w(df2)[,j], j]
}
})
# user system elapsed
# 8.508 0.000 8.523 colN_OF<- ncol(df3[,grep("OF",colnames(df3))])
system.time({
indx1<- unlist(df3[,grep("OF",colnames(df3))],use.names=FALSE)
indx1[rep(seq(nrow(df3)),colN_OF) %in%
1995:2000][indx1[rep(seq(nrow(d
- val1[indx1+seq_along(indx1)]
colnames(df3)[602:901]<- colnames(df2)[602:901]
})
# user system elapsed
# 0.568 0.000 0.569
identical(df2,df3)
#[1] TRUE
A.K.
- Original Message -
From: arun
To: Ira Sharenow
Cc:
Sent: Sunday, September 22, 2013 1:28 AM
Subject: Re:
d your approach.
Ira
On 9/21/2013 8:30 PM, arun wrote:
Hi Ira, This info was not provided before. In fact, I wanted to ask about this
in cases where the last row entries are not 0. So, you wanted the newPrices to
be NA's for the corresponding rows, right??
Fro
ot;,colnames(df3))],use.names=FALSE)
df3[,602:901]<- val1[indx1+seq_along(indx1)]
colnames(df3)[602:901]<- colnames(df2)[602:901]
})
# user system elapsed
# 0.600 0.016 0.616
identical(df2,df3)
#[1] TRUE
A.K.
________________
From: Ira Sharenow
To: arun
Sent:
be confusing as: with(df1,P1+OF1)
# [1] 13 14 16 18 16 17 18 18 19 19
with(df1,P2+OF2)
# [1] 105 104 106 105 105 107 108 108 109 109 which is the same as:
df1$newPrice1
# [1] 13 14 16 18 16 17 18 18 19 19
df1$newPrice2
# [1] 105 104 106 105 105 107 108 108 109 109 - Original Message -
From:
tfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Ira Sharenow
> Sent: Saturday, September 21, 2013 8:31 AM
> To: r-help@r-project.org
> Subject: [R] Obtaining data
I have a large data frame with 2,000 rows and 600 columns. I can write
loops to solve a smaller problem, but I need a better strategy for this
data frame.
Below is a simple example with just two stocks.
In the data frame, each row represents a trading day. The first column
is dates. The next g
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