Hello,
You're right, I carelessly coded this.
which.max returns the index to the first maximum of a vector, while the
comparison of a vector with its max() returns an index to all vector
elements.
Às 23:27 de 03/12/21, Bert Gunter escreveu:
Perhaps you meant to point this out, but the cfs[which.max(cfs)] and
cfs == ... are not the same:
x <- rep(1:2,3)
x
[1] 1 2 1 2 1 2
x[which.max(x)]
[1] 2
x[x==max(x)]
[1] 2 2 2
So maybe your point is: which does the OP want (in case there are
repeated maxes)? I suspect the == forms, but ...?
Bert Gunter
On Fri, Dec 3, 2021 at 2:56 PM Rui Barradas <ruipbarra...@sapo.pt> wrote:
Hello,
Inline.
Às 22:08 de 03/12/21, Rich Shepard escreveu:
On Fri, 3 Dec 2021, Rich Shepard wrote:
I find solutions when the data_frame is grouped, but none when it's not.
Thanks, Bert. ?which.max confirmed that's all I need to find the maximum
value.
Now I need to read more than ?filter to learn why I'm not getting the
relevant row with:
which.max(pdx_disc$cfs)
[1] 8054
This is the *index* for which cfs is the first maximum, not the maximum
value itself.
filter(pdx_disc, cfs == 8054)
Therefore, you probably want any of
Should be "one of", not "any of"
Rui Barradas
filter(pdx_disc, cfs == cfs[8054])
filter(pdx_disc, cfs == cfs[which.max(cfs)])
filter(pdx_disc, cfs == max(cfs)) # I find this one better, simpler
Hope this helps,
Rui Barradas
# A tibble: 0 × 9
# … with 9 variables: site_nbr <chr>, year <int>, mon <int>, day <int>,
# hr <dbl>, min <dbl>, tz <chr>, cfs <dbl>, sampdt <dttm>
Regards,
Rich
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