Download economic official data from a Central Bank web page

Question

I've been searching for a while any answer to my question. I read this, this and this and some others related but I still get not answer.

My problem is quite simple (I hope it is) but the answer is not (at least for myself), I want to import some economic data from this web which is an indicator for Nicaraguan economic activity measured each month, so far I've tried this:

library(XML)
u <- "http://www.bcn.gob.ni/estadisticas/trimestrales_y_mensuales/siec/datos/4.IMAE.htm"
u <- htmlParse(u,encoding="UTF-8")
imae <- readHTMLTable(doc=u, header=T)
imae

library(httr)
u2 <- "http://www.bcn.gob.ni/estadisticas/trimestrales_y_mensuales/siec/datos/4.IMAE.htm"
page <- GET(u2, user_agent("httr"))
x <- readHTMLTable(text_content(page), as.data.frame=TRUE)

with no success as you can imagine. The first chunk of code gave me this output

   $`NULL`
                                  BANCO CENTRAL DE NICARAGUA    NA    NA    NA   NA   NA   NA   NA   NA   NA   NA    NA    NA       NA
1                                                             <NA>  <NA>  <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA>  <NA>  <NA>     <NA>
2 <U+633C><U+3E64>ndice Mensual de Actividad Económica(IMAE)  <NA>  <NA>  <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA>  <NA>  <NA>     <NA>
3                                           (Base: 1994=100)  <NA>  <NA>  <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA>  <NA>  <NA>     <NA>
4                                                             <NA>  <NA>  <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA>  <NA>  <NA>     <NA>
5                                                        Año   Ene   Feb   Mar  Abr  May  Jun  Jul  Ago  Sep  Oct   Nov   Dic Promedio
6                                                             <NA>  <NA>  <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA>  <NA>  <NA>     <NA>
7                                                       1994 101.6 107.6 100.1 95.7 94.7 92.8 92.1 96.8 98.5 97.4 101.7 121.1    100.0
8                                               Fuente: BCN.  <NA>  <NA>  <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA>  <NA>  <NA>     <NA>

I tried using skip.rows=1:5 but it doesn't really change the main result which is too much NA. Is there anybody who can shed some light on this question?

The expected result is a data.frame with the information shown in this web

I don't know much about web scraping, but the page you're trying to scrape is very poorly written. For instance, there are only 9 `` tags and 27 `` tags. Perhaps that is part of the problem. — A5C1D2H2I1M1N2O1R2T1, Oct 05 '12 at 19:46

A5C1D2H2I1M1N2O1R2T1 · Accepted Answer · 2012-10-06T18:41:52.877

As I mentioned in my comment, the problem is most likely arising because of the poorly coded table.

You can try an approach something like the following (tested on Ubuntu using RStudio). It requires that you have wget and html tidy installed. If you don't want to install these useful programs, jump to the updated part of this answer.

Download the page and "tidy" it up.

system("wget http://www.bcn.gob.ni/estadisticas/trimestrales_y_mensuales/siec/datos/4.IMAE.htm")
system("tidy 4.IMAE.htm > new.html")

Proceed with R as you normally would

library(XML)
u <- htmlParse("new.html")
imae <- readHTMLTable(u)

If we view the output of the above readHTMLTable, we would see that we need to skip a few rows. Let's run it again:

imae <- readHTMLTable(u, skip.rows=c(1:5, 7, 27, 28), header=TRUE)
imae
# $`NULL`
#     Año   Ene   Feb   Mar   Abr   May   Jun   Jul   Ago   Sep   Oct   Nov   Dic Promedio
# 1  1994 101.6 107.6 100.1  95.7  94.7  92.8  92.1  96.8  98.5  97.4 101.7 121.1    100.0
# 2  1995 113.2 105.0 113.6  98.0 100.9  95.4  99.8 101.5 108.3 107.1 107.6 133.2    107.0
# 3  1996 123.6 116.0 109.1 107.3  94.8 101.2 100.7 115.3 110.6 112.7 117.5 137.7    112.2
# 4  1997 133.4 115.9 117.4 118.8 120.4 108.2 107.4 111.1 120.3 117.7 119.5 142.3    119.4
# 5  1998 131.4 120.4 127.9 118.4 130.2 116.5 122.1 129.7 127.3 127.5 112.7 156.6    126.7
# 6  1999 146.0 139.6 146.9 134.8 140.6 131.8 130.6 128.3 128.9 131.8 142.7 172.6    139.5
# 7  2000 157.8 142.1 147.3 138.5 137.7 135.7 128.9 131.2 141.7 143.0 156.6 191.2    146.0
# 8  2001 163.3 143.8 154.8 141.5 147.6 134.0 135.7 143.3 138.2 138.8 145.3 187.3    147.8
# 9  2002 152.1 144.7 143.3 142.1 143.1 131.9 136.1 145.7 146.4 147.8 157.5 185.0    148.0
# 10 2003 159.3 151.4 149.1 142.7 139.7 139.1 145.6 147.8 154.9 158.4 157.8 195.7    153.5
# 11 2004 172.8 157.1 166.9 153.6 161.2 150.5 155.3 153.3 156.6 155.6 167.7 213.0    163.6
# 12 2005 183.1 170.6 173.6 158.7 160.8 158.5 158.8 168.7 165.8 165.4 178.4 218.8    171.8
# 13 2006 187.7 177.8 185.6 161.8 166.4 163.2 164.7 175.1 175.1 185.3 189.6 231.2    180.3
# 14 2007 200.1 184.1 196.5 180.1 169.7 171.4 181.6 180.9 173.0 182.8 202.0 236.7    188.2
# 15 2008 205.4 194.4 193.1 205.9 171.0 174.8 181.3 190.7 183.1 182.7 182.5 244.7    192.5
# 16 2009 195.7 191.0 190.8 177.0 168.1 172.6 179.2 185.6 178.9 181.4 191.3 241.4    187.7
# 17 2010 195.2 193.7 205.1 185.2 179.3 190.1 191.6 190.0 193.5 197.6 210.9 266.0    199.8
# 18 2011 213.9 207.4 217.3 198.7 196.1 198.8 191.9 210.0 203.7 207.9 217.3 274.5    211.5
# 19 2012 233.6                                                                      233.6

Update: A little function to help out

If you can live with having to do some text cleanup for the accented characters, the W3C offers an online implementation of html tidy. This allows you to write a basic function like the following:

tidyHTML <- function(URL) {
  require(XML)
  URL = gsub("/", "%2F", URL)
  URL <- gsub(":", "%3A", URL)
  URL <- paste("http://services.w3.org/tidy/tidy?docAddr=", URL, sep = "")
  htmlParse(URL)
}

Usage is simple:

u <- tidyHTML("http://www.bcn.gob.ni/estadisticas/trimestrales_y_mensuales/siec/datos/4.IMAE.htm")
readHTMLTable(u)

It works perfect!!! thank you. Nice solution using `tidyHTML` I found very useful services.w3.org — Jilber Urbina, Oct 07 '12 at 16:59

ako · Answer 2 · 2012-10-05T22:15:05.630

This is a bit of a hack job which sort of works if the table is not well enough structured as in the other responses you linked to. But it is really more of a one-off that works if the format doesn't change, but beware--can be risky. There are likely more general solutions that folks can add.

require(RCurl)
require(XML)
u <- "http://www.bcn.gob.ni/estadisticas/trimestrales_y_mensuales/siec/datos/4.IMAE.htm"

webpage <- getURL(u)
lines <- readLines(tc <- textConnection(webpage)); close(tc)
pagetree <- htmlTreeParse(lines, error=function(...){}, useInternalNodes = TRUE)

# parse tree by any tables
x <- xpathSApply(pagetree, "//*/table", xmlValue)  

# remove white space and such w/ regexes
unlisted <- unlist(strsplit(x, "\n"))
notabs <- gsub("\t","",unlisted)
nowhitespace <- sub("^[[:space:]]*(.*?)[[:space:]]*$", "\\1", notabs, perl=TRUE)
data <- nowhitespace[!(nowhitespace %in% c("", "|"))]

here comes the dodgy part:

months<-data[5:16]
data_out<-data[18:(length(data)-4)] #omits 2012 data to easily fit structure argument

finalhack<-data.frame(t(structure(
data_out,dim = c(14,18),.Dimnames =     
list(c('year',months,'index'),seq(1994,2011)))))

Note also that this omits data in the end to conform to the dimensions of the `structure()` call, requiring a vector of the proper length, in this case, 14 by 18. The `seq(1995,2012)` is obviously not ideal, too. — ako, Oct 05 '12 at 21:24

Download economic official data from a Central Bank web page

2 Answers2

Update: A little function to help out