When I trying print table with knitr::kable function "id" word apperas in the column names. How can I change it?
Example:
> x <- structure(c(42.3076923076923, 53.8461538461538, 96.1538461538462,
2.56410256410256, 1.28205128205128, 3.84615384615385,
44.8717948717949, 55.1282051282051, 100),
.Dim = c(3L, 3L),
.Dimnames = structure(list(Condition1 = c("Yes", "No", "Sum"),
Condition2 = c("Yes", "No", "Sum")),
.Names = c("Condition1", "Condition2")), class = c("table", "matrix"))
> print(x)
Condition2
Condition1 Yes No Sum
Yes 42,31 2,56 44,87
No 53,85 1,28 55,13
Sum 96,15 3,85 100,00
> library(knitr)
> kable(x)
|id | Yes| No| Sum|
|:----|-----:|-----:|------:|
|Yes | 42,3| 2,56| 44,9|
|No | 53,8| 1,28| 55,1|
|Sum | 96,2| 3,85| 100,0|
Edit: I find reason of this behavior in the knitr:::kable_mark function. But now I not understand how to make it more flexible.
An alternative to kable might be the general S3 method of pander:
> library(pander)
> pander(x, style = 'rmarkdown')
| | Yes | No | Sum |
|:---------:|:-----:|:-----:|:-----:|
| **Yes** | 42.31 | 2.564 | 44.87 |
| **No** | 53.85 | 1.282 | 55.13 |
| **Sum** | 96.15 | 3.846 | 100 |
If you need to set the decimal mark to comma, then set the relevant option before and use that in your R session:
> panderOptions('decimal.mark', ',')
> pander(x, style = 'rmarkdown')
| | Yes | No | Sum |
|:---------:|:-----:|:-----:|:-----:|
| **Yes** | 42,31 | 2,564 | 44,87 |
| **No** | 53,85 | 1,282 | 55,13 |
| **Sum** | 96,15 | 3,846 | 100 |
There are also some other possible tweaks: http://rapporter.github.io/pander/#pander-options
I think the easiest way is to rip out and replace kable_mark completely. Note: this is quite dirty – but it seems to work, and there is no current way to customise how kable_mark works (you could submit a patch to knitr though).
km <- edit(knitr:::kable_mark)
# Now edit the code and remove lines 7 and 8.
unlockBinding('kable_mark', environment(knitr:::kable_mark))
assign('kable_mark', km, envir=environment(knitr:::kable_mark))
Explanation: First we edit the function and store the amended definition in a temporary variable. We remove the two lines
if (grepl("^\\s*$", cn[1L]))
cn[1L] = "id"
… of course you can also hard-code the amended function rather than editing it, or change the function around completely.
Next we use unlockBinding to make knitr:::kable_mark overridable. If we don’t do this, the next assign command wouldn’t work.
Finally, we assign the patched function back to knitr:::kable_mark. Done.
Related
I have two dataframes. One is a set of ≈4000 entries that looks similar to this:
| grade_col1 | grade_col2 |
| --- | --- |
| A-| A-|
| B | 86|
| C+| C+|
| B-| D |
| A | A |
| C-| 72|
| F | 96|
| B+| B+|
| B | B |
| A-| A-|
The other is a set of ≈700 entries that look similar to this:
| grade | scale |
| --- | --- |
| A+|100|
| A+| 99|
| A+| 98|
| A+| 97|
| A | 96|
| A | 95|
| A | 94|
| A | 93|
| A-| 92|
| A-| 91|
| A-| 90|
| B+| 89|
| B+| 88|
...and so on.
What I'm trying to do is create a new column that shows whether grade_col2 matches grade_col1 with a binary, 0-1 output (0 = no match, 1 = match). Most of grade_col2 is shown by letter grade. But every once in awhile an entry in grade_col2 was accidentally entered as a numeric grade instead. I want this match column to give me a "1" even when grade_col2 is a numeric grade instead of a letter grade. In other words, if grade_col1 is B and grade_col2 is 86, I want this to still be read as a match. Only when grade_col1 is F and grade_col2 is 96 would this not be a match (similar to when grade_col1 is B- and grade_col2 is D = not a match).
The second data frame gives me the information I need to translate between one and the other (entries between 97-100 are A+, between 93-96 are A, and so on). I just don't know how to run a script that uses this information to find matches through all ≈4000 entries. Theoretically, I could do this manually, but the real dataset is so lengthy that this isn't realistic.
I had been thinking of using nested if_else statements with dplyr. But once I got past the first "if" statement, I got stuck. I'd appreciate any help with this people can offer.
You can do this using a join.
Let your first dataframe be grades_df and your second dataframe be lookup_df, then you want something like the following:
output = grades_df %>%
# join on look up, keeping everything grades table
left_join(lookup_df, by = c(grade_col2 = "scale")) %>%
# combine grade_col2 from grades_df and grade from lookup_df
mutate(grade_col2b = ifelse(is.na(grade), grade_col2, grade)) %>%
# indicator column
mutate(indicator = ifelse(grade_col1 == grade_col2b, 1, 0))
I am trying to parse the below data in Kusto. Need help.
[[ObjectCount][LinkCount][DurationInUs]]
[ChangeEnumeration][[88][9][346194]]
[ModifyTargetInLive][[3][6][595903]]
Need generic implementation without any hardcoding.
ideally - you'd be able to change the component that produces source data in that format to use a standard format (e.g. CSV, Json, etc.) instead.
The following could work, but you should consider it very inefficient
let T = datatable(s:string)
[
'[[ObjectCount][LinkCount][DurationInUs]]',
'[ChangeEnumeration][[88][9][346194]]',
'[ModifyTargetInLive][[3][6][595903]]',
];
let keys = toscalar(
T
| where s startswith "[["
| take 1
| project extract_all(#'\[([^\[\]]+)\]', s)
);
T
| where s !startswith "[["
| project values = extract_all(#'\[([^\[\]]+)\]', s)
| mv-apply with_itemindex = i keys on (
extend Category = tostring(values[0]), p = pack(tostring(keys[i]), values[i + 1])
| summarize b = make_bag(p) by Category
)
| project-away values
| evaluate bag_unpack(b)
--->
| Category | ObjectCount | LinkCount | DurationInUs |
|--------------------|-------------|-----------|--------------|
| ChangeEnumeration | 88 | 9 | 346194 |
| ModifyTargetInLive | 3 | 6 | 595903 |
I have a table that looks like this:
+-----------------------------------+-------+--------+------+
| | Male | Female | n |
+-----------------------------------+-------+--------+------+
| way more than my fair share | 2,4 | 21,6 | 135 |
| a little more than my fair share | 5,4 | 38,1 | 244 |
| about my fair share | 54,0 | 35,3 | 491 |
| a littles less than my fair share | 25,1 | 3,0 | 153 |
| way less than my fair share | 8,7 | 0,7 | 51 |
| Can't say | 4,4 | 1,2 | 31 |
| n | 541,0 | 564,0 | 1105 |
+-----------------------------------+-------+--------+------+
Everything is fine but what I would like to do is to show no digits in the last row at all since they show the margins (real cases). Is there any chance in R I can manipulate specific cells and their digits?
Thanks!
You could use ifelse to output the numbers in different formats in different rows, as in the example below. However, it will take some additional finagling to get the values in the last row to line up by place value with the previous rows:
library(knitr)
library(tidyverse)
# Fake data
set.seed(10)
dat = data.frame(category=c(LETTERS[1:6],"n"), replicate(3, rnorm(7, 100,20)))
dat %>%
mutate_if(is.numeric, funs(sprintf(ifelse(category=="n", "%1.0f", "%1.1f"), .))) %>%
kable(align="lrrr")
|category | X1| X2| X3|
|:--------|-----:|-----:|-----:|
|A | 100.4| 92.7| 114.8|
|B | 96.3| 67.5| 101.8|
|C | 72.6| 94.9| 80.9|
|D | 88.0| 122.0| 96.1|
|E | 105.9| 115.1| 118.5|
|F | 107.8| 95.2| 109.7|
|n | 76| 120| 88|
The huxtable package makes it easy to decimal-align the values (see the Vignette for more on table formatting):
library(huxtable)
tab = dat %>%
mutate_if(is.numeric, funs(sprintf(ifelse(category=="n", "%1.0f", "%1.1f"), .))) %>%
hux %>% add_colnames()
align(tab)[-1] = "."
tab
Here's what the PDF output looks like when knitted to PDF from an rmarkdown document:
I want to create a calculated field to use with the rpivotTable package, similar to the functionality seen in excel.
For instance, consider the following table:
+--------------+--------+---------+-------------+-----------------+
| Manufacturer | Vendor | Shipper | Total Units | Defective Units |
+--------------+--------+---------+-------------+-----------------+
| A | P | X | 173247 | 34649 |
| A | P | Y | 451598 | 225799 |
| A | P | Z | 759695 | 463414 |
| A | Q | X | 358040 | 225565 |
| A | Q | Y | 102068 | 36744 |
| A | Q | Z | 994961 | 228841 |
| A | R | X | 454672 | 231883 |
| A | R | Y | 275994 | 124197 |
| A | R | Z | 691100 | 165864 |
| B | P | X | 755594 | 302238 |
| . | . | . | . | . |
| . | . | . | . | . |
+--------------+--------+---------+-------------+-----------------+
(my actual table has many more columns, both dimensions and measures, time, etc. and I need to define multiple such "calculated columns")
If I want to calculate defect rate (which would be Defective Units/Total Units) and I want to aggregate by either of the first three columns, I'm not able to.
I tried assignment by reference (:=), but that still didn't seem to work and summed up defect rates (i.e., sum(Defective_Units/Total_Units)), instead of sum(Defective_Units)/sum(Total_Units):
myData[, Defect.Rate := Defective_Units / Total_Units]
This ended up giving my defect rates greater than 1. Is there anywhere I can declare a calculated field, which is just a formula evaluated post aggregation?
You're lucky - the creator of pivottable.js foresaw cases like yours (and mine, earlier today) by implementing an aggregator called "Sum over Sum" and a few more, likewise, cf. https://github.com/nicolaskruchten/pivottable/blob/master/src/pivot.coffee#L111 and https://github.com/nicolaskruchten/pivottable/blob/master/src/pivot.coffee#L169.
So we'll use "Sum over Sum" as parameter "aggregatorName", and the columns whose quotient we want in the "vals" parameter.
Here's a meaningless usage example from the mtcars data for reproducibility:
require(rpivotTable)
data(mtcars)
rpivotTable(mtcars,rows="gear", cols=c("cyl","carb"),
aggregatorName = "Sum over Sum",
vals =c("mpg","disp"),
width="100%", height="400px")
I'm using a double apply function to get a list of p-values for cor.test between any two columns of two tables.
hel_plist<-apply(bc, 2, function(x) { apply(otud, 2, function(y) { if (cor.test(x,y,method="spearman", exact=FALSE)$p.value<0.05){cor.test(x,y,method="spearman", exact=FALSE)$p.value}}) })
The otud data.frame is 90X11 (90rows,11 colums or to say dim(otud) 90 11) and will be used with different data.frames.
bc and hel - are both 90X2 data.frame-s - so for both I get 2*11=22 p-values out of functions
bc_plist<-apply(bc, 2, function(x) { apply(otud, 2, function(y) { if (cor.test(x,y,method="spearman", exact=FALSE)$p.value<0.05){cor.test(x,y,method="spearman", exact=FALSE)$p.value}}) })
hel_plist<-apply(hel, 2, function(x) { apply(otud, 2, function(y) { if (cor.test(x,y,method="spearman", exact=FALSE)$p.value<0.05){cor.test(x,y,method="spearman", exact=FALSE)$p.value}}) })
For bc I will have an output with dim=NULL a list of elements of otunames$bcnames$ p-value (a format that I have always got from these scripts and are happy with)
But for hel I will get and output of dim(hel) 11 2 - an 11X2 table with p-values written inside.
Shortened examples of output.
hel_plist
+--------+--------------+--------------+
| | axis1 | axis2 |
+--------+--------------+--------------+
| Otu037 | 1.126362e-18 | 0.01158251 |
| Otu005 | 3.017458e-2 | NULL |
| Otu068 | 0.00476002 | NULL |
| Otu070 | 1.27646e-15 | 5.252419e-07 |
+--------+--------------+--------------+
bc_plist
$axis1
$axis1$Otu037
[1] 1.247717e-06
$axis1$Otu005
[1] 1.990313e-05
$axis1$Otu068
[1] 5.664597e-07
Why is it like that when the input formats are all the same? (Shortened examples)
bc
+-------+-----------+-----------+
| group | axis1 | axis2 |
+-------+-----------+-----------+
| 1B041 | 0.125219 | 0.246319 |
| 1B060 | -0.022412 | -0.030227 |
| 1B197 | -0.088005 | -0.305351 |
| 1B222 | -0.119624 | -0.144123 |
| 1B227 | -0.148946 | -0.061741 |
+-------+-----------+-----------+
hel
+-------+---------------+---------------+
| group | axis1 | axis2 |
+-------+---------------+---------------+
| 1B041 | -0.0667782322 | -0.1660606406 |
| 1B060 | 0.0214470932 | -0.0611351008 |
| 1B197 | 0.1761876858 | 0.0927570627 |
| 1B222 | 0.0681058251 | 0.0549292399 |
| 1B227 | 0.0516864361 | 0.0774155225 |
| 1B235 | 0.1205676221 | 0.0181712761 |
+-------+---------------+---------------+
How could I force my scripts to always produce "flat" outputs as in the case of bc
OK different output-s are caused because of the NULL results from conditional function in bc_plist case. If I'd to modify code to replace possible NULL-s with NA-s I'd get 2d tables in any case.
So to keep things constant :
bc_nmds_plist<-apply(bc_nmds, 2, function(x) { apply(stoma_otud, 2, function(y) { if (cor.test(x,y,method="spearman", exact=FALSE)$p.value<0.05){cor.test(x,y,method="spearman", exact=FALSE)$p.value}else NA}) })
And I get a 2d tabel out for bc_nmds_plist too.
So I guess this thing can be called solved - as I now have a piece of code that produces predictable output on any correct input.
If anyone has any idea how to force the output to conform to previos bc_plist format instead I would still be interested as I do actually prefer that form:
$axis1
$axis1$Otu037
[1] 1.247717e-06
$axis1$Otu005
[1] 1.990313e-05
$axis1$Otu068
[1] 5.664597e-07