差异来自“超额分配”，来自源代码：

/* This over-allocates proportional to the list size, making room
     * for additional growth.  The over-allocation is mild, but is
     * enough to give linear-time amortized behavior over a long
     * sequence of appends() in the presence of a poorly-performing
     * system realloc().
     * Add padding to make the allocated size multiple of 4.
     * The growth pattern is:  0, 4, 8, 16, 24, 32, 40, 52, 64, 76, ...
     * Note: new_allocated won't overflow because the largest possible value
     *       is PY_SSIZE_T_MAX * (9 / 8) + 6 which always fits in a size_t.

但确切的行为是实现细节！
在Python 3.10中：

import sys

lst1 = list(range(1, 10))
lst2 = [item for item in range(1, 10)]
lst3 = [1, 2, 3, 4, 5, 6, 7, 8, 9]
lst4 = []
for i in range(1, 10):
    lst4.append(i)

print(sys.getsizeof(lst1)) # 136
print(sys.getsizeof(lst2)) # 184
print(sys.getsizeof(lst3)) # 136
print(sys.getsizeof(lst4)) # 184

在Python 3.5.1中：

import sys

lst1 = list(range(1, 10))
lst2 = [item for item in range(1, 10)]
lst3 = [1, 2, 3, 4, 5, 6, 7, 8, 9]
lst4 = []
for i in range(1, 10):
    lst4.append(i)

print(sys.getsizeof(lst1)) # 192
print(sys.getsizeof(lst2)) # 192
print(sys.getsizeof(lst3)) # 136
print(sys.getsizeof(lst4)) # 192

我很确定当.append()被调用时会发生这种情况（列表理解也是如此），这就是为什么在两个版本中，lst2和lst4都有最大的大小。
看起来像是在Python 3.10中，对于lst1，解释器从range对象说OK，我知道（__len__ and __length_hint__）它需要10个元素，所以我构建了一个10个元素的列表，不需要过度分配。