最快的方法-使用* like list1 = [False] * 1_000_000

比较所有常用方法(列表追加、预分配、for和while)，我发现使用*可以获得最高效的执行时间。

import time

large_int = 10_000_000
start_time = time.time()

# Test 1: List comprehension
l1 = [False for _ in range(large_int)]
end_time_1 = time.time()

# Test 2: Using *
l2 = [False] * large_int
end_time_2 = time.time()

# Test 3: Using append with for loop & range
l3 = []
for _ in range(large_int):
    l3.append(False)
end_time_3 = time.time()

# Test 4: Using append with while loop
l4, i = [], 0
while i < large_int:
    l4.append(False)
    i += 1
end_time_4 = time.time()

# Results
diff_1 = end_time_1 - start_time
diff_2 = end_time_2 - end_time_1
diff_3 = end_time_3 - end_time_2
diff_4 = end_time_4 - end_time_3
print(f"Test 1. {diff_1:.4f} seconds")
print(f"Test 2. {diff_2:.4f} seconds")
print(f"Test 3. {diff_3:.4f} seconds")
print(f"Test 4. {diff_4:.4f} seconds")

print("\nTest 2 is faster than - ")
print(f"            Test 1 by - {(diff_1 / diff_2 * 100 - 1):,.0f}%")
print(f"            Test 3 by - {(diff_3 / diff_2 * 100 - 1):,.0f}%")
print(f"            Test 4 by - {(diff_4 / diff_2 * 100 - 1):,.0f}%")

Python列表没有内置的预分配。如果你真的需要做一个列表，并且需要避免附加的开销(并且你应该验证你做了)，你可以这样做:

l = [None] * 1000 # Make a list of 1000 None's
for i in xrange(1000):
    # baz
    l[i] = bar
    # qux

也许你可以通过使用生成器来避免列表:

def my_things():
    while foo:
        #baz
        yield bar
        #qux

for thing in my_things():
    # do something with thing

这样，列表就不会全部存储在内存中，而只是根据需要生成。

最快的方法-使用* like list1 = [False] * 1_000_000

比较所有常用方法(列表追加、预分配、for和while)，我发现使用*可以获得最高效的执行时间。

import time

large_int = 10_000_000
start_time = time.time()

# Test 1: List comprehension
l1 = [False for _ in range(large_int)]
end_time_1 = time.time()

# Test 2: Using *
l2 = [False] * large_int
end_time_2 = time.time()

# Test 3: Using append with for loop & range
l3 = []
for _ in range(large_int):
    l3.append(False)
end_time_3 = time.time()

# Test 4: Using append with while loop
l4, i = [], 0
while i < large_int:
    l4.append(False)
    i += 1
end_time_4 = time.time()

# Results
diff_1 = end_time_1 - start_time
diff_2 = end_time_2 - end_time_1
diff_3 = end_time_3 - end_time_2
diff_4 = end_time_4 - end_time_3
print(f"Test 1. {diff_1:.4f} seconds")
print(f"Test 2. {diff_2:.4f} seconds")
print(f"Test 3. {diff_3:.4f} seconds")
print(f"Test 4. {diff_4:.4f} seconds")

print("\nTest 2 is faster than - ")
print(f"            Test 1 by - {(diff_1 / diff_2 * 100 - 1):,.0f}%")
print(f"            Test 3 by - {(diff_3 / diff_2 * 100 - 1):,.0f}%")
print(f"            Test 4 by - {(diff_4 / diff_2 * 100 - 1):,.0f}%")

根据我的理解，Python列表已经非常类似于数组列表。但如果你想调整这些参数，我在互联网上找到了这篇文章，可能会很有趣(基本上，只需要创建自己的ScalableList扩展):

http://mail.python.org/pipermail/python-list/2000-May/035082.html

对于某些应用程序，字典可能是您正在寻找的。例如，在find_totient方法中，我发现使用字典更方便，因为我没有零索引。

def totient(n):
    totient = 0

    if n == 1:
        totient = 1
    else:
        for i in range(1, n):
            if math.gcd(i, n) == 1:
                totient += 1
    return totient

def find_totients(max):
    totients = dict()
    for i in range(1,max+1):
        totients[i] = totient(i)

    print('Totients:')
    for i in range(1,max+1):
        print(i,totients[i])

这个问题也可以用预分配的列表来解决:

def find_totients(max):
    totients = None*(max+1)
    for i in range(1,max+1):
        totients[i] = totient(i)

    print('Totients:')
    for i in range(1,max+1):
        print(i,totients[i])

我觉得这不是很优雅，而且容易产生错误，因为我存储的是None，如果我不小心错误地使用它们，它可能会抛出异常，而且因为我需要考虑映射让我避免的边缘情况。

没错，字典的效率不会那么高，但正如其他人评论的那样，速度上的微小差异并不总是值得冒重大维护风险。

警告:这个答案有争议。看到评论。

def doAppend( size=10000 ):
    result = []
    for i in range(size):
        message= "some unique object %d" % ( i, )
        result.append(message)
    return result

def doAllocate( size=10000 ):
    result=size*[None]
    for i in range(size):
        message= "some unique object %d" % ( i, )
        result[i]= message
    return result

结果。(计算每个函数144次，平均时间)

simple append 0.0102
pre-allocate  0.0098

结论。这无关紧要。

过早的优化是万恶之源。