[R] Help R

[Jose Iparraguirre]

As said by arun, the code is not clear.
Ma Teresa, what is it that you actually want to do?
Regards,
José


-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of arun
Sent: 29 August 2013 15:12
To: R help
Subject: Re: [R] Help R



HI,
Your code is not clear:
 for( i in 1: nrow(D))
 {
   for( j in 5:ncol(D))
   {
    D[((D[i,j]/D[i,2])>1.5)]15999)]<-1000000
##  "....1.5)]15999)]"
   } ^^^
  
 } 

D<- structure(list(X. = c(1108L, 1591L, 3408L, 3872L, 5823L, 8099L, 10640L, 12600L, 14680L, 14698L), media = c(22, 86.6, 807, 103.25, 73, 125.16667, 67.33333, 2417, 38, 698.16667), IE.2003 = c(32, 92, 166, 237.2, 160.6, 169, 494.8, 1958, 142.2, 505), IE.2004 = c(63, 122.8, 290, 117.8, 73.6, 206, 168.2, 1871, 46, 482), IE.2005 = c(60, 276.6, 302, 154.4, 133.6, 196, 424.8, 1960, 30, 553), IE.2006 = c(39L, 87L, 322L, 113L, 70L, 161L, 476L, 2383L, 61L, 664L), IE.2007 = c(4, 95, 621, 116, 80, 150, 670.6, 2453, 404, 847), IE.2008 = c(8L, 94L, 1071L, 90L, 74L, 94L, 74L, 2506L, 42L, 800L), IE.2009 = c(16L, 81L, 1301L, 94L, 69L, 72L, 77L, 2758L, 19L, 679L), IE.2010 = c(5, 76, 1225, 1911.2, 72, 78, 51, 2442, 243.8, 646)), .Names = c("X.", "media", "IE.2003", "IE.2004", "IE.2005", "IE.2006", "IE.2007", "IE.2008", "IE.2009", "IE.2010"), class = "data.frame", row.names = c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10")) D[,-c(1:4)][D[,-c(1:4)]/D[,2]>1.5]
# [1]   60.0  276.6  133.6  196.0  424.8   39.0  476.0   61.0  670.6  404.0 #[11] 1301.0 1225.0 1911.2  243.8


A.K.
 






Hi to everyone, 
  
I would like to replace some values in a data.frame (D) 

> str(D)
'data.frame':   116 obs. of  10 variables: 
 $ X.     : int  1108 1591 3408 3872 5823 8099 10640 12600 14680 14698 ... 
 $ media  : num  22 86.6 807 103.2 73 ... 
 $ IE.2003: num  32 92 166 237 161 ... 
 $ IE.2004: num  63 122.8 290 117.8 73.6 ... 
 $ IE.2005: num  60 277 302 154 134 ... 
 $ IE.2006: num  39 87 322 113 70 ... 
 $ IE.2007: num  4 95 621 116 80 ... 
 $ IE.2008: num  8 94 1071 90 74 ... 
 $ IE.2009: num  16 81 1301 94 69 ... 
 $ IE.2010: num  5 76 1225 1911 72 ... 

D 

    X.      media IE.2003 IE.2004 IE.2005 IE.2006 IE.2007 IE.2008 IE.2009 IE.2010
1   1108   22.00000    32.0    63.0    60.0      39     4.0       8      16     5.0
2   1591   86.60000    92.0   122.8   276.6      87    95.0      94      81    76.0
3   3408  807.00000   166.0   290.0   302.0     322   621.0    1071    1301  1225.0
4   3872  103.25000   237.2   117.8   154.4     113   116.0      90      94  1911.2
5   5823   73.00000   160.6    73.6   133.6      70    80.0      74      69    72.0
6   8099  125.16667   169.0   206.0   196.0     161   150.0      94      72    78.0
7  10640   67.33333   494.8   168.2   424.8     476   670.6      74      77    51.0
8  12600 2417.00000  1958.0  1871.0  1960.0    2383  2453.0    2506    2758  2442.0
9  14680   38.00000   142.2    46.0    30.0      61   404.0      42      19   243.8
10 14698  698.16667   505.0   482.0   553.0     664   847.0     800     679   646.0 



WHat I really want to do is: 

 for( i in 1: nrow(D))
 {
   for( j in 5:ncol(D))
   {
    D[((D[i,j]/D[i,2])>1.5)]15999)]<-1000000
   } 
  
 }

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