exp

Oracle数据导入导出imp/exp 功能：Oracle数据导入导出imp/exp就相当与oracle数据还原与备份。 大多情况都可以用Oracle数据导入导出完成数据的备份和还原（不会造成数据的丢失）。 Oracle有个好处，虽然你的电脑不是服务器，但是你装了oracle客户端，并建立了连接（通过Net Configuration Assistant添加正确的服务命名，其实你可以想成是客户端与服务器端 修了条路，然后数据就可以被拉过来了）这样你可以把数据导出到本地，虽然可能服务器离你很远。 你同样可以把dmp文件从本地导入到远处的数据库服务器中。 利用这个功能你可以构建俩个相同的数据库，一个用来测试，一个用来正式使用。 执行环境：可以在SQLPLUS.EXE或者DOS（命令行）中执行，DOS中可以执行时由于 在oracle 8i 中 安装目录\$ora10g\BIN被设置为全局路径，该目录下有EXP.EXE与IMP.EXE文件被用来执行导入导出。 Oracle用java编写，我想SQLPLUS.EXE、EXP.EXE、IMP.EXE这俩个文件是被包装后的类文件。SQLPLUS.EXE调用EXP.EXE、IMP.EXE他们所包裹的类，完成导入导出功能。 下面介绍的是导入导出的实例，向导入导出看实例基本上就可以完成，因为导入导出很简单。 查看帮助： exp -help | imp

I have a table with 2 column: No. Name Serial 1 Tom 1 2 Bob 5 3 Don 3 4 Jim 6 I want to add a column whose a content is multiply Serial column like this: No. Name Serial Multiply 1 Tom 2 2 2 Bob 5 10 3 Don 3 30 4 Jim 6 180 How can i do that? Oh, this is a pain. Most databases do not support a product aggregation function. You can emulate it with logs and powers. So, something like this might work: select t.*, (select exp(sum(log(serial))) from table t2 where t2.no <= t.no ) as cumeProduct from table t; Note that log() might be called ln() in some databases. Also, this works for positive

I'm trying to generate a beautiful legend in R plots. I have a factor=1e-5 , that should appear nicely formatted in the legend. I found a nice function in the package sfsmisc , that transforms numbers to expressions. To add this expression to my bquote command, it seems that I need to transform itto a call. unfortunately, there are braces added at the end of the string ( 10^-5() ). Is there a way to avoid the addition of thoses braces? Or is there even an easier way to transform numbers to plotmaths commands for their use in legends? (without doing it manually) factor = 1e-5 alpha = 1:10 omega

If I want to take the product of a list of floating point numbers, what's the worst-case/average-case precision lost by adding their logs and then taking exp of the sum as opposed to just multiplying them. Is there ever a case when this is actually more precise? tmyklebu Absent any overflow or underflow shenanigans, if a and b are floating-point numbers, then the product a*b will be computed to within a relative error of 1/2 ulp. A crude bound on the relative error after multiplying a chain of N double s therefore results in answer off by a factor of at most (1 - epsilon/2) -N , which is about

As in title, I need to perform numpy.exp on a very large ndarray, let's say ar , and store the result in ar itself. Can this operation be performed in-place? You can use the optional out argument of exp : a = np.array([3.4, 5]) res = np.exp(a, a) print(res is a) print(a) Output: True [ 29.96410005 148.4131591 ] exp(x[, out]) Calculate the exponential of all elements in the input array. Returns out : ndarray Output array, element-wise exponential of x . Here all elements of a will be replaced by the result of exp . The return value res is the same as a . No new array is created kmario23 Mike