小编Eas*_*sun的帖子

将lmplot在seaborn拟合回归模型的截距.但是,有时我想要在没有截距的情况下拟合回归模型,即通过原点进行回归.

例如:

In [1]: import numpy as np
   ...: import pandas as pd
   ...: import seaborn as sns
   ...: import matplotlib.pyplot as plt
   ...: import statsmodels.formula.api as sfa
   ...: 

In [2]: %matplotlib inline
In [3]: np.random.seed(2016)
In [4]: x = np.linspace(0, 10, 32)
In [5]: y = 0.3 * x + np.random.randn(len(x))
In [6]: df = pd.DataFrame({'x': x, 'y': y})
In [7]: r = sfa.ols('y ~ x + 0', data=df).fit()
In [8]: sns.lmplot(x='x', y='y', data=df, fit_reg=True) …

我有一个由matlab创建的struct数组,并存储在v7.3格式的mat文件中:

struArray = struct('name', {'one', 'two', 'three'}, 
                   'id', {1,2,3}, 
                   'data', {[1:10], [3:9], [0]})
save('test.mat', 'struArray', '-v7.3')

现在我想通过python使用h5py读取这个文件:

data = h5py.File('test.mat')
struArray = data['/struArray']

我不知道如何逐个获取结构数据struArray:

for index in range(<the size of struArray>):
    elem = <the index th struct in struArray>
    name = <the name of elem>
    id = <the id of elem>
    data = <the data of elem>

I want to process stock level-2 data in pandas. Suppose there are four kinds data in each row for simplicity:

• millis: timestamp, int64
• last_price: the last trade price, float64,
• ask_queue: the volume of ask side, a fixed size (200) array of int32
• bid_queue: the volume of bid side, a fixed size (200) array of int32

Which can be easily defined as a structured dtype in numpy:

dtype = np.dtype([
   ('millis', 'int64'), 
   ('last_price', 'float64'), 
   ('ask_queue', ('int32', 200)), 
   ('bid_queue', ('int32', 200))
]) …

我想在没有任何配置文件的情况下使用log4j .我要做的是:

logger = (Logger) LogManager.getLogger(this.getClass());
String pattern = "[%level] %m%n";
//do something to make this logger output to an local file "/xxx/yyy/zzz.log"

我找到了这个答案:以 编程方式配置Log4j记录器.

但是文档Logger#addAppender说: 这种方法不是通过公共API公开的,主要用于单元测试.

我不确定在我的代码中使用此方法是否正确,或者还有其他更好的解决方案来解决我的问题.

我想写一个Tuple2 [A,B]到Seq [C]的隐式转换,其中C是A和B的超类型.我的第一个尝试如下:

implicit def t2seq[A,B,C](t: (A,B))(implicit env: (A,B) <:< (C,C)): Seq[C] = {
    val (a,b) = env(t)
    Seq(a,b)
}

但它不起作用:

scala> (1,2): Seq[Int]
<console>:7: error: type mismatch;
 found   : (Int, Int)
 required: Seq[Int]
       (1,2): Seq[Int]
       ^

虽然这个工作:

class Tuple2W[A,B](t: (A,B)) {
    def toSeq[C](implicit env: (A,B) <:< (C,C)): Seq[C] = {
        val (a,b) = env(t)
        Seq(a,b)
    }
}
implicit def t2tw[A,B](t: (A,B)): Tuple2W[A,B] = new Tuple2W(t)

使用案例:

scala> (1,2).toSeq
res0: Seq[Int] = List(1, 2)

我不知道为什么第一个解决方案没有按预期工作. Scala版本2.8.0.r22634-b20100728020027(Java HotSpot(TM)Client VM,Java 1.6.0_20).

例如,我想绘制函数

f(x,y) =sin(x^2+y^2)/(x^2+y^2),       x^2+y^2 <=4?

在Mathematica中,我可以这样做:

Plot3D[Sin[x^2 + y^2]/(x^2 + y^2), {x, -4, 4}, {y, -4, 4}, 
 RegionFunction -> (#1^2 + #2^2 <= 4 Pi &)]

其中RegionFunction指定x,y的区域来绘制.

我有一个名为的矩阵xs:

array([[1, 1, 1, 1, 1, 0, 1, 0, 0, 2, 1],
       [2, 1, 0, 0, 0, 1, 2, 1, 1, 2, 2]])

现在我想用同一行中最近的前一个元素替换零(假设第一列必须非零.).粗略的解决方案如下:

In [55]: row, col = xs.shape

In [56]: for r in xrange(row):
   ....:     for c in xrange(col):
   ....:         if xs[r, c] == 0:
   ....:             xs[r, c] = xs[r, c-1]
   ....: 

In [57]: xs
Out[57]: 
array([[1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1],
       [2, 1, 1, 1, 1, 1, 2, 1, 1, …

在python 3.4.3和Pandas 0.16中,如何指定dtype索引为str？以下代码是我尝试过的:

In [1]: from io import StringIO

In [2]: import pandas as pd

In [3]: import numpy as np

In [4]: fra = pd.read_csv(StringIO('date,close\n20140101,10.2\n20140102,10.5'), index_col=0, dtype={'date': np.str_, 'close': np.float})

In [5]: fra.index
Out[5]: Int64Index([20140101, 20140102], dtype='int64')

我希望在RDD性能方面采取行动,reduce但不需要操作员可交换.即我希望result跟随永远是"123456789".

scala> val rdd = sc.parallelize(1 to 9 map (_.toString))
rdd: org.apache.spark.rdd.RDD[String] = ParallelCollectionRDD[24] at parallelize at <console>:24

scala> val result = rdd.someAction{ _+_ }

首先,我找到了fold.文件RDD#fold说:

def fold(zeroValue:T)(op:(T,T)⇒T):T使用给定的关联函数和中性"零值" 聚合每个分区的元素,然后聚合所有分区的结果

请注意,doc中不需要交换.但是,结果并不像预期的那样:

scala> rdd.fold(""){ _+_ }
res10: String = 312456879

编辑我试过@ dk14提到的,没有运气:

scala> val rdd = sc.parallelize(1 to 9 map (_.toString))
rdd: org.apache.spark.rdd.RDD[String] = ParallelCollectionRDD[48] at parallelize at <console>:24

scala> rdd.zipWithIndex.map(x => (x._2, x._1)).sortByKey().map(_._2).fold(""){ _+_ …

编译以下代码失败,但如果我删除specialized方法中的注释,它就会通过dot.

Scala代码运行器版本2.12.0-RC2 - 版权所有2002-2016,LAMP/EPFL和Lightbend,Inc.

abstract class Op[@specialized Left, @specialized Right] {
  @specialized
  type Result

  def r: Numeric[Result]
  def times(left: Left, right: Right): Result
}


object Op {

  implicit object IntDoubleOp extends Op[Int, Double] {
    type Result = Double
    val r = implicitly[Numeric[Double]]
    def times(left: Int, right: Double): Double = left * right
  }
}


object calc {

  def dot[@specialized Left, @specialized Right](xs: Array[Left], ys: Array[Right])
          (implicit op: Op[Left, Right]): op.Result = {
    var total = op.r.zero …