小编bcf*_*bcf的帖子

Consider this bit of code:

#include <iostream>                                                   
#include <string>

int main()
{
  std::string str;
  std::cout << "Enter a string: \n";
  getline(std::cin, str);
}

Why must I use std:: for string, cin and cout, but not getline()? Is getline() not in the standard library? I'm actually somewhat confused why I can't just write using namespace std; and not have to #include anything in the standard library, too. Thanks in advance!

考虑Series obj：

In [50]: obj = Series(np.arange(6,10), index = ['a', 'b', 'c', 'd'])

In [51]: obj
Out[51]: 
a    6
b    7
c    8
d    9
dtype: int64

我想切一块obj，我可以通过以下几种方式做到这一点：

In [52]: obj[1:3]
Out[52]: 
b    7
c    8
dtype: int64

In [53]: obj['b' : 'c']
Out[53]: 
b    7
c    8
dtype: int64

现在考虑DataFrame param_estimates_good：

In [54]: param_estimates_good
Out[54]: 
             a         b     sigma      a_se      b_se  sigma_se  success
1968  0.648508  1.803889  0.498017  0.784340  0.082366  0.529649        1
1972  0.539485  1.733304 …

考虑一下这段代码:

#include <iostream>
#include <complex>

int main()
{
  std::complex<double> z1 = 5;
  std::cout << z1 - 1 << "\n";  // must change to z1 - 1.0 to compile

  std::complex<int> z2 = 5;
  std::cout << z2 - 1.0 << "\n";  // must change to z2 - 1 to compile
}

这会产生编译错误,因为operator-表达式中的类型没有找到z1 - 1或z2 - 1.0.另一方面,更改这些表达式以使基本类型匹配工作正常.

天真,对z1 - 1我期望int1被提升到double和预期的z2,与基本类型int,在z2 - 1.0晋升为一个complex<double>.这是怎么回事？

考虑两个ndarrays长度n,arr1和arr2.我正在计算以下产品总数,并按时num_runs进行基准测试:

import numpy as np
import time

num_runs = 1000
n = 100

arr1 = np.random.rand(n)
arr2 = np.random.rand(n)

start_comp = time.clock()
for r in xrange(num_runs):
    sum_prods = np.sum( [arr1[i]*arr2[j] for i in xrange(n) 
                         for j in xrange(i+1, n)] )

print "total time for comprehension = ", time.clock() - start_comp

start_loop = time.clock()
for r in xrange(num_runs):
    sum_prod = 0.0
    for i in xrange(n):
        for j in xrange(i+1, n):
            sum_prod …

我很惊讶,调用np.inner计算平方和比调用np.sum预先计算的正方形数组快5倍:

对此行为的任何见解？我实际上对一个正方形的快速实现感兴趣,所以这些想法也是受欢迎的.

我有点惊讶的是，该hashCode()方法的默认（本机）实现比以下基准测试的方法的简单覆盖慢了大约 50 倍。

考虑一个Book不覆盖的基本类hashCode()：

public class Book {
private int id;
private String title;
private String author;
private Double price;

public Book(int id, String title, String author, Double price) {
    this.id = id;
    this.title = title;
    this.author = author;
    this.price = price;
}
}

或者，考虑一个完全相同的Book类，BookWithHash，它hashCode()使用 Intellij 的默认实现覆盖该方法：

public class BookWithHash {
private int id;
private String title;
private String author;
private Double price;


public …

我希望向量迭代器指向一个向量元素.我有

#include <iostream>
#include <vector>

int main() {
  std::vector<int> vec = {1,2,3,4,5};
  std::vector<int>::iterator it;

  // want "it" to point to the "3" element, so something like
  //   it = &prices[2];
  //   it = &prices.at(2);

}

但这些尝试都不起作用.我想我需要一些返回迭代器的向量函数,而不是地址(？)

在Windows上启动Python 2.7解释器后的第一行:

>>> dir()
['__builtins__', '__doc__', '__name__', '__package__']

输入dir()命令后,_应定义特殊变量:

>>> _
['__builtins__', '__doc__', '__name__', '__package__']

但是,即使在输入之后_,当我尝试使用以下内容列出交互式命名空间中的所有名称时,它也不会显示dir():

>>> dir()
['__builtins__', '__doc__', '__name__', '__package__']

如果解释器不在解释器的命名空间中,解释器如何识别它？

我认为!在IPython中添加一个shell命令会使系统shell实际执行命令,但似乎并非如此.考虑以下内容,我开始使用/home/Documents/Rx,启动IPython,使用命令cd(没有!),然后退出IPython并查看我所在的目录:

$ pwd
/home/Documents/Rx
$ ipython
Python 2.7.11 |Anaconda 2.5.0 (64-bit)| (default, Dec  6 2015, 18:08:32) 
Type "copyright", "credits" or "license" for more information.

IPython 4.0.3 -- An enhanced Interactive Python.
?         -> Introduction and overview of IPython's features.
%quickref -> Quick reference.
help      -> Python's own help system.
object?   -> Details about 'object', use 'object??' for extra details.

In [1]: cd Papers/
/home/Documents/Rx/Papers

In [2]: pwd
Out[2]: u'/home/Documents/Rx/Papers'

In [3]: exit() …

我意识到有些方法应该用 来调用()，而另一些则不能。我如何使用 IPython 检查是否使用括号？例如下面的文件scratch.py

import numpy as np

arr = np.random.randn(5)

print arr.sort, "\n"
print arr.sort(), "\n";
print arr.shape, "\n";
print arr.shape(), "\n";

产生这个输出：

<built-in method sort of numpy.ndarray object at 0x7fb4b5312300> 

None 

(5,) 

Traceback (most recent call last):
  File "scratch.py", line 8, in <module>
    print arr.shape(), "\n";
TypeError: 'tuple' object is not callable