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Dear R colleagues,
consider my data.frame named "df" with 3 columns - being level,
prevalence and sensitivity - and 7 rows of data (see dump below).
df <-
structure(list(level = structure(1:7, .Label = c("0", "1", "10",
"100", "1010", "11", "110"), class = "factor"), prevalence =
structure(c(4L,
2L, 3L, 5L, 6L, 1L, 7L), .Label = c("0.488", "0.5", "0.754",
"0.788", "0.803", "0.887", "0.905"), class = "factor"), sensitivity =
structure(c(6L,
1L, 5L, 4L, 3L, 2L, 1L), .Label = c("0", "0.05", "0.091", "0.123",
"0.327", "0.933"), class = "factor")), .Names = c("level", "prevalence",
"sensitivity"), class = "data.frame", row.names = c(NA, -7L))
I'd like to order df by a vector which is NOT contained in the
data.frame. Let's call this vector desiredOrder (see dump below).
desiredOrder <- c("0", "1", "10", "100", "11", "110", "1010")
So after sorting, the order of the level column (df$level) should be in
the order of the vector desiredOrder (as well a the associated data in
the other columns).
I know that this is not an easy task to achieve by order(...) as the
order of desiredOrder isn't a natural one. But I would expect both of
the following to work:
## using match
df[match(df$level,desiredOrder),]
## using factor
df[factor(df$level,levels=desiredOrder),]
Unfortunately the result isn't what I expected: I get a data.frame with
the level column in the order 0,1,10,100,110,1010,11 instead of the
order in desiredOrder (0,1,10,100,11,110,1010).
Does anybody see, what I am doing wrong?
I'd appreciate any kind of help very much!
Best regards, Felix
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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Your desiredOrder vector is a vector of strings. Convert it to numeric and it should work.
---------------------------------------------------------------------------
Jeff Newmiller The ..... ..... Go Live...
DCN:< [hidden email]> Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---------------------------------------------------------------------------
Sent from my phone. Please excuse my brevity.
drflxms < [hidden email]> wrote:
>Dear R colleagues,
>
>consider my data.frame named "df" with 3 columns - being level,
>prevalence and sensitivity - and 7 rows of data (see dump below).
>
>df <-
>structure(list(level = structure(1:7, .Label = c("0", "1", "10",
>"100", "1010", "11", "110"), class = "factor"), prevalence =
>structure(c(4L,
>2L, 3L, 5L, 6L, 1L, 7L), .Label = c("0.488", "0.5", "0.754",
>"0.788", "0.803", "0.887", "0.905"), class = "factor"), sensitivity =
>structure(c(6L,
>1L, 5L, 4L, 3L, 2L, 1L), .Label = c("0", "0.05", "0.091", "0.123",
>"0.327", "0.933"), class = "factor")), .Names = c("level",
>"prevalence",
>"sensitivity"), class = "data.frame", row.names = c(NA, -7L))
>
>I'd like to order df by a vector which is NOT contained in the
>data.frame. Let's call this vector desiredOrder (see dump below).
>
>desiredOrder <- c("0", "1", "10", "100", "11", "110", "1010")
>
>So after sorting, the order of the level column (df$level) should be in
>the order of the vector desiredOrder (as well a the associated data in
>the other columns).
>I know that this is not an easy task to achieve by order(...) as the
>order of desiredOrder isn't a natural one. But I would expect both of
>the following to work:
>
>## using match
>df[match(df$level,desiredOrder),]
>
>## using factor
>df[factor(df$level,levels=desiredOrder),]
>
>Unfortunately the result isn't what I expected: I get a data.frame with
>the level column in the order 0,1,10,100,110,1010,11 instead of the
>order in desiredOrder (0,1,10,100,11,110,1010).
>
>Does anybody see, what I am doing wrong?
>I'd appreciate any kind of help very much!
>Best regards, Felix
>
>______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help>PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html>and provide commented, minimal, self-contained, reproducible code.
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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|
Jeff,
thanks a lot for your quick reply and the hint!
Meanwhile I found a solution that works - at least for my case ;)
The code to get the job done is
df[order(match(df$level,desiredOrder)),]
So we seem in need of one order statement more. I found this solution
doing it stepwise:
## sorting the levels of the level column in the data.frame
df$level <- factor(df$level,levels=desiredOrder)
## sorting the data frame by the newly sorted level column
df[order(df$level),]
Maybe this solution is of a help for someone else as well?
But honestly I still do not exactly understand why
df[match(df$level,desiredOrder),] doesn't work...
Cheers, Felix
Am 29.12.11 10:58, schrieb Jeff Newmiller:
> Your desiredOrder vector is a vector of strings. Convert it to numeric and it should work.
> ---------------------------------------------------------------------------
> Jeff Newmiller The ..... ..... Go Live...
> DCN:< [hidden email]> Basics: ##.#. ##.#. Live Go...
> Live: OO#.. Dead: OO#.. Playing
> Research Engineer (Solar/Batteries O.O#. #.O#. with
> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> ---------------------------------------------------------------------------
> Sent from my phone. Please excuse my brevity.
>
> drflxms < [hidden email]> wrote:
>
>> Dear R colleagues,
>>
>> consider my data.frame named "df" with 3 columns - being level,
>> prevalence and sensitivity - and 7 rows of data (see dump below).
>>
>> df <-
>> structure(list(level = structure(1:7, .Label = c("0", "1", "10",
>> "100", "1010", "11", "110"), class = "factor"), prevalence =
>> structure(c(4L,
>> 2L, 3L, 5L, 6L, 1L, 7L), .Label = c("0.488", "0.5", "0.754",
>> "0.788", "0.803", "0.887", "0.905"), class = "factor"), sensitivity =
>> structure(c(6L,
>> 1L, 5L, 4L, 3L, 2L, 1L), .Label = c("0", "0.05", "0.091", "0.123",
>> "0.327", "0.933"), class = "factor")), .Names = c("level",
>> "prevalence",
>> "sensitivity"), class = "data.frame", row.names = c(NA, -7L))
>>
>> I'd like to order df by a vector which is NOT contained in the
>> data.frame. Let's call this vector desiredOrder (see dump below).
>>
>> desiredOrder <- c("0", "1", "10", "100", "11", "110", "1010")
>>
>> So after sorting, the order of the level column (df$level) should be in
>> the order of the vector desiredOrder (as well a the associated data in
>> the other columns).
>> I know that this is not an easy task to achieve by order(...) as the
>> order of desiredOrder isn't a natural one. But I would expect both of
>> the following to work:
>>
>> ## using match
>> df[match(df$level,desiredOrder),]
>>
>> ## using factor
>> df[factor(df$level,levels=desiredOrder),]
>>
>> Unfortunately the result isn't what I expected: I get a data.frame with
>> the level column in the order 0,1,10,100,110,1010,11 instead of the
>> order in desiredOrder (0,1,10,100,11,110,1010).
>>
>> Does anybody see, what I am doing wrong?
>> I'd appreciate any kind of help very much!
>> Best regards, Felix
>>
>> ______________________________________________
>> [hidden email] mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html>> and provide commented, minimal, self-contained, reproducible code.
>
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
|
|
|
drflxms wrote
Dear R colleagues,
consider my data.frame named "df" with 3 columns - being level,
prevalence and sensitivity - and 7 rows of data (see dump below).
df <-
structure(list(level = structure(1:7, .Label = c("0", "1", "10",
"100", "1010", "11", "110"), class = "factor"), prevalence =
structure(c(4L,
2L, 3L, 5L, 6L, 1L, 7L), .Label = c("0.488", "0.5", "0.754",
"0.788", "0.803", "0.887", "0.905"), class = "factor"), sensitivity =
structure(c(6L,
1L, 5L, 4L, 3L, 2L, 1L), .Label = c("0", "0.05", "0.091", "0.123",
"0.327", "0.933"), class = "factor")), .Names = c("level", "prevalence",
"sensitivity"), class = "data.frame", row.names = c(NA, -7L))
I'd like to order df by a vector which is NOT contained in the
data.frame. Let's call this vector desiredOrder (see dump below).
desiredOrder <- c("0", "1", "10", "100", "11", "110", "1010")
So after sorting, the order of the level column (df$level) should be in
the order of the vector desiredOrder (as well a the associated data in
the other columns).
I know that this is not an easy task to achieve by order(...) as the
order of desiredOrder isn't a natural one. But I would expect both of
the following to work:
## using match
df[match(df$level,desiredOrder),]
## using factor
df[factor(df$level,levels=desiredOrder),]
Unfortunately the result isn't what I expected: I get a data.frame with
the level column in the order 0,1,10,100,110,1010,11 instead of the
order in desiredOrder (0,1,10,100,11,110,1010).
Does anybody see, what I am doing wrong?
Try this:
df[match(desiredOrder,df$level),]
Berend
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drflxms wrote
Jeff,
thanks a lot for your quick reply and the hint!
Meanwhile I found a solution that works - at least for my case ;)
The code to get the job done is
df[order(match(df$level,desiredOrder)),]
So we seem in need of one order statement more. I found this solution
doing it stepwise:
## sorting the levels of the level column in the data.frame
df$level <- factor(df$level,levels=desiredOrder)
## sorting the data frame by the newly sorted level column
df[order(df$level),]
Maybe this solution is of a help for someone else as well?
But honestly I still do not exactly understand why
df[match(df$level,desiredOrder),] doesn't work...
Read carefully
?match
and then do
match(df$level,desiredOrder)
match(desiredOrder,df$level)
and look carefully at the results. Then it should be clear.
Berend
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