df = data.frame(a = c(1, 1, 2, 2), b = c(1, 1, 2, 3), value = 1:4)
df
a b value
1 1 1 1
2 1 1 2
3 2 2 3
4 2 3 4
The approach to drop duplicates in the first and second columns has as a consequence the arbitrary choice of 'value' for the duplicate entries -- why chose a value of '1' rather than '2' (or the average of 1 and 2, or a list containing all possible values, or...) for the rows duplicated in columns a and b?
df[!duplicated(df[,1:2]),]
a b value
1 1 1 1
3 2 2 3
4 2 3 4
In base R one might
aggregate(value ~ a + b, df, mean)
a b value
1 1 1 1.5
2 2 2 3.0
3 2 3 4.0
aggregate(value ~ a + b, df, list)
a b value
1 1 1 1, 2
2 2 2 3
3 2 3 4
but handling several value-like columns would be hard(?)
Using library(dplyr), I have
group_by(df, a, b) %>% summarize(mean_value = mean(value))
# A tibble: 3 x 3
# Groups: a [2]
a b mean_value
<dbl> <dbl> <dbl>
1 1 1 1.5
2 2 2 3
3 2 3 4
or
group_by(df, a, b) %>% summarize(values = list(value))
# A tibble: 3 x 3
# Groups: a [2]
a b values
<dbl> <dbl> <list>
1 1 1 <int [2]>
2 2 2 <int [1]>
3 2 3 <int [1]>
summarizing multiple columns with dplyr
df$v1 = 1:4
df$v2 = 4:1
group_by(df, a, b) %>% summarize(v1_mean = mean(v1), v2_median = median(v2))
# A tibble: 3 x 4
# Groups: a [2]
a b v1_mean v2_median
<dbl> <dbl> <dbl> <dbl>
1 1 1 1.5 3.5
2 2 2 3 2
3 2 3 4 1
I do not know how performant this would be with data of your size.
Martin Morgan
?On 11/8/19, 1:39 PM, "R-help on behalf of Ana Marija" <r-help-bounces at r-project.org on behalf of sokovic.anamarija at gmail.com> wrote:
Thank you so much!!!
On Fri, Nov 8, 2019 at 11:40 AM Bert Gunter <bgunter.4567 at gmail.com> wrote:
>
> Correction:
> df <- data.frame(a = 1:3, b = letters[c(1,1,2)], d = LETTERS[c(1,1,2)])
> df[!duplicated(df[,2:3]), ] ## Note the ! sign
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Fri, Nov 8, 2019 at 7:59 AM Bert Gunter <bgunter.4567 at gmail.com> wrote:
>>
>> Sorry, but you ask basic questions.You really need to spend some more time with an R tutorial or two. This list is not meant to replace your own learning efforts.
>>
>> You also do not seem to be reading the docs carefully. Under ?unique, it links ?duplicated and tells you that it gives indices of duplicated rows of a data frame. These then can be used by subscripting to remove those rows from the data frame. Here is a reproducible example:
>>
>> df <- data.frame(a = 1:3, b = letters[c(1,1,2)], d = LETTERS[c(1,1,2)])
>> df[-duplicated(df[,2:3]), ] ## Note the - sign
>>
>> If you prefer, the "Tidyverse" world has what are purported to be more user-friendly versions of such data handling functionality that you can use instead.
>>
>>
>> Bert
>>
>> On Fri, Nov 8, 2019 at 7:38 AM Ana Marija <sokovic.anamarija at gmail.com> wrote:
>>>
>>> would you know how would I extract from my original data frame, just
>>> these unique rows?
>>> because this gives me only those 3 columns, and I want all columns
>>> from the original data frame
>>>
>>> chr pos gene_id
>>> 1 chr1 54490 ENSG00000227232
>>> 2 chr1 58814 ENSG00000227232
>>> 3 chr1 60351 ENSG00000227232
>>> 4 chr1 61920 ENSG00000227232
>>> 5 chr1 63671 ENSG00000227232
>>> 6 chr1 64931 ENSG00000227232
>>>
>>> chr pos gene_id pval_nominal pval_ret wl wr META
>>> 1: chr1 54490 ENSG00000227232 0.608495 0.783778 31.62278 21.2838 0.7475480
>>> 2: chr1 58814 ENSG00000227232 0.295211 0.897582 31.62278 21.2838 0.6031214
>>> 3: chr1 60351 ENSG00000227232 0.439788 0.867959 31.62278 21.2838 0.6907182
>>> 4: chr1 61920 ENSG00000227232 0.319528 0.601809 31.62278 21.2838 0.4032200
>>> 5: chr1 63671 ENSG00000227232 0.237739 0.988039 31.62278 21.2838 0.7482519
>>> 6: chr1 64931 ENSG00000227232 0.276679 0.907037 31.62278 21.2838 0.5974800
>>>
>>> On Fri, Nov 8, 2019 at 9:30 AM Ana Marija <sokovic.anamarija at gmail.com> wrote:
>>> >
>>> > Thank you so much! Converting it to data frame resolved the issue!
>>> >
>>> > On Fri, Nov 8, 2019 at 9:19 AM Gerrit Eichner
>>> > <gerrit.eichner at math.uni-giessen.de> wrote:
>>> > >
>>> > > It seems as if dt is not a (base R) data frame but a
>>> > > data table. I assume, you will have to transform dt
>>> > > into a data frame (maybe with as.data.frame) to be
>>> > > able to apply unique in the suggested way. However,
>>> > > I am not familiar with data tables. Perhaps somebody
>>> > > else can provide a more profound guess.
>>> > >
>>> > > Regards -- Gerrit
>>> > >
>>> > > ---------------------------------------------------------------------
>>> > > Dr. Gerrit Eichner Mathematical Institute, Room 212
>>> > > gerrit.eichner at math.uni-giessen.de Justus-Liebig-University Giessen
>>> > > Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
>>> > > http://www.uni-giessen.de/eichner
>>> > > ---------------------------------------------------------------------
>>> > >
>>> > > Am 08.11.2019 um 16:02 schrieb Ana Marija:
>>> > > > I tried it but I got this error:
>>> > > >> udt <- unique(dt[c("chr", "pos", "gene_id")])
>>> > > > Error in `[.data.table`(dt, c("chr", "pos", "gene_id")) :
>>> > > > When i is a data.table (or character vector), the columns to join by
>>> > > > must be specified using 'on=' argument (see ?data.table), by keying x
>>> > > > (i.e. sorted, and, marked as sorted, see ?setkey), or by sharing
>>> > > > column names between x and i (i.e., a natural join). Keyed joins might
>>> > > > have further speed benefits on very large data due to x being sorted
>>> > > > in RAM.
>>> > > >
>>> > > > On Fri, Nov 8, 2019 at 8:58 AM Gerrit Eichner
>>> > > > <gerrit.eichner at math.uni-giessen.de> wrote:
>>> > > >>
>>> > > >> Hi, Ana,
>>> > > >>
>>> > > >> doesn't
>>> > > >>
>>> > > >> udt <- unique(dt[c("chr", "pos", "gene_id")])
>>> > > >> nrow(udt)
>>> > > >>
>>> > > >> get close to what you want?
>>> > > >>
>>> > > >> Hth -- Gerrit
>>> > > >>
>>> > > >> ---------------------------------------------------------------------
>>> > > >> Dr. Gerrit Eichner Mathematical Institute, Room 212
>>> > > >> gerrit.eichner at math.uni-giessen.de Justus-Liebig-University Giessen
>>> > > >> Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
>>> > > >> http://www.uni-giessen.de/eichner
>>> > > >> ---------------------------------------------------------------------
>>> > > >>
>>> > > >> Am 08.11.2019 um 15:38 schrieb Ana Marija:
>>> > > >>> Hello,
>>> > > >>>
>>> > > >>> I have a data frame like this:
>>> > > >>>
>>> > > >>> chr pos gene_id pval_nominal pval_ret wl wr
>>> > > >>> 1: chr1 54490 ENSG00000227232 0.6084950 0.7837780 31.62278 21.2838
>>> > > >>> 2: chr1 58814 ENSG00000227232 0.2952110 0.8975820 31.62278 21.2838
>>> > > >>> 3: chr1 60351 ENSG00000227232 0.4397880 0.8679590 31.62278 21.2838
>>> > > >>> 4: chr1 61920 ENSG00000227232 0.3195280 0.6018090 31.62278 21.2838
>>> > > >>> 5: chr1 63671 ENSG00000227232 0.2377390 0.9880390 31.62278 21.2838
>>> > > >>> 6: chr1 64931 ENSG00000227232 0.2766790 0.9070370 31.62278 21.2838
>>> > > >>> 7: chr1 81587 ENSG00000227232 0.6057930 0.6167630 31.62278 21.2838
>>> > > >>> 8: chr1 115746 ENSG00000227232 0.4078770 0.7799110 31.62278 21.2838
>>> > > >>> 9: chr1 135203 ENSG00000227232 0.4078770 0.9299130 31.62278 21.2838
>>> > > >>> 10: chr1 138593 ENSG00000227232 0.8464560 0.5696060 31.62278 21.2838
>>> > > >>>
>>> > > >>> it is very big,
>>> > > >>> [1] 73719122 8
>>> > > >>>
>>> > > >>> To count number of unique rows for all 3 columns: chr, pos and gene_id
>>> > > >>> I could just join those 3 columns and than count. But how would I find
>>> > > >>> unique number of rows for these 4 columns without joining them?
>>> > > >>>
>>> > > >>> Thanks
>>> > > >>> Ana
>>> > > >>>
>>> > > >>> ______________________________________________