該函式的輸出是洗掉前 n 個字串包含在另一行中的所有行（包括文本后面的 list1）。
我試過的代碼寫得非常糟糕而且速度非常慢，但我不知道如何做到這一點。有沒有更簡單快捷的方法來做到這一點？
行中每個元素的順序很重要。
list1 = [["0","0","0","0","0"],
["0","0","0","0","0","0"],
["0","0","0","0","0","275"],
["0","0","0","0","0","275","275"],
["0","0","0","0","275"],
["0","0","0","0","275","275"],
["0","0","0","0","275","990"],
["0","0","0","0","275","990","990"],
["0","0","0","0","275","990","2761"],
["0","0","0","0","275","990","2761","2761"],
["0","0","0","0","688"],
["0","0","0","0","688","688"],
["0","0","0","0","688","1940"],
["0","0","0","0","688","1940","1940"],
["0","0","0","0","688","1940","5041"],
["0","0","0","0","688","1940","5041","5041"],
["0","0","0","165","165","165"],
["0","0","0","165","165","165","165"]]
remove_lines = []
index_to_be_removed = []
for x in range(len(list1)):
for y in range(len(list1)):
if len(list1[x]) > len(list1[y]):
c = 0
for i in range(len(list1[y])):
if list1[x][i] == list1[y][i]:
c = c 1
if c == len(list1[y]):
if list1[x] not in remove_lines:
remove_lines.append(list1[x])
index_to_be_removed.append(x)
for x in range(len(index_to_be_removed)):
print("lines to be removed:", list1[index_to_be_removed[x]])
Output of the lines which we want to remove:
lines to be removed: ['0', '0', '0', '0', '0', '0']
lines to be removed: ['0', '0', '0', '0', '0', '275']
lines to be removed: ['0', '0', '0', '0', '0', '275', '275']
lines to be removed: ['0', '0', '0', '0', '275', '275']
lines to be removed: ['0', '0', '0', '0', '275', '990']
lines to be removed: ['0', '0', '0', '0', '275', '990', '990']
lines to be removed: ['0', '0', '0', '0', '275', '990', '2761']
lines to be removed: ['0', '0', '0', '0', '275', '990', '2761', '2761']
lines to be removed: ['0', '0', '0', '0', '688', '688']
lines to be removed: ['0', '0', '0', '0', '688', '1940']
lines to be removed: ['0', '0', '0', '0', '688', '1940', '1940']
lines to be removed: ['0', '0', '0', '0', '688', '1940', '5041']
lines to be removed: ['0', '0', '0', '0', '688', '1940', '5041', '5041']
lines to be removed: ['0', '0', '0', '165', '165', '165', '165']
uj5u.com熱心網友回復：
預計此代碼將是通用且快速的。
Lists = [["0","0","0","0","0"], ["0","0","0","0","0","0"], ["0","0","0","0","0","275"], ["0","0","0","0","0","275","275"], ["0","0","0","0","275"], ["0","0","0","0","275","275"], ["0","0","0","0","275","990"], ["0","0","0","0","275","990","990"], ["0","0","0","0","275","990","2761"], ["0","0","0","0","275","990","2761","2761"], ["0","0","0","0","688"], ["0","0","0","0","688","688"], ["0","0","0","0","688","1940"], ["0","0","0","0","688","1940","1940"], ["0","0","0","0","688","1940","5041"], ["0","0","0","0","688","1940","5041","5041"], ["0","0","0","165","165","165"], ["0","0","0","165","165","165","165"]]
def Get_Uniques(Lists):
Selected, Rejected = [], []
for List in Lists :
lenL, selected = len(List), True
for ele in Selected:
lenE = len(ele)
if lenE > lenL: continue
if List[:lenE] == ele:
Rejected.append(List)
selected = False
break
if selected: Selected.append(List)
return Selected, Rejected
Selected, Rejected = Get_Uniques(Lists)
print("Selected Lists : ", *Selected, "\nRejected Lists : ", *Rejected, sep="\n")
輸出
Selected Lists :
['0', '0', '0', '0', '0']
['0', '0', '0', '0', '275']
['0', '0', '0', '0', '688']
['0', '0', '0', '165', '165', '165']
Rejected Lists :
['0', '0', '0', '0', '0', '0']
['0', '0', '0', '0', '0', '275']
['0', '0', '0', '0', '0', '275', '275']
['0', '0', '0', '0', '275', '275']
['0', '0', '0', '0', '275', '990']
['0', '0', '0', '0', '275', '990', '990']
['0', '0', '0', '0', '275', '990', '2761']
['0', '0', '0', '0', '275', '990', '2761', '2761']
['0', '0', '0', '0', '688', '688']
['0', '0', '0', '0', '688', '1940']
['0', '0', '0', '0', '688', '1940', '1940']
['0', '0', '0', '0', '688', '1940', '5041']
['0', '0', '0', '0', '688', '1940', '5041', '5041']
['0', '0', '0', '165', '165', '165', '165']
希望這可以幫助！
uj5u.com熱心網友回復：
我會通過成對比較 while 回圈來做到這一點。基本上，我首先存盤所有串列對，（n 個串列變成0.5*n*(n-1)個串列對）。然后，一次一個，我查看每一對，找到較小的串列，并使用分隔字串比較檢查一個串列是否包含在另一個串列中。這是代碼：
from itertools import combinations
lists = [["0","0","165","165"],...]
pairs = combinations(lists)
for pair in pairs:
smaller = pair[0] if len(pair[0]) <= len(pair[1]) else pair[1]
bigger = pair[1] if smaller == pair[0] else pair[0]
smaller_str = ",".join(smaller)
bigger_str = ",".join(bigger)
if smaller_str in bigger_str and smaller in lists:
lists.remove(smaller)
print(lists)
這應該只留下那些不包含在任何其他陣列中的陣列。我不知道當你分解它的組件操作時它是否一定比你的演算法更有效，但它肯定更容易理解。
uj5u.com熱心網友回復：
您可以對各種前綴長度使用一組元組作為對可見串列前綴的控制
keep = dict() # lists to keep (as tuples)
minLen = min(map(len,list1)) # shortest prefix length
for i,L in enumerate(list1):
prefixes = (tuple(L[:n]) for n in range(minLen,len(L) 1))
if not any(sl in keep for sl in prefixes): # check all prefix lengths
keep[tuple(L)]=i # track lists that remain
else:
print("list to remove",L) # list at index i will be dropped
# clean up
list1 = [*map(list,keep)]
輸出：
list to remove ['0', '0', '0', '0', '0', '0']
list to remove ['0', '0', '0', '0', '0', '275']
list to remove ['0', '0', '0', '0', '0', '275', '275']
list to remove ['0', '0', '0', '0', '275', '275']
list to remove ['0', '0', '0', '0', '275', '990']
list to remove ['0', '0', '0', '0', '275', '990', '990']
list to remove ['0', '0', '0', '0', '275', '990', '2761']
list to remove ['0', '0', '0', '0', '275', '990', '2761', '2761']
list to remove ['0', '0', '0', '0', '688', '688']
list to remove ['0', '0', '0', '0', '688', '1940']
list to remove ['0', '0', '0', '0', '688', '1940', '1940']
list to remove ['0', '0', '0', '0', '688', '1940', '5041']
list to remove ['0', '0', '0', '0', '688', '1940', '5041', '5041']
list to remove ['0', '0', '0', '165', '165', '165', '165']
print(list1)
[['0', '0', '0', '0', '0'],
['0', '0', '0', '0', '275'],
['0', '0', '0', '0', '688'],
['0', '0', '0', '165', '165', '165']]
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