创建dataframe R时合并两个列表

Question

对于两个list，我有以下情况：Path和UTMZones

> Path
[[1]]
[1] "/home/rus/S1A_IW_GRDH_1SDV_20190824T003615_20190824T003640_028704_033FD2_7CC8.SAFE/"

[[2]]
[1] "/home/rus/S2A_MSIL2A_20190827T105621_N0213_R094_T30TVK_20190827T141656.SAFE/"

[[3]]
[1] "/home/rus/S2B_MSIL2A_20190826T153819_N0213_R011_T18TXL_20190826T195901.SAFE/"

第二个列表(空号没问题)

> UTMZones
[[1]]
NULL

[[2]]
[1] "30"

[[3]]
[1] "18"

使用此输入，我将使用以下代码创建一个df：

df<-enframe(Path, name = "number", value = "uri") %>%
        unnest %>%
        mutate(plugin = case_when(substr(uri, 11, 12) == "S1" ~ "class org.esa.s1tbx.io.sentinel1.Sentinel1ProductReaderPlugIn", TRUE ~ "class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM18N_ReaderPlugIn"))

代码可以工作，但现在我需要插入一个小的修改。在代码的最后一部分中，当创建我正在做的数据访问时

 ~ paste0("class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM", UTMZones, "N_ReaderPlugIn", collapse = "")))

当然，这个代码不起作用。我想要做的是，在创建df时，对于Path中的[i]位置，UTMZones的第一个位置(例如，在paste函数中应该使用[j] )

我一直在尝试使用一个双变量for循环，但没有得到正确的结果：

for (i in seq_along(Path)){
      for(j in seq_along(UTMZones)){

        df<-enframe(Path[[i]], name = "number", value = "uri") %>%
        unnest %>%
        mutate(plugin = case_when(substr(uri, 11, 12) == "S1" ~ "class org.esa.s1tbx.io.sentinel1.Sentinel1ProductReaderPlugIn", TRUE ~ paste0("class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM", UTMZones[[j]], "N_ReaderPlugIn", collapse = "")))

      }
    }

-编辑--

输出应该如下所示。请注意UTM是如何使用UTMZones作为参考的顺序变化的。

> df
# A tibble: 5 x 3
  number uri                                                                     plugin                                                                      
   <int> <chr>                                                                   <chr>                                                                       
1      1 /home/rus/S1A_IW_GRDH_1SDV_20190824T003615_20190824T003640_028704_033F? class org.esa.s1tbx.io.sentinel1.Sentinel1ProductReaderPlugIn               
2      2 /home/rus/S2A_MSIL2A_20190827T105621_N0213_R094_T30TVK_20190827T141656? class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM30...
3      3 /home/rus/S2B_MSIL2A_20190826T153819_N0213_R011_T18TXL_20190826T195901? class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM18...

-编辑2 --

这是使用@Ronak Shah解决方案运行的代码

>     UTMZones[lengths(UTMZones) == 0] <- ""
> library(tidyverse)
> df<-enframe(Path, name = "number", value = "uri") %>%
+       mutate(UTM  = UTMZones) %>%
+       unnest %>%  
+       mutate(plugin = ifelse(substr(uri, 11, 12) == "S1", 
+                              "class org.esa.s1tbx.io.sentinel1.Sentinel1ProductReaderPlugIn", 
+                              paste0("class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM", 
+                                     UTM, "N_ReaderPlugIn", collapse = "")))
> df$plugin[[3]]
[1] "class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTMN_ReaderPlugInclass org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM30N_ReaderPlugInclass org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM18N_ReaderPlugIn"

Answer 1

我们可以在base R中做到这一点

UTMZones <- lapply(UTMZones, function(x) replace(x, is.null(x), ""))
within(stack(setNames(Path, seq_along(Path)))[2:1],{ UTM <- unlist(UTMZones);plugin <- ifelse(substr(values, 11, 12) == "S1", 
 "class org.esa.s1tbx.io.sentinel1.Sentinel1ProductReaderPlugIn", 
    paste0("class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM", 
         UTM, "N_ReaderPlugIn"))})

如果我们不需要“UTM”列

transform(stack(setNames(Path, seq_along(Path)))[2:1], 
    plugin= ifelse(substr(values, 11, 12) == "S1", 
 "class org.esa.s1tbx.io.sentinel1.Sentinel1ProductReaderPlugIn", 
    paste0("class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM", 
         unlist(UTMZones), "N_ReaderPlugIn")))

Answer 2

代码中很少有变化。首先将NULL元素替换为空白元素

UTMZones[lengths(UTMZones) == 0] <- ""

然后在dataframe中包括UTMZones，这样就可以很容易地替换值。

library(tidyverse)

enframe(Path, name = "number", value = "uri") %>%
    mutate(UTM  = UTMZones) %>%
    unnest %>%  
    mutate(plugin = ifelse(substr(uri, 11, 12) == "S1", 
    "class org.esa.s1tbx.io.sentinel1.Sentinel1ProductReaderPlugIn", 
paste0("class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM", 
            UTM, "N_ReaderPlugIn")))

Answer 3

虽然@akrun和@Ronak的回答效率要高得多，这绝对是我想要的，但我会把我的--不是完美的--尝试写下来，这是更基本的尝试，但如果有人感兴趣的话，我会说很容易跟踪。

由于我无法正确地迭代dataframe，一旦为plugin列创建了带有错误内容的dataframe，我使用下面的代码来更正它。

for (i in seq_along(Path)){
      if (substr(Path[[i]], 11,12) == 'S1') {
        df$plugin[[i]] <- 'class org.esa.s1tbx.io.sentinel1.Sentinel1ProductReaderPlugIn'
      } else {
        df$plugin[[i]] <- paste0("class org.esa.s2tbx.dataio.s2.ortho.plugins.Sentinel2L1CProduct_Multi_UTM", UTMZones[[i]], "N_ReaderPlugIn", collapse = "")
      }
    }