我第二次参加Stackoverflow的活动。
我有一个叫bw_test的函数,有几个args是这样的。
bw_test <- function(localip, remoteip, localspeed, remotespeed , duracion =30,direction ="both"){
comando <- str_c("ssh usuario@", localip ," /tool bandwidth-test direction=", direction," remote-tx-speed=",remotespeed,"M local-tx-speed=",localspeed,"M protocol=udp user=usuario password=mipasso duration=",duracion," ",remoteip)
resultado <- system(comando,intern = T,ignore.stderr = T)
# resultado pull from a ssh server a vector like this:
# head(resultado)
#[1] " status: connecting\r" " tx-current: #0bps\r" " tx-10-second-average: 0bps\r"
#[4] " tx-total-average: 0bps\r" " rx-current: #0bps\r" " rx-10-second-average: 0bps\r"
resultado %<>%
replace("\r","") %>%
tail(17) %>%
trimws("both") %>%
as_tibble %>%
mutate(local=localip, remote=remoteip) %>%
separate(value,sep=":", into=c("parametro","valor")) %>%
head(15)
resultado$valor %<>%
trimws() %>%
str_replace("Mbps","") %>% str_replace("%","") %>% str_replace("s","")
resultado %<>%
spread(parametro,valor)
resultado %<>%
mutate(`tx-percentaje`=as.numeric(resultado$`tx-total-average`)/localspeed) %>%
mutate(`rx-percentaje`=as.numeric(resultado$`rx-total-average`)/remotespeed)
return(resultado)
}
这个函数返回一个像这样的tibble。
A tibble: 1 x 19
local remote `connection-cou… direction duration `local-cpu-load` `lost-packets` `random-data` `remote-cpu-loa…
<chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
1 192.… 192.1… 1 both 4 13 0 no 12
# … with 10 more variables: `rx-10-second-average` <chr>, `rx-current` <chr>, `rx-size` <chr>,
# `rx-total-average` <chr>, `tx-10-second-average` <chr>, `tx-current` <chr>, `tx-size` <chr>,
# `tx-total-average` <chr>, `tx-percentaje` <dbl>, `rx-percentaje` <dbl>
所以,当我在rbind里面调用这个函数的时候,得到了每一次运行在tibble上的结果。
rbind(bw_test("192.168.105.10" ,"192.168.105.18", 75,125),
bw_test("192.168.133.11","192.168.133.9", 5 ,50),
bw_test("192.168.254.251","192.168.254.250", 25,150))
我的结果是针对这个例子的
# A tibble: 3 x 19
local remote `connection-cou… direction duration `local-cpu-load` `lost-packets` `random-data` `remote-cpu-loa…
<chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
1 192.… 192.1… 20 both 28 63 232 no 48
2 192.… 192.1… 20 both 29 4 0 no 20
3 192.… 192.1… 20 both 29 15 0 no 22
# … with 10 more variables: `rx-10-second-average` <chr>, `rx-current` <chr>, `rx-size` <chr>,
# `rx-total-average` <chr>, `tx-10-second-average` <chr>, `tx-current` <chr>, `tx-size` <chr>,
# `tx-total-average` <chr>, `tx-percentaje` <dbl>, `rx-percentaje` <dbl>
我的问题是要把这个函数应用到像这样的tibble上
aps <- tribble(
~name, ~ip, ~remoteip , ~bw_test, ~localspeed,~remotespeed,
"backbone_border_core","192.168.253.1", "192.168.253.3", 1,200,200,
"backbone_2_site2","192.168.254.251", "192.168.254.250", 1, 25,150
}
我试着使用map,但得到的结果是:"我不知道该怎么做。
map(c(aps$ip,aps$remoteip,aps$localspeed,aps$remotespeed), bw_test)
el argumento "remotespeed" está ausente, sin valor por omisión
我相信是因为c(aps$ip,aps$remoteip,aps$localspeed,aps$remotespeed)先输入所有aps$ip的情况,然后输入所有aps$remoteip的情况,以此类推。
我使用的策略是正确的吗?
我做错了什么?
¿我如何应用函数到每一行以获得所需的tibble?
我将感谢您的友好帮助。
问候。
尝试使用 pmap_df
.
output <- purrr::pmap_df(list(aps$ip, aps$remoteip, aps$localspeed,
aps$remotespeed), bw_test)