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Contando valores únicos/distintos por grupo en un marco de datos (10)

Aquí hay un punto de referencia de la solución de @David Arenburg, así como un resumen de algunas soluciones publicadas aquí ( @mnel , @Sven Hohenstein , @Henrik ):

library(dplyr) library(data.table) library(microbenchmark) library(tidyr) library(ggplot2) df <- mtcars DT <- as.data.table(df) DT_32k <- rbindlist(replicate(1e3, mtcars, simplify = FALSE)) df_32k <- as.data.frame(DT_32k) DT_32M <- rbindlist(replicate(1e6, mtcars, simplify = FALSE)) df_32M <- as.data.frame(DT_32M) bench <- microbenchmark( base_32 = aggregate(hp ~ cyl, df, function(x) length(unique(x))), base_32k = aggregate(hp ~ cyl, df_32k, function(x) length(unique(x))), base_32M = aggregate(hp ~ cyl, df_32M, function(x) length(unique(x))), dplyr_32 = summarise(group_by(df, cyl), count = n_distinct(hp)), dplyr_32k = summarise(group_by(df_32k, cyl), count = n_distinct(hp)), dplyr_32M = summarise(group_by(df_32M, cyl), count = n_distinct(hp)), data.table_32 = DT[, .(count = uniqueN(hp)), by = cyl], data.table_32k = DT_32k[, .(count = uniqueN(hp)), by = cyl], data.table_32M = DT_32M[, .(count = uniqueN(hp)), by = cyl], times = 10 )

Resultados:

print(bench) # Unit: microseconds # expr min lq mean median uq max neval cld # base_32 816.153 1064.817 1.231248e+03 1.134542e+03 1263.152 2430.191 10 a # base_32k 38045.080 38618.383 3.976884e+04 3.962228e+04 40399.740 42825.633 10 a # base_32M 35065417.492 35143502.958 3.565601e+07 3.534793e+07 35802258.435 37015121.086 10 d # dplyr_32 2211.131 2292.499 1.211404e+04 2.370046e+03 2656.419 99510.280 10 a # dplyr_32k 3796.442 4033.207 4.434725e+03 4.159054e+03 4857.402 5514.646 10 a # dplyr_32M 1536183.034 1541187.073 1.580769e+06 1.565711e+06 1600732.034 1733709.195 10 b # data.table_32 403.163 413.253 5.156662e+02 5.197515e+02 619.093 628.430 10 a # data.table_32k 2208.477 2374.454 2.494886e+03 2.448170e+03 2557.604 3085.508 10 a # data.table_32M 2011155.330 2033037.689 2.074020e+06 2.052079e+06 2078231.776 2189809.835 10 c

Trama:

as_tibble(bench) %>% group_by(expr) %>% summarise(time = median(time)) %>% separate(expr, c("framework", "nrow"), "_", remove = FALSE) %>% mutate(nrow = recode(nrow, "32" = 32, "32k" = 32e3, "32M" = 32e6), time = time / 1e3) %>% ggplot(aes(nrow, time, col = framework)) + geom_line() + scale_x_log10() + scale_y_log10() + ylab("microseconds")

Información de la sesión:

sessionInfo() # R version 3.4.1 (2017-06-30) # Platform: x86_64-pc-linux-gnu (64-bit) # Running under: Linux Mint 18 # # Matrix products: default # BLAS: /usr/lib/atlas-base/atlas/libblas.so.3.0 # LAPACK: /usr/lib/atlas-base/atlas/liblapack.so.3.0 # # locale: # [1] LC_CTYPE=fr_FR.UTF-8 LC_NUMERIC=C LC_TIME=fr_FR.UTF-8 # [4] LC_COLLATE=fr_FR.UTF-8 LC_MONETARY=fr_FR.UTF-8 LC_MESSAGES=fr_FR.UTF-8 # [7] LC_PAPER=fr_FR.UTF-8 LC_NAME=C LC_ADDRESS=C # [10] LC_TELEPHONE=C LC_MEASUREMENT=fr_FR.UTF-8 LC_IDENTIFICATION=C # # attached base packages: # [1] stats graphics grDevices utils datasets methods base # # other attached packages: # [1] ggplot2_2.2.1 tidyr_0.6.3 bindrcpp_0.2 stringr_1.2.0 # [5] microbenchmark_1.4-2.1 data.table_1.10.4 dplyr_0.7.1 # # loaded via a namespace (and not attached): # [1] Rcpp_0.12.11 compiler_3.4.1 plyr_1.8.4 bindr_0.1 tools_3.4.1 digest_0.6.12 # [7] tibble_1.3.3 gtable_0.2.0 lattice_0.20-35 pkgconfig_2.0.1 rlang_0.1.1 Matrix_1.2-10 # [13] mvtnorm_1.0-6 grid_3.4.1 glue_1.1.1 R6_2.2.2 survival_2.41-3 multcomp_1.4-6 # [19] TH.data_1.0-8 magrittr_1.5 scales_0.4.1 codetools_0.2-15 splines_3.4.1 MASS_7.3-47 # [25] assertthat_0.2.0 colorspace_1.3-2 labeling_0.3 sandwich_2.3-4 stringi_1.1.5 lazyeval_0.2.0 # [31] munsell_0.4.3 zoo_1.8-0

Digamos que tengo el siguiente marco de datos:

> myvec name order_no 1 Amy 12 2 Jack 14 3 Jack 16 4 Dave 11 5 Amy 12 6 Jack 16 7 Tom 19 8 Larry 22 9 Tom 19 10 Dave 11 11 Jack 17 12 Tom 20 13 Amy 23 14 Jack 16

Quiero contar la cantidad de valores order_no diferentes para cada name . Debe producir el siguiente resultado:

name number_of_distinct_orders Amy 2 Jack 3 Dave 1 Tom 2 Larry 1

¿Cómo puedo hacer eso?


Aquí hay una solución con sqldf

library("sqldf") myvec <- read.table(header=TRUE, text= " name order_no 1 Amy 12 2 Jack 14 3 Jack 16 4 Dave 11 5 Amy 12 6 Jack 16 7 Tom 19 8 Larry 22 9 Tom 19 10 Dave 11 11 Jack 17 12 Tom 20 13 Amy 23 14 Jack 16") sqldf("SELECT name,COUNT(distinct(order_no)) as number_of_distinct_orders FROM myvec GROUP BY name") # > sqldf("SELECT name,COUNT(distinct(order_no)) as number_of_distinct_orders FROM myvec GROUP BY name") # name number_of_distinct_orders # 1 Amy 2 # 2 Dave 1 # 3 Jack 3 # 4 Larry 1 # 5 Tom 2


En dplyr puede usar n_distinct para " contar el número de valores únicos ":

library(dplyr) myvec %>% group_by(name) %>% summarise(n_distinct(order_no))


Esta es una solución simple con la función aggregate :

aggregate(order_no ~ name, myvec, function(x) length(unique(x)))


Esto debería funcionar:

ddply(myvec,~name,summarise,number_of_distinct_orders=length(unique(order_no)))

Esto requiere paquete plyr.


Esto también funcionaría, pero es menos elocuente que la solución plyr:

x <- sapply(split(myvec, myvec$name), function(x) length(unique(x[, 2]))) data.frame(names=names(x), number_of_distinct_orders=x, row.names = NULL)


Solo puede usar las funciones R incorporadas de forma tapply con la length

tapply(myvec$order_no, myvec$name, FUN = function(x) length(unique(x)))


Un enfoque de data.table

library(data.table) DT <- data.table(myvec) DT[, .(number_of_distinct_orders = length(unique(order_no))), by = name]

data.table v> = 1.9.5 tiene una función integrada uniqueN ahora

DT[, .(number_of_distinct_orders = uniqueN(order_no)), by = name]


Usando la table :

library(magrittr) myvec %>% unique %>% ''[''(1) %>% table %>% as.data.frame %>% setNames(c("name","number_of_distinct_orders")) # name number_of_distinct_orders # 1 Amy 2 # 2 Dave 1 # 3 Jack 3 # 4 Larry 1 # 5 Tom 2


my.1 <- table(myvec) my.1[my.1 != 0] <- 1 rowSums(my.1)