这里,
http://www.ffiec.gov/census/report.aspx?year=2011&state=01&report=demographic&msa=11500
这里有张桌子.我的目标是提取表并将其保存到csv文件中.我写了一个代码:
import urllib
import os
web = urllib.urlopen("http://www.ffiec.gov/census/report.aspx?year=2011&state=01&report=demographic&msa=11500")
s = web.read()
web.close()
ff = open(r"D:\ex\python_ex\urllib\output.txt", "w")
ff.write(s)
ff.close()
我从这里输了.谁可以提供帮助呢？谢谢!
. sysuse auto, clear
(1978 Automobile Data)
. di "I am getting some summary statistics for PRICE"
I am getting some summary statistics for PRICE
. su price
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
price | 74 6165.257 2949.496 3291 15906
.
end of do-file
我想隐藏命令行,并只显示如下结果:
I am getting some summary statistics for PRICE
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------------------
price | 74 6165.257 2949.496 3291 15906
我怎样才能做到这一点？谢谢.
请看下面的小工作示例:
#### Pseudo data
nobs1 <- 4000
nobs2 <- 5000
mylon1 <- runif(nobs1, min=0, max=1)-76
mylat1 <- runif(nobs1, min=0, max=1)+37
mylon2 <- runif(nobs2, min=0, max=1)-76
mylat2 <- runif(nobs2, min=0, max=1)+37
#### define a distance function
thedistance <- function(lon1, lat1, lon2, lat2) {
R <- 6371 # Earth mean radius [km]
delta.lon <- (lon2 - lon1)
delta.lat <- (lat2 - lat1)
a <- sin(delta.lat/2)^2 + cos(lat1) * cos(lat2) * sin(delta.lon/2)^2
c <- 2 * asin(min(1,sqrt(a)))
d = R * c
return(d) …我有一个简单的一阶微分方程系统要在Matlab中求解.它看起来如下:
function passing
y0 = [0; 0];
t = 0:0.05:1;
y = myprocedure(@myfunc,0.05,t,y0);
myans = y'
end
function f = myfunc(t,y)
f = [-y(1) + y(2);
-0.8*t + y(2)];
end
function y=myprocedure(f,h,t,y0)
n = length(t);
y = zeros(length(y0),n);
y(:,1) = y0;
for i=1:n-1
k1 = f(t(i),y(:,i));
k2 = f(t(i)+h/2, y(:,i)+h/2*k1);
k3 = f(t(i)+h/2, y(:,i)+h/2*k2);
k4 = f(t(i)+h, y(:,i)+h*k3);
y(:,i+1) = y(:,i)+h/6*(k1+2*k2+2*k3+k4);
end
end
所以,运行pass.m文件后,我得到了结果:
>> passing
myans =
0 0
-0.0000 -0.0010
-0.0001 -0.0041
-0.0005 -0.0095
-0.0011 -0.0171
-0.0021 -0.0272 …在d:\文件夹中,我有很多Stata文件,例如data_aa_1.dta,data_aa_2.dta,data_aa_3.dta data_bb_1.dta,data_bb_2.dta,data_bb_3.dta,data_cc_1.dta ....我想转换这些文件获得与R中的dta文件一样多的数据帧.所以,我相信我必须循环c("aa","bb","cc")和c(1:3).我尝试过以下内容:
library(foreign)
for(i in c("aa","bb","cc"){
for (j in 1:3){
data_[i]_[j] <-read.dta("d:/folder/data_[i]_[j].dta")
}
}
但是,这看起来不对 - 当然.
任何帮助将不胜感激.
谢谢!
我是Fortran编程的新手.我有两个.f90文件.
fmat.f90
function fmat(t,y)
implicit none
real::t
real::y(2)
real::fmat(2)
fmat(1) = -2*t+y(1)
fmat(2) = y(1)-y(2)
end function fmat
并且,main.f90看起来像:
program main
implicit none
real::t
real::y(2)
real::fmat(2)
real::k(2)
t=0.1
y(1)=0.5
y(2)=1.4
k=fmat(t,y)
write(*,*) k
end program main
所以,我期待0.3-0.9.但我不断收到以下错误消息:
ifort fmat.f90 main.f90
main.f90(13): error #6351: The number of subscripts is incorrect. [FMAT]
k=fmat(t,y)
--^
compilation aborted for main.f90 (code 1)
任何帮助表示赞赏!
!====编辑====
我感谢马克的答案.我实际上可以使用"子程序"方法编译单独的文件而没有任何错误.
main.f90时
program main
implicit none
real::t
real::y(2)
real::k(2)
t=0.1
y(1)=0.5
y(2)=1.4
call fmat_sub(t,y,k)
write(*,*) k
end program main
fmat_sub.f90 …
我正在尝试编写一个简单的Fortran代码来计算h = g(f(x)).x是长度= 2的向量.
module m1
implicit none
contains
function f(x)
implicit none
real::f(2),x(2)
f(1)=x(1)-x(2)
f(2)=exp(x(1))-x(2)**2
end function f
function g(ff)
implicit none
real::g(2),x1(2),ffreslt(2)
interface
function ff(x)
implicit none
real::x(2),ff(2)
end function ff
end interface
ffreslt=ff(x1)
g(1)=1-ffreslt(1)
g(2)=2*ffreslt(1)**2-3*ffreslt(2)+4.2
end function g
end module m1
program hgf
use m1
implicit none
real::x1(2),h(2)
x1 = (/0.55,2.47/)
h = g(f(x1))
write(*,*) h
end program hgf
但是,我收到此错误消息:
h = g(f(x1))
1
Error: Actual parameter 'ff' at <1> is not a PROCEDURE
我错过了什么吗？谢谢.