 2, 3, 4, 5, 6], 8: [8, 9, 10, 11], 14: [14], 16: [16], 18: [18], 26: [26, 27], 29: [29, 30], 34: [34, 35], 39: [39], 48: [48], 64: [64]}
條件/限制:
3
,不要創建列表3,4
,而是將3
附加到現有列表[1,2]
我當前的迭代工作正常,但是由於for x,e in c.items()
現有列表檢查中的for x,e in c.items()
列表越大,它就會指數越慢。
如何在實現相同結果的同時加快速度?
新解決方案(使用19,000個整數的輸入列表,從13秒到0.03秒):
c = {}
i = 0
last_list = None
while i < len(ints):
cur = ints[i]
if last_list is None:
c[cur] = [cur]
last_list = c[cur]
else:
if last_list[-1] == cur-1:
last_list.append(cur)
else:
c[cur] = [cur]
last_list = c[cur]
i += 1
由於你有連續數字的列表,我建議你使用range
對象而不是list
s:
d, head = {}, None
for x in l:
if head is None or x != d[head].stop:
head = x
d[head] = range(head, x+1)
如果您使用for
循環並且只是跟蹤當前列表,解決方案很簡單。 當您發現差距時,不要忘記制作新的清單:
result = {}
cl = None
for i in ints:
if cl is None or i - 1 != cl[-1]:
cl = result.setdefault(i, [])
cl.append(i)
有一個很棒的庫叫做more_itertools
,它有一個名為: consecutive_groups()
:
import more_itertools as mit
x = [1,2,3,4,5,6,8,9,10,11,14,34,14,35,16,18,39,10,29,30,14,26,64,27,48,65]
x = [list(j) for j in mit.consecutive_groups(sorted(list(set(x))))]
# [[1, 2, 3, 4, 5, 6], [8, 9, 10, 11], [14], [16], [18], [26, 27], [29, 30], [34, 35], [39], [48], [64, 65]]
dct_x = {i[0]: i for i in x}
print(dct_x)
輸出:
{1: [1, 2, 3, 4, 5, 6], 8: [8, 9, 10, 11], 14: [14], 16: [16], 18: [18], 26: [26, 27], 29: [29, 30], 34: [34, 35], 39: [39], 48: [48], 64: [64, 65]}
還有一個注釋,您希望在轉換為集合之后進行排序,因為集合是無序的。
可以用O(n)
(線性)復雜度來解決這個任務。 保持簡單:
integers = [1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 14, 16, 18, 26, 27, 29, 30, 34, 35, 39, 48, 64, 65]
helper = []
counter = 0
while counter < len(integers):
if not helper or helper[-1] + 1 != integers[counter]:
print('gap found', integers[counter]) # do your logic
helper.append(integers[counter])
counter += 1
上面的算法假定輸入列表已經排序。 它給了我們巨大的優勢。 同時,在運行此算法之前,可以明確地對整數列表進行排序。 然后,解的總復雜度為: O(n * log n) + O(n)
,其有效地為O(n * log n)
。 O(n * log n)
是排序過程的復雜性。
我建議記住這個非常有用的技巧,即在接近未來用途的任務之前使用排序。
這是一個簡單的實現,使用列表切片實現您的目標:
integers = [1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 14, 16, 18, 26, 27, 29, 30, 34, 35, 39, 48, 64, 65]
for i, integer in enumerate(integers):
if i == 0:
out_dict = {}
start = 0
else:
if integer != prev_integer + 1:
out_dict[integers[start]] = integers[start:i]
start = i
if i == len(integers) - 1:
out_dict[integers[start]] = integers[start:]
prev_integer = integer
>>>out_dict = {1: [1, 2, 3, 4, 5, 6], 8: [8, 9, 10, 11], 14: [14], 16: [16], 18: [18], 26: [26, 27], 29: [29, 30], 34: [34, 35], 39: [39], 48: [48], 64: [64]}
注意:字典可能不會按升序鍵排序,因為dict
類型沒有排序。
您可以嘗試使用itertools,但我想嘗試遞歸:
input_dta=[1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 14, 16, 18, 26, 27, 29, 30, 34, 35, 39, 48, 64, 65]
final_=[]
def consecutives(data):
sub_final=[]
if not data:
return 0
else:
for i,j in enumerate(data):
try:
if abs(data[i]-data[i+1])==1:
sub_final.extend([data[i],data[i+1]])
else:
if sub_final:
final_.append(set(sub_final))
return consecutives(data[i+1:])
except IndexError:
pass
final_.append(set(sub_final))
consecutives(input_dta)
print(final_)
輸出:
[{1, 2, 3, 4, 5, 6}, {8, 9, 10, 11}, {26, 27}, {29, 30}, {34, 35}, {64, 65}]
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.