I don't think I understand you entirely, but here are a couple
things.
You need to shorten the loop so you don't run off the end.
You basically just write the code as you've stated it except that
"and" is && for single values and & for vectors
You don't say what should happen if the test is not true.
You can probably use "ifelse" instead of a loop.
iseq <- 2:(length(x.dif) - 2)
test <- x.dif[iseq-1] <= 0 & x.dif[iseq] > 0 & x.dif[iseq + 2]
> 0
ifelse(test, iseq + 2, NA)
I suspect that the last term in test should be i+1 and not i+2 ?
You can see S Poetry for details.
Patrick Burns
Burns Statistics
patrick@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")
ebashi wrote:
>I'll appreciate if some one can help me with the
>following loop. This is the logic of the loop,
>if we have the following data;
>
>
>>x.df
>>
>>
> x.dif
> . .
> . .
>102 0.00
>103 0.42
>104 0.08
>105 0.00
>106 0.00
>107 0.00
>108 -0.16
>109 -0.34
>110 0.00
>111 -0.17
>112 -0.33
>113 0.00
>114 0.00
>115 0.00
>116 0.33
>117 0.17
>118 0.00
> . .
> . .
>I'm trying to find i's where
> for (i in 2:length(x.dif))
> if (x.dif[i-1]<=0 and x.dif[i]>0 and x.dif[i+2]>0)
> it would return i+2 to me,
>How can I turn this to a right format as a loop.(I
>can't figure out the syntax)
>
>Cheers,
>
>Sean
>
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