Question

数据框结构

> str(df)
    'data.frame':   459 obs. of  6 variables:    
     $ Source     : chr  "Mumbai" "Mumbai" "Bangalore" "Bangalore" ...    
     $ Destination: chr  "Bangalore" "Bangalore" "Chennai" "Cochin" ...    
     $ src_loc    :'data.frame':    459 obs. of  2 variables:    
      ..$ lon: num  72.9 72.9 77.6 77.6 73.9 ...    
      ..$ lat: num  19.1 19.1 13 13 18.5 ...    
     $ dest_loc   :'data.frame':    459 obs. of  2 variables:    
      ..$ lon: num  77.6 77.6 80.3 76.3 78.5 ...    
      ..$ lat: num  12.97 12.97 13.08 9.93 17.39 ...    
    $ route_line:List of 459    
      ..$ :'data.frame':    219 obs. of  2 variables:    
      .. ..$ lat: num  19.1 19.1 19.1 19.1 19.1 ...    
      .. ..$ lon: num  72.9 72.9 72.9 72.9 73 ...    
      ..$ :'data.frame':    219 obs. of  2 variables:    
      .. ..$ lat: num  19.1 19.1 19.1 19.1 19.1 ...    
      .. ..$ lon: num  72.9 72.9 72.9 72.9 73 ...    
      ..$ :'data.frame':    244 obs. of  2 variables:    
      .. ..$ lat: num  13 13 13 13 13 ...    
      .. ..$ lon: num  77.6 77.6 77.6 77.6 77.6 ...
      ..$ :'data.frame':    228 obs. of  2 variables:    
      .. ..$ lat: num  13 13 13 12.9 12.9 ...    
      .. ..$ lon: num  77.6 77.6 77.6 77.6 77.6 ...    
      ..$ :'data.frame':    232 obs. of  2 variables:    
      .. ..$ lat: num  18.5 18.5 18.5 18.5 18.5 ...    
      .. ..$ lon: num  73.9 73.9 73.9 73.9 73.9 ...    
      ..$ :'data.frame':    234 obs. of  2 variables:    
      .. ..$ lat: num  15.4 15.4 15.4 15.4 15.4 ...    
      .. ..$ lon: num  75.1 75.1 75.1 75.1 75.1 ...    
      ..$ :'data.frame':    218 obs. of  2 variables:    
      .. ..$ lat: num  17.4 17.4 17.4 17.5 17.5 ...    
      .. ..$ lon: num  78.5 78.5 78.5 78.5 78.5 ...

等......

> df$route_line[[1]] #gives a data frame



            lat      lon
    1   19.07597 72.87765
    2   19.06575 72.89918
    3   19.06331 72.91443
    4   19.05159 72.93661
    5   19.06758 72.98437
    6   19.06653 73.02000
    7   19.04099 73.02868
    8   19.02309 73.04452
    9   19.03844 73.07676
    10  18.99688 73.13215
    11  18.98191 73.14718
    12  18.96049 73.15789
    13  18.94201 73.15694
    14  18.92484 73.16662
    15  18.89439 73.20433
    16  18.84075 73.24026
    17  18.81434 73.27669
    18  18.79409 73.29148
    19  18.77373 73.32182
    20  18.77023 73.33760
    21  18.76414 73.34698
    22  18.77114 73.36076
    23  18.76580 73.35765
    24  18.77090 73.36348
    25  18.75822 73.37283
    26  18.76368 73.38653
    27  18.76939 73.40145
    28  18.76301 73.41848
    29  18.75766 73.42920
    30  18.73973 73.42921

我想创建一个新列（名称为route_str），其中包含通过连接上面获取的df中每行的所有纬度和经度获得的字符串

例如，

> df$route_str[1] #should give
[1] "19.07597 72.87765, 19.06575 72.89918, 19.06331 72.91443,19.05159 72.93661..." so  on till 30

我试过这个

> fun <- function(ip)
+ {
+ a <- ip[[1]]
+ a[3] <- paste(a[1],a[2]," ")
+ op <- paste(a[3],collapse = ",")
+ return(op)
+ }
> df$route_str <- lapply(df$route_line,fun)

但我得到的输出是

> unique_routes$route_str[1]
[[1]]
[1] "19.0759696960449 19.0657501220703  "

我尝试使用以下代码创建可重现的数据，但结构不一样

df <- data.frame(src=c("chennai","Mumbai","Bangalore"),dest=c("Mumbai","Bangalore","Mumbai"),route=list(list(lat=c(19,20,21),lon=c(72,73,74)),data.frame(lat=c(19,20,21),lon=c(72,73,74)),data.frame(lat=c(19,20,21),lon=c(72,73,74))))

但上述创建数据的结构如下

> str(df)
'data.frame':   3 obs. of  8 variables:
 $ src        : Factor w/ 3 levels "Bangalore","chennai",..: 2 3 1
 $ dest       : Factor w/ 2 levels "Bangalore","Mumbai": 2 1 2
 $ route.lat  : num  19 20 21
 $ route.lon  : num  72 73 74
 $ route.lat.1: num  19 20 21
 $ route.lon.1: num  72 73 74
 $ route.lat.2: num  19 20 21
 $ route.lon.2: num  72 73 74

我在Windows 10上使用R版本3.3.1帮助！

编辑：

这就是我最终得到那个复杂的数据框架

初始数据框架就像这样

> df <- data.frame(source=c("chennai","Mumbai","Bangalore"),destination=c("Mumbai","Bangalore","Mumbai"))

> df
     source destination
1   chennai      Mumbai
2    Mumbai   Bangalore
3 Bangalore      Mumbai

我想要一个包含单个字符串的列，其中源和目标之间的所有路径点（lat lon）用逗号分隔我使用googleway包来获取航路点

> library(googleway)
> res <- function(src,dest,key) #key is google maps API key
+ {
+ polylinex <- google_directions(origin = src,destination = dest,key = key)
+ return(polylinex$routes$overview_polyline$points)
+ } 

> df$source <- as.character(df$source)
> df$destination <- as.character(df$destination)
> df$x <- mapply(res,df$source,df$destination,key)
> df$route_line <- lapply(df$x,function(y) googleway::decode_pl(y))
> df <- df[,!(names(df)=="x")]
> str(df)
'data.frame':   3 obs. of  3 variables:
 $ source     : chr  "chennai" "Mumbai" "Bangalore"
 $ destination: chr  "Mumbai" "Bangalore" "Mumbai"
 $ route_line :List of 3
  ..$ :'data.frame':    219 obs. of  2 variables:
  .. ..$ lat: num  13.1 13.1 13.1 13.1 13.1 ...
  .. ..$ lon: num  80.3 80.2 80.2 80.2 80.2 ...
  ..$ :'data.frame':    219 obs. of  2 variables:
  .. ..$ lat: num  19.1 19.1 19.1 19.1 19.1 ...
  .. ..$ lon: num  72.9 72.9 72.9 72.9 73 ...
  ..$ :'data.frame':    218 obs. of  2 variables:
  .. ..$ lat: num  13 13 13 13 13 ...
  .. ..$ lon: num  77.6 77.6 77.6 77.6 77.5 ...

Answer 1

对lapply稍加修改为sapply，稍微更改paste序列就会让您想要

df$route_str <- sapply(df$x, function(y){
    df_coords <- decode_pl(y)
    paste0(t(sapply(df_coords, paste0)), collapse = ",")
})


str(df)

'data.frame':   3 obs. of  4 variables:
    $ source     : chr  "chennai" "Mumbai" "Bangalore"
$ destination: chr  "Mumbai" "Bangalore" "Mumbai"
$ x          : chr  "weznA{z|hNjrAlkDue@vsDtVnhD|dAnkErSbdI~kGzmRtmLjrNldI|iWnjBbuDf^duJgPzqNsiCtaIyLpnOyXzrKe{AvaG|JxpF~VpkCga@tkG_sBp|Cev@fvDpI|gF"| __truncated__ "ywlsBi|x{Lz~@qeCfNi~AfhAsiC}bBoiHpEu}Er~Cgu@znB_bB}~AohEvbGeyIp|A}|AzdC}aAnrB|DhjBo{@h}DujFfnIq_F`dDubFp}Bm{Af~Bs|DzTsaB`e@uy@w"| __truncated__ "oodnA}drxMkcAhKggApm@s}A|uAey@|rAi~BdjF{fDpaLgxB||F}`DvxE{sDdmDgkGthKmlK|vJmgIbzJa`BrjCssC|aBw`Dvw@osBrkCutNpbIigD|sCk`Ft_C}iPv"| __truncated__
$ route_str  : chr  "13.0826797485352,80.2706985473633,13.0693397521973,80.2431106567383,13.0755300521851,80.2141876220703,13.0717391967773,80.18706"| __truncated__ "19.0759696960449,72.8776473999023,19.0657501220703,72.8991775512695,19.0633087158203,72.9144287109375,19.0515899658203,72.93660"| __truncated__ "12.9715995788574,77.5945510864258,12.9825401306152,77.5925750732422,12.9940996170044,77.5851287841797,13.0092391967773,77.57122"| __truncated__

注意：我是googleway作者，感谢您使用该软件包

如何连接作为数据帧列表R的列中的所有值

1 个答案: