为什么Rust比我的类似Python慢？

我有以下Rust程序（rustc 1.0.0-nightly（44a287e6e 2015-01-08 17:03:40 -0800））：

use std::io::BufferedReader;
use std::io::File;

fn main() {
    let path = Path::new("nc.txt");
    let mut file = BufferedReader::new(File::open(&path));
    let lines: Vec<String> = file.lines().map(|x| x.unwrap()).collect();
    println!("{}", lines[500]);
}

根据http://doc.rust-lang.org/std/io/上的示例，以上是将文件的行拉入字符串向量的方法。我已经输入了第500行的输出。

为了解决Python中的相同任务，我编写了以下代码：

#!/usr/local/bin/python3

def main():
    with open('nc.txt', 'r') as nc:
        lines = nc.read().split('\n')
    print("{}".format(lines[500]))

if __name__ == '__main__':
    main()

当我运行编译的Rust并计时时，我得到以下信息：

rts@testbed $ time ./test
A declaration of independence by Kosovo will likely bring a similar declaration from Georgia's breakaway Abkhazia region, which Russia could well recognize.

./test  1.09s user 0.02s system 99% cpu 1.120 total

运行Python可以：

rts@testbed $ time ./test.py 
A declaration of independence by Kosovo will likely bring a similar declaration from Georgia's breakaway Abkhazia region, which Russia could well recognize.
./test.py  0.05s user 0.03s system 90% cpu 0.092 total

我知道这println!是一个宏，它会扩展为更复杂的宏

::std::io::stdio::println_args(::std::fmt::Arguments::new({
    #[inline]
    #[allow(dead_code)]
    static __STATIC_FMTSTR: &'static [&'static str] = &[""];
    __STATIC_FMTSTR
},
&match (&lines[500],) {
    (__arg0,) => [::std::fmt::argument(::std::fmt::String::fmt, __arg0)],
}));

尽管如此，这似乎并不会导致超过一秒钟的额外执行时间。这些代码片段实际上是否相似？我是否误解了将线读入向量并输出其中之一的最有效方法？

供参考，nc.txt具有以下属性：

rts@testbed $ du -hs nc.txt 
7.5M    nc.txt
rts@testbed $ wc -l nc.txt
   60219 nc.txt