Displaying 6 results from an estimated 6 matches for "c_d".
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2013 Mar 02
3
if value is in vector, perform this function
...that makes sense. Here's what I have (and it doesn't work):
params = c(1, 0.15, 0.164, 1)
init = c(1.5, 0.05)
t=seq(1,60, by=0.5) #all time values, experiment ran for 60 days
#feeding sequence - every "3.5 days"
feed_days = seq(1,60,by=3.5)
Daphnia <- function(t,x,params){
C_D = x[2];
C_A = 0;
for(t %in% feed_days){
if t == TRUE {
C_A = 1.5
}
else{
C_A = 0
}}
gamma = params[1]; m_D = params[2]; K_q = params[3]; q_max = params[4];
M_D = m_D * C_D
I_A = (C_D * q_max * C_A) / (K_q + C_A)
r_D = gamma * I_A
return(
list(c(
- I_A,
r_D - M_D
)))
}
libra...
2012 Aug 09
1
Zoo object problem: problem when I attempt to create a zoo object of only one column
...ct "NULL"? It should be one, shouldn't it? (The following
example does not involve the creation of a zoo object; in reality, similar
problems are encountered when I create a zoo object)
> temp5<-read.csv("A_Consumption.csv", header=TRUE)> temp5[1:3,] TIME C C_D C_ND
1 196101 70345 1051 69294
2 196102 61738 905 60833
3 196103 63838 860 62978> temp6<-temp5[,2:ncol(temp5)]> temp6[1:3,]
C C_D C_ND
1 70345 1051 69294
2 61738 905 60833
3 63838 860 62978> colnames(temp6)[1] "C" "C_D" "C_ND">
temp7<-re...
2012 Aug 10
2
Zoo object problem: Find the column name of a univariate zoo object
...4 1
>
> R> dim(z) <- c(NROW(z), NCOL(z))
> R> z
>
> 1 0.8414710
> 2 0.9092974
> 3 0.1411200
> 4 -0.7568025
> R> dim(z)
> [1] 4 1
>
>
> temp5<-read.csv("A_**Consumption.csv", header=TRUE)> temp5[1:3,] TIME
>>> C C_D C_ND
>>>
>> 1 196101 70345 1051 69294
>> 2 196102 61738 905 60833
>> 3 196103 63838 860 62978> temp6<-temp5[,2:ncol(temp5)]> temp6[1:3,]
>> C C_D C_ND
>> 1 70345 1051 69294
>> 2 61738 905 60833
>> 3 63838 860 62978> colnames(te...
2010 Oct 26
4
divide column in a dataframe based on a character
Hello,
If I have a dataframe:
example(data.frame)
zz<-c("aa_bb","bb_cc","cc_dd","dd_ee","ee_ff","ff_gg","gg_hh","ii_jj","jj_kk","kk_ll")
ddd <- cbind(dd, group = zz)
and I want to divide the column named group by the "_", how would I do this?
so instead of the first row being
x y fa...
2010 Aug 19
1
Help with Vectors and conditional functions
Good morning,
I have something like this: names(coint_tests) <- apply(b,2,paste, collapse="_") which prints 15 names like: A_B, C_D, E_F, ...
AA,B,C,D.. Are time series. Then there is a vector called coint_tests of length 15 which yields "yes" or "no".
I need to add a function to plot the time series Ai_Bi if the coint_tests vectors gives me a "YES".
I tried: for (i in 1:(length(coint_tests))...
2009 Aug 28
1
problems with strsplit using a split of ' \\\ ' : a regex problem
...a simple little function to do this:
get.first.id.func <- function(vec, splitter){
vec.lst <- strsplit(vec, splitter)
first.func <- function(vec1){vec1[1]}
vec.out <- sapply(vec.lst, first.func)
vec.out
}
For a trivial example, this works:
> a <- c("a_b", "c_d")
> get.first.id.func(a, "_")
[1] "a" "c"
I am running into problems, however, with the real world split of ' \\\ '
I'm not even able to construct a sample vector of my own! Here is what I
get:
> a <- c('a \\\ b', 'a \\\ b')
&...