CODEWITHSUNDEEP
rdataframe
kslayerr
kslayerr

Reputation: 819

Calculate mean, standard deviation, n, etc. across columns and create new data frame

I am trying to calculate the number of samples, mean, standard deviation, coefficient of variation, lower and upper 95% confidence limits, and quartiles of this data set across each column and put it into a new data frame.

The numbers below are not necessarily all correct & I didn't fill them all in, just provides an example. These values will be used to create a box plot, hence the need for the quartiles. Rows and columns would be headers in the end. See example below.

Here is the structure:

B1 <- c(8, 6, 13, 6, 27, 104, 18, 3)
B2 <- c(2, 13, 1, 64, 127, 24, 4, 3)
B3 <- c(8, 16, 113, 680, 227, 310, 138, 30)
B4 <- c(238, 46, 613, 69, 7, 14, 4, 8)

x <- data.frame(B1, B2, B3, B4)

> head(x)
    B1  B2  B3  B4
1    8   2   8 238
2    6  13  16  46
3   13   1 113 613
4    6  64 680  69
5   27 127 227   7
6  104  24 310  14

Desired output:

> y
                   B1    B2   B3    B4
n                  8     8     8    8 
mean               23   30    190   125
Stand dev          5    2     34     2
CoeffofVariation   0.3   0.4  0.7   1.3
LowerConfInterval  2    20    35    45
UpperConfInterval  50    120  122   120
LowerQuartile
Median
Upper Quantile
Inter Quartile Range
Minimum
Maximum 
Regression equation

Upvotes: 5

Views: 39838

Answers (3)

gregV
gregV

Reputation: 1107

in base R:

> summary(x)
       B1               B2               B3              B4        
 Min.   :  3.00   Min.   :  1.00   Min.   :  8.0   Min.   :  4.00  
 1st Qu.:  6.00   1st Qu.:  2.75   1st Qu.: 26.5   1st Qu.:  7.75  
 Median : 10.50   Median :  8.50   Median :125.5   Median : 30.00  
 Mean   : 23.12   Mean   : 29.75   Mean   :190.2   Mean   :124.88  
 3rd Qu.: 20.25   3rd Qu.: 34.00   3rd Qu.:247.8   3rd Qu.:111.25  
 Max.   :104.00   Max.   :127.00   Max.   :680.0   Max.   :613.00

this is more what you asked for:

> library(skimr)
> skim(x)
-- Data Summary ------------------------
                           Values
Name                       x     
Number of rows             8     
Number of columns          4     
_______________________          
Column type frequency:           
  numeric                  4     
________________________         
Group variables            None  

-- Variable type: numeric ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
# A tibble: 4 x 11
  skim_variable n_missing complete_rate  mean    sd    p0   p25   p50   p75  p100 hist 
* <chr>             <int>         <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr>
1 B1                    0             1  23.1  33.6     3  6     10.5  20.2   104 ▇▁▁▁▁
2 B2                    0             1  29.8  44.6     1  2.75   8.5  34     127 ▇▁▁▁▁
3 B3                    0             1 190.  225.      8 26.5  126.  248.    680 ▇▂▂▁▂
4 B4                    0             1 125.  212.      4  7.75  30   111.    613 ▇▁▁▁▁

Upvotes: 2

Patricio Moracho
Patricio Moracho

Reputation: 717

As lmo mentioned, you could use sapply, like this:

sapply(x, function(x) c( "Stand dev" = sd(x), 
                         "Mean"= mean(x,na.rm=TRUE),
                         "n" = length(x),
                         "Median" = median(x),
                         "CoeffofVariation" = sd(x)/mean(x,na.rm=TRUE),
                         "Minimum" = min(x),
                         "Maximun" = max(x),
                         "Upper Quantile" = quantile(x,1),
                         "LowerQuartile" = quantile(x,0)
                    )
)

Output:

                            B1         B2         B3         B4
Stand dev            33.604581  44.592600 224.722527 212.086531
Mean                 23.125000  29.750000 190.250000 124.875000
n                     8.000000   8.000000   8.000000   8.000000
Median               10.500000   8.500000 125.500000  30.000000
CoeffofVariation      1.453171   1.498911   1.181196   1.698391
Minimum               3.000000   1.000000   8.000000   4.000000
Maximun             104.000000 127.000000 680.000000 613.000000
Upper Quantile.100% 104.000000 127.000000 680.000000 613.000000
LowerQuartile.0%      3.000000   1.000000   8.000000   4.000000

Upvotes: 6

DataTx
DataTx

Reputation: 1869

You could use something like this:

B1 <- c(8, 6, 13, 6, 27, 104, 18, 3)
B2 <- c(2, 13, 1, 64, 127, 24, 4, 3)
B3 <- c(8, 16, 113, 680, 227, 310, 138, 30)
B4 <- c(238, 46, 613, 69, 7, 14, 4, 8)

combDF <- data.frame(cbind(B1,B2,B3,B4))

data_long <- gather(combDF, factor_key=TRUE)

data_long%>% group_by(key)%>%
  summarise(mean= mean(value), sd= sd(value), max = max(value),min = min(value))

and the output would be:

    # A tibble: 4 x 5
     key    mean        sd   max   min
  <fctr>   <dbl>     <dbl> <dbl> <dbl>
1     B1  23.125  33.60458   104     3
2     B2  29.750  44.59260   127     1
3     B3 190.250 224.72253   680     8
4     B4 124.875 212.08653   613     4

You have not specified which confidence level you are looking but the code I posted can be adapted to your problem.

Upvotes: 8

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