def histeq(im,nbr_bins=256):

"""对一幅灰度图像进行直方图均衡化"""

#计算图像的直方图

imhist,bins = histogram(im.flatten(),nbr_bins,normed=True)

cdf = imhist.cumsum() #累计分布函数

cdf = 255 * cdf / cdf[-1] #归一化

#使用累计分布函数的线性插值，计算新的像素

im2 = interp(im.flatten(),bins[:-1],cdf)

return im2.reshape(im.shape),cdf

以上代码我定义在imtools.py文件里并且放在了python2.7里

然后我在num.py里引用他

Python code?

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from PIL import Image

from pylab import *

from numpy import *

import imtools

im= array(Image.open('E:\\daima\\pydaima\\shijue\\tupian1\\gang2.jpg').convert('L'))

im2,cdf =imtools.histeq(im)

出现以下错误：

Traceback (most recent call last):

File "<pyshell#56>", line 1, in <module>

a=imtools.histeq(im)

File "E:\daima\pydaima\shijue\imtools.py", line 32, in histeq

NameError: global name 'histogram' is not defined

形态参数

While a general continuous random variable can be shifted and scaled

with the loc and scale parameters, some distributions require additional

shape parameters. For instance, the gamma distribution, with density

γ(x,a)=λ(λx)a−1Γ(a)e−λx,

requires the shape parameter a. Observe that setting λ can be obtained by setting the scale keyword to 1/λ.

虽然一个一般的连续随机变量可以被位移和伸缩通过loc和scale参数，但一些分布还需要额外的形态参数。作为例子，看到这个伽马分布，这是它的密度函数

γ(x,a)=λ(λx)a−1Γ(a)e−λx,

要求一个形态参数a。注意到λ的设置可以通过设置scale关键字为1/λ进行。

Let’s check the number and name of the shape parameters of the gamma

distribution. (We know from the above that this should be 1.)

让我们检查伽马分布的形态参数的名字的数量。（我们知道从上面知道其应该为1）

>>>

>>>from scipy.stats import gamma

>>>gamma.numargs

1

>>>gamma.shapes

'a'

Now we set the value of the shape variable to 1 to obtain the

exponential distribution, so that we compare easily whether we get the

results we expect.

现在我们设置形态变量的值为1以变成指数分布。所以我们可以容易的比较是否得到了我们所期望的结果。

>>>

>>> gamma(1, scale=2.).stats(moments="mv")

(array(2.0), array(4.0))

Notice that we can also specify shape parameters as keywords:

注意我们也可以以关键字的方式指定形态参数：

>>>

>>>gamma(a=1, scale=2.).stats(moments="mv")

(array(2.0), array(4.0))

Freezing a Distribution

冻结分布

Passing the loc and scale keywords time and again can become quite

bothersome. The concept of freezing a RV is used to solve such problems.

不断地传递loc与scale关键字最终会让人厌烦。而冻结RV的概念被用来解决这个问题。

>>>

>>>rv = gamma(1, scale=2.)

By using rv we no longer have to include the scale or the shape

parameters anymore. Thus, distributions can be used in one of two ways,

either by passing all distribution parameters to each method call (such

as we did earlier) or by freezing the parameters for the instance of the

distribution. Let us check this:

通过使用rv我们不用再更多的包含scale与形态参数在任何情况下。显然，分布可以被多种方式使用，我们可以通过传递所有分布参数给对方法的每次调用（像我们之前做的那样）或者可以对一个分布对象冻结参数。让我们看看是怎么回事：

>>>

>>>rv.mean(), rv.std()

(2.0, 2.0)

This is indeed what we should get.

这正是我们应该得到的。

Broadcasting

广播

The basic methods pdf and so on satisfy the usual numpy broadcasting

rules. For example, we can calculate the critical values for the upper

tail of the t distribution for different probabilites and degrees of

freedom.

像pdf这样的简单方法满足numpy的广播规则。作为例子，我们可以计算t分布的右尾分布的临界值对于不同的概率值以及自由度。

>>>

>>>stats.t.isf([0.1, 0.05, 0.01], [[10], [11]])

array([[ 1.37218364, 1.81246112, 2.76376946],

[ 1.36343032, 1.79588482, 2.71807918]])

Here, the first row are the critical values for 10 degrees of freedom

and the second row for 11 degrees of freedom (d.o.f.). Thus, the

broadcasting rules give the same result of calling isf twice:

这里，第一行是以10自由度的临界值，而第二行是以11为自由度的临界值。所以，广播规则与下面调用了两次isf产生的结果相同。

>>>

>>>stats.t.isf([0.1, 0.05, 0.01], 10)

array([ 1.37218364, 1.81246112, 2.76376946])

>>>stats.t.isf([0.1, 0.05, 0.01], 11)

array([ 1.36343032, 1.79588482, 2.71807918])

If the array with probabilities, i.e, [0.1, 0.05, 0.01] and the array of

degrees of freedom i.e., [10, 11, 12], have the same array shape, then

element wise matching is used. As an example, we can obtain the 10% tail

for 10 d.o.f., the 5% tail for 11 d.o.f. and the 1% tail for 12 d.o.f.

by calling

但是如果概率数组，如[0.1,0.05,0.01]与自由度数组,如[10,11,12]具有相同的数组形态，则元素对应捕捉被作用，我们可以分别得到10%，5%，1%尾的临界值对于10，11,12的自由度。

>>>

>>>stats.t.isf([0.1, 0.05, 0.01], [10, 11, 12])

array([ 1.37218364, 1.79588482, 2.68099799])

Specific Points for Discrete Distributions

离散分布的特殊之处

Discrete distribution have mostly the same basic methods as the

continuous distributions. However pdf is replaced the probability mass

function pmf, no estimation methods, such as fit, are available, and

scale is not a valid keyword parameter. The location parameter, keyword

loc can still be used to shift the distribution.

离散分布的简单方法大多数与连续分布很类似。当然像pdf被更换为密度函数pmf，没有估计方法，像fit是可用的。而scale不是一个合法的关键字参数。Location参数，关键字loc则仍然可以使用用于位移。

The computation of the cdf requires some extra attention. In the case of

continuous distribution the cumulative distribution function is in most

standard cases strictly monotonic increasing in the bounds (a,b) and

has therefore a unique inverse. The cdf of a discrete distribution,

however, is a step function, hence the inverse cdf, i.e., the percent

point function, requires a different definition:

ppf(q) = min{x : cdf(x) >= q, x integer}

Cdf的计算要求一些额外的关注。在连续分布的情况下，累积分布函数在大多数标准情况下是严格递增的，所以有唯一的逆。而cdf在离散分布，无论如何，是阶跃函数，所以cdf的逆，分位点函数，要求一个不同的定义：

ppf(q) = min{x : cdf(x) >= q, x integer}

For further info, see the docs here.

为了更多信息可以看这里。

We can look at the hypergeometric distribution as an example

>>>

>>>from scipy.stats import hypergeom

>>>[M, n, N] = [20, 7, 12]

我们可以看这个超几何分布的例子

>>>

>>>from scipy.stats import hypergeom

>>>[M, n, N] = [20, 7, 12]

If we use the cdf at some integer points and then evaluate the ppf at

those cdf values, we get the initial integers back, for example

如果我们使用在一些整数点使用cdf，它们的cdf值再作用ppf会回到开始的值。

>>>

>>>x = np.arange(4)*2

>>>x

array([0, 2, 4, 6])

>>>prb = hypergeom.cdf(x, M, n, N)

>>>prb

array([ 0.0001031991744066, 0.0521155830753351, 0.6083591331269301,

0.9897832817337386])

>>>hypergeom.ppf(prb, M, n, N)

array([ 0., 2., 4., 6.])

If we use values that are not at the kinks of the cdf step function, we get the next higher integer back:

如果我们使用的值不是cdf的函数值，则我们得到一个更高的值。

>>>

>>>hypergeom.ppf(prb + 1e-8, M, n, N)

array([ 1., 3., 5., 7.])

>>>hypergeom.ppf(prb - 1e-8, M, n, N)

array([ 0., 2., 4., 6.])

直接去 http://www.lfd.uci.edu/~gohlke/pythonlibs/#jpype 下载对应的whl文件，使用

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pip install x.whl

安装即可，可能需要安其他的库，都可以在这个链接中找到

CentOS（7）:

出离愤怒啊，搞了一大堆编译的东西都解决不了 - -!

要先安装epel这个yum源，netCDF4和相关的在这个源中可以获得

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yum -y install epel-release

检测是否安装成功可以通过下面的命令检查，如果有版本信息的打印说明安装成功

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rpm -q epel-release

接下来就可以安装netCDF4-python和相关的库了

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yum install netcdf4-python

Ubuntu ：

可以安完的，在stackoverflow上可以找到一些，试的试的就可以了，两次各花半天，下次遇到再上图

有 apt-get install Python-netcdf4

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sudo ldconfig on all relevant

sudo apt-get install libhdf5-serial-dev

sudo apt-get install libnetcdf-dev

sudo apt-get install libnetcdf4

sudo apt-get install python-netcdf

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</pre><pre code_snippet_id="1705134" snippet_file_name="blog_20160602_7_674880" name="code" class="plain">

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祝大家：

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>>>import netCDF4

>>>