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bring back antibiogram()
, without deps
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@ -157,7 +157,7 @@ Using the `left_join()` function from the `dplyr` package, we can 'map' the gend
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data <- data %>% left_join(patients_table)
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```
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The resulting data set contains `r format(nrow(data), big.mark = ",")` blood culture isolates. With the `head()` function we can preview the first 6 rows of this data set:
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The resulting data set contains `r format(nrow(data), big.mark = " ")` blood culture isolates. With the `head()` function we can preview the first 6 rows of this data set:
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```{r preview data set 1, eval = FALSE}
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head(data)
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@ -251,7 +251,7 @@ data_1st <- data %>%
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filter_first_isolate()
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```
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So we end up with `r format(nrow(data_1st), big.mark = ",")` isolates for analysis. Now our data looks like:
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So we end up with `r format(nrow(data_1st), big.mark = " ")` isolates for analysis. Now our data looks like:
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```{r preview data set 3, eval = FALSE}
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head(data_1st)
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@ -362,7 +362,7 @@ data_1st %>%
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data_1st %>%
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group_by(hospital) %>%
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summarise(amoxicillin = resistance(AMX)) %>%
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knitr::kable(align = "c", big.mark = ",")
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knitr::kable(align = "c", big.mark = " ")
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```
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Of course it would be very convenient to know the number of isolates responsible for the percentages. For that purpose the `n_sir()` can be used, which works exactly like `n_distinct()` from the `dplyr` package. It counts all isolates available for every group (i.e. values S, I or R):
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@ -382,7 +382,7 @@ data_1st %>%
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amoxicillin = resistance(AMX),
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available = n_sir(AMX)
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) %>%
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knitr::kable(align = "c", big.mark = ",")
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knitr::kable(align = "c", big.mark = " ")
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```
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These functions can also be used to get the proportion of multiple antibiotics, to calculate empiric susceptibility of combination therapies very easily:
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@ -404,7 +404,7 @@ data_1st %>%
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gentamicin = susceptibility(GEN),
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amoxiclav_genta = susceptibility(AMC, GEN)
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) %>%
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knitr::kable(align = "c", big.mark = ",")
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knitr::kable(align = "c", big.mark = " ")
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```
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Or if you are curious for the resistance within certain antibiotic classes, use a antibiotic class selector such as `penicillins()`, which automatically will include the columns `AMX` and `AMC` of our data:
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@ -49,7 +49,7 @@ As said, SPSS is easier to learn than R. But SPSS, SAS and Stata come with major
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* **R has a huge community.**
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Many R users just ask questions on websites like [StackOverflow.com](https://stackoverflow.com), the largest online community for programmers. At the time of writing, [`r format(suppressWarnings(read.csv("https://data.stackexchange.com/stackoverflow/csv/1674647", quote = '"'))[[1]], big.mark = ",")` R-related questions](https://stackoverflow.com/questions/tagged/r?sort=votes) have already been asked on this platform (that covers questions and answers for any programming language). In my own experience, most questions are answered within a couple of minutes.
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Many R users just ask questions on websites like [StackOverflow.com](https://stackoverflow.com), the largest online community for programmers. At the time of writing, [`r format(suppressWarnings(read.csv("https://data.stackexchange.com/stackoverflow/csv/1674647", quote = '"'))[[1]], big.mark = " ")` R-related questions](https://stackoverflow.com/questions/tagged/r?sort=votes) have already been asked on this platform (that covers questions and answers for any programming language). In my own experience, most questions are answered within a couple of minutes.
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* **R understands any data type, including SPSS/SAS/Stata.**
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@ -126,7 +126,7 @@ run_it <- microbenchmark(mo_name(x),
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print(run_it, unit = "ms", signif = 3)
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```
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So getting official taxonomic names of `r format(length(x), big.mark = ",")` (!!) items consisting of `r n_distinct(x)` unique values only takes `r round(median(run_it$time, na.rm = TRUE) / 1e9, 3)` seconds. That is `r round(median(run_it$time, na.rm = TRUE) / length(x), 0)` nanoseconds on average. You only lose time on your unique input values.
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So getting official taxonomic names of `r format(length(x), big.mark = " ")` (!!) items consisting of `r n_distinct(x)` unique values only takes `r round(median(run_it$time, na.rm = TRUE) / 1e9, 3)` seconds. That is `r round(median(run_it$time, na.rm = TRUE) / length(x), 0)` nanoseconds on average. You only lose time on your unique input values.
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### Precalculated results
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@ -30,7 +30,7 @@ options(knitr.kable.NA = "")
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structure_txt <- function(dataset) {
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paste0(
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"A data set with ",
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format(nrow(dataset), big.mark = ","), " rows and ",
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format(nrow(dataset), big.mark = " "), " rows and ",
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ncol(dataset), " columns, containing the following column names: \n",
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AMR:::vector_or(colnames(dataset), quotes = "*", last_sep = " and ", sort = FALSE), "."
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)
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@ -142,7 +142,7 @@ Included (sub)species per taxonomic kingdom:
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```{r, echo = FALSE}
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microorganisms %>%
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count(kingdom) %>%
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mutate(n = format(n, big.mark = ",")) %>%
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mutate(n = format(n, big.mark = " ")) %>%
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setNames(c("Kingdom", "Number of (sub)species")) %>%
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print_df()
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```
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