Correct Syntax for a Loop

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
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)
In this sort of case you can do it without a loop. The
following code (which as written is specific to the precise
question you have asked) is the sort of thing you can use:

  n<-length(x.dif)
  y2<-x.dif[4:n]
  y0<-x.dif[2:(n-2)]
  y1<-x.dif[1:(n-3)]
  which((y1<=0)&(y0>0)&(y2>0))+3

Hoping this helps,
Ted.

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E-Mail: (Ted Harding) <Ted.Harding at nessie.mcc.ac.uk>
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Date: 28-Nov-04                                       Time: 12:10:00
------------------------------ XFMail ------------------------------
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)
Solution to the question:

a) you cannot loop to length(x.dif), x.dif[i+2] does not exist for 
i==length(x.dif).
b) Use && instead of "and" in this case
c) vector operations are much mor

Hint: If you lengthen the vectors appropriately, you can apply the 
comparisons vectorized  rather than in a loop...

Uwe Ligges
  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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ebashi <arshia22 <at> yahoo.com> writes:

: 
: 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)

One way is to convert it to a ts time series so you can use the lag
operator and have everything automatically aligned for you:

x.dif.ts <- ts(x.dif)
bool <- lag(x.dif.ts,-3) <= 0 & lag(x.dif.ts,-2) > 0 & x.dif.ts > 0
time(bool)[bool]