在 Python 中，我如何編寫代碼來移出列表的最后一個元素并在開頭添加一個新元素 - 以便在執行時盡可能快地運行？有一些很好的解決方案涉及使用追加、旋轉等，但并非所有解決方案都可以轉化為快速執行。
4 回答
嗶嗶one
TA貢獻1854條經驗 獲得超8個贊
不要使用列表。
列表只能在其末尾快速插入和刪除項目。你會使用pop(-1)and append，你最終會得到一個堆棧。
相反，請使用collections.deque，它專為在兩端進行高效添加和刪除而設計。在雙端隊列的“前端”工作使用popleftandappendleft方法。請注意，“deque”的意思是“雙端隊列”，發音為“deck”。
神不在的星期二
TA貢獻1963條經驗 獲得超6個贊
L = [1, 2, 3]
L.pop() # returns 3, L is now [1, 2]
L.append(4) # returns None, L is now [1, 2, 4]
L.insert(0, 5) # returns None, L is now [5, 1, 2, 4]
L.remove(2) # return None, L is now [5, 1, 4]
del(L[0]) # return None, L is now [1, 4]
L.pop(0) # return 1, L is now [4]
慕容3067478
TA貢獻1773條經驗 獲得超3個贊
我為您運行了一些基準測試。這是結果。
TL;DR 您可能想使用deque. 否則，insert/append或pop/del都可以。
添加到最后
from collections import deque
import perfplot
# Add to end
def use_append(n):
"adds to end"
a = [1,2,3,4,5,6,7,8,9,10]*n
a.append(7)
return 1
def use_insert_end(n):
"adds to end"
a = [1,2,3,4,5,6,7,8,9,10]*n
a.insert(len(a),7)
return 1
def use_add_end(n):
"adds to end"
a = [1,2,3,4,5,6,7,8,9,10]*n
a = a + [7]
return 1
perfplot.show(
setup=lambda n: n, # or simply setup=numpy.random.rand
kernels=[
lambda a: use_append(a),
lambda a: use_insert_end(a),
lambda a: use_add_end(a),
],
labels=["use_append", "use_insert_end", "use_add_end"],
n_range=[2 ** k for k in range(15)],
xlabel="len(a)",
)
從末尾刪除
# Removing from the end
def use_pop(n):
"removes from end"
a = [1,2,3,4,5,6,7,8,9,10]*n
a.pop()
return 1
def use_del_last(n):
"removes from end"
a = [1,2,3,4,5,6,7,8,9,10]*n
del(a[-1])
return 1
def use_index_to_end(n):
"removes from end"
a = [1,2,3,4,5,6,7,8,9,10]*n
a = a[:-1]
return 1
perfplot.show(
setup=lambda n: n,
kernels=[
lambda a: use_pop(a),
lambda a: use_del_last(a),
lambda a: use_index_to_end(a),
],
labels=["use_pop", "use_del_last", "use_index_to_end"],
n_range=[2 ** k for k in range(20)],
xlabel="len(a)",
)
添加到開頭
# Add to beginning
def use_insert(n):
"adds to beginning"
a = [1,2,3,4,5,6,7,8,9,10]*n
a.insert(0,7)
return 1
def use_deque_appendleft(n):
"adds to beginning"
a = [1,2,3,4,5,6,7,8,9,10]*n
a = deque(a)
a.appendleft(7)
return 1
def use_add_start(n):
"adds to beginning"
a = [1,2,3,4,5,6,7,8,9,10]*n
a = [7] + a
return 1
perfplot.show(
setup=lambda n: n, # or simply setup=numpy.random.rand
kernels=[
lambda a: use_insert(a),
lambda a: use_deque_appendleft(a),
lambda a: use_add_start(a),
],
labels=["use_insert", "use_deque_appendleft","use_add_start"],
n_range=[2 ** k for k in range(15)],
xlabel="len(a)",
)
從頭開始刪除
# Remove from beginning
def use_del_first(n):
"removes from beginning"
a = [1,2,3,4,5,6,7,8,9,10]*n
del(a[0])
return 1
def use_deque_popleft(n):
"removes from beginning"
a = [1,2,3,4,5,6,7,8,9,10]*n
a = deque(a)
a.popleft()
return 1
def use_index_start(n):
"removes from beginning"
a = [1,2,3,4,5,6,7,8,9,10]*n
a = a[1:]
return 1
perfplot.show(
setup=lambda n: n, # or simply setup=numpy.random.rand
kernels=[
lambda a: use_del_first(a),
lambda a: use_deque_popleft(a),
lambda a: use_index_start(a),
],
labels=["use_del_first", "use_deque_popleft", "use_index_start"],
n_range=[2 ** k for k in range(15)],
xlabel="len(a)",
)
BIG陽
TA貢獻1859條經驗 獲得超6個贊
您可以使用 python 列表對象的插入方法。
l = [1, 2, 3]
#lets insert 10 at biggining, which means at index 0
l.insert(0, 10)
print(l) # this will print [10, 1, 2, 3]
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