将函数应用于tibble的每一种情况

Question

我的第二次参与是在Stackoverflow中。

我有一个名为bw_test的函数，其中有几个参数如下：

  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)


    }

此函数返回如下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，当我调用rbind中的函数时，得到了tibble上每次运行的结果：

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))

我的结果是为了这个例子：

# 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>

我的问题是将该函数应用于tibble这样的情况。

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
}

我试着使用地图，但是我得到了：

map(c(aps$ip,aps$remoteip,aps$localspeed,aps$remotespeed), bw_test)

el argumento "remotespeed" está ausente, sin valor por omisión 

我相信c(aps$ip，aps$remoteip，aps$localspeed，aps$remotespeed)首先提供所有aps$ip，然后是所有aps$remoteip，依此类推。

我用的是正确的策略？这是一种合适的方式

我哪里做错了？

如何对每一行应用函数以获取请求的tibble？

我将非常感谢你的帮助。

欢迎光临。

Answer 1

尝试使用pmap_df。

output <- purrr::pmap_df(list(aps$ip, aps$remoteip, aps$localspeed, 
                              aps$remotespeed), bw_test)