如何連接所有vlaues在一列是數據列表畫面R

結構數據幀如何連接所有vlaues在一列是數據列表畫面R

> 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

我想創建一個新列（以名稱route_str），其中包含通過連接上述獲得的數據幀中的所有經度和經度而獲得的字符串的字符串df

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

我想這

> 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)

但輸出我得到的是

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

我試着用下面的代碼來創建可重複的數據，但結構是不一樣的

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))))

但是上面創建的數據的結構如下

> 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請幫助！

編輯：

這是我結束了與複雜的數據幀

初始數據幀是這樣

> 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

我想有一個包含單絃一柱全部源和目的地之間用逗號隔開的航點（lat lon）我使用googleway包獲取航點

> 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 ...

來源

2017-03-02 sruthi sripathi

讓我們保持簡單，使用汽車的簡單數據框我可以創建一個新的列，例如'cars $ test < - paste0（cars $ speed，「，」，cars $ dist）'。所以我認爲使用類似的東西會無用嗎？ 'df $ route_str < - paste0（df $ route_line [[1]] [1]，「，」，df $ route_line [[1]] [2]）或者我錯過了解釋中的某些內容？如果是這樣，你能提供一些樣本數據，以便我們可以幫助你。 –

我編輯了這篇文章，解釋瞭如何獲取樣本數據。請檢查它。謝謝！ :) –

少許修改您的lapply爲sapply，並且改變paste順序稍有會得到你想要你想要

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|[email protected]|dAnkErSbdI~kGzmRtmLjrNldI|iWnjBbuDf^duJgPzqNsiCtaIyLpnOyXzrKe{AvaG|[email protected]_sBp|[email protected]|gF"| __truncated__ "ywlsBi|x{[email protected]~AfhAsiC}bBoiHpEu}[email protected]_bB}~AohEvbGeyIp|A}|AzdC}aAnrB|DhjBo{@h}DujFfnIq_F`dDubFp}Bm{Af~Bs|DzTsaB`[email protected]@w"| __truncated__ "oodnA}[email protected]}A|[email protected]|rAi~BdjF{fDpaLgxB||F}`DvxE{sDdmDgkGthKmlK|vJmgIbzJa`BrjCssC|aBw`[email protected]|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：我使用包googleway作者，謝謝！

來源

2017-03-02 20:59:42 SymbolixAU

非常感謝！這工作！我想要這樣的結構，因爲我試圖使用api調用來獲取特定路由中的收費價格，並將路標列表作爲其參數中的字符串。我使用的其他軟件包是ggmap。但googleway似乎給了更多的路標。所以謝謝你的包裝！:) –

@sruthisripathi - 啊，現在有道理:)如果這回答你的問題，請隨時按下投票箭頭下的'打勾':) – SymbolixAU

是可以在python中使用googleway軟件包嗎？ –

如何連接所有vlaues在一列是數據列表畫面R

回答

相關問題