在 R 中组合两个具有不同结构的列表

Question

我有vendor_list

vendor_list[57:59]
[[1]]
[1] "ibm"

[[2]]
[1] "apache"    "canonical" "apple"     "novell"   

[[3]]
[1] "gnu"    "oracle"

我有problemtype_list

problemtype_list[57:59]
[[1]]
[1] "NVD-CWE-Other"

[[2]]
[1] "NVD-CWE-Other"

[[3]]
[1] "CWE-824"

我需要将它们组合起来制作一个数据框，使得

A              B
ibm       NVD-CWE-Other
apache    NVD-CWE-Other
canonical NVD-CWE-Other
apple     NVD-CWE-Other
novelle   NVD-CWE-Other 
gnu       CWE-824
oracle    CWE-824

我看过类似的问题

但它给我错误

do.call(rbind, Map(data.frame, A=problemtype_list, B=vendor_list))
Error in (function (..., row.names = NULL, check.rows = FALSE, check.names = TRUE,  : 
  arguments imply differing number of rows: 1, 0

编辑

我的每个列表结构

str(vendor_list)
 $ : chr "cisco"
 $ : NULL
 $ : chr [1:5] "redhat" "novell" "debian" "oracle" ...
 $ : chr [1:4] "redhat" "novell" "debian" "google"
 $ : chr [1:4] "redhat" "novell" "debian" "google"

 str(problemtype_list)
 $ : chr "CWE-254"
 $ : chr "CWE-79"
 $ : chr "NVD-CWE-Other"
 $ : chr "NVD-CWE-Other"
 $ : chr "CWE-254"
 $ : chr "CWE-189"
 $ : chr "CWE-119"

Answer 1

你说的代码对我不起作用 - 我已经调用了 p 和 v:

> v = list("ibm",c("apache","canonical","apple","novelle"),c("gnu","oracle"))

> p = list("NVD-CWE-Other","NVD-CWE-Other","CWE-824")

> p
[[1]]
[1] "NVD-CWE-Other"

[[2]]
[1] "NVD-CWE-Other"

[[3]]
[1] "CWE-824"

> v
[[1]]
[1] "ibm"

[[2]]
[1] "apache"    "canonical" "apple"     "novelle"  

[[3]]
[1] "gnu"    "oracle"

然后你的代码：

> do.call(rbind, Map(data.frame, A=p, B=v))
              A         B
1 NVD-CWE-Other       ibm
2 NVD-CWE-Other    apache
3 NVD-CWE-Other canonical
4 NVD-CWE-Other     apple
5 NVD-CWE-Other   novelle
6       CWE-824       gnu
7       CWE-824    oracle

所以您的数据结构可能不同。

或者：

> do.call(rbind.data.frame,mapply(cbind,v,p))
         V1            V2
1       ibm NVD-CWE-Other
2    apache NVD-CWE-Other
3 canonical NVD-CWE-Other
4     apple NVD-CWE-Other
5   novelle NVD-CWE-Other
6       gnu       CWE-824
7    oracle       CWE-824
>

Answer 2

我猜你的列表中有一个零长度元素。

vendor_list <- list("ibm", c("apache", "canonical", "apple", "novell"), c("gnu", "oracle"))
problemtype_list <- list("NVD-CWE-Other", "NVD-CWE-Other", "CWE-824")
do.call(rbind.data.frame, Map(data.frame, A=vendor_list, B=problemtype_list))
#           A             B
# 1       ibm NVD-CWE-Other
# 2    apache NVD-CWE-Other
# 3 canonical NVD-CWE-Other
# 4     apple NVD-CWE-Other
# 5    novell NVD-CWE-Other
# 6       gnu       CWE-824
# 7    oracle       CWE-824

但是，如果我们提供一个空位：

vendor_list[[3]] <- character(0)
vendor_list
# [[1]]
# [1] "ibm"
# [[2]]
# [1] "apache"    "canonical" "apple"     "novell"   
# [[3]]
# character(0)

...和快速测试：

any(lengths(vendor_list) == 0)
# [1] TRUE
any(lengths(problemtype_list) == 0)
# [1] FALSE

...然后合并失败：

do.call(rbind.data.frame, Map(data.frame, A=vendor_list, B=problemtype_list))
# Error in (function (..., row.names = NULL, check.rows = FALSE, check.names = TRUE,  (from pit-roads.R!8460QVH#21) : 
#   arguments imply differing number of rows: 0, 1

您可以将有问题的条目替换为有意义的内容（例如 NA），也可以删除它们。使用哪种方法完全取决于您的用途。

替换：

vendor_list[lengths(vendor_list) == 0] <- NA
problemtype_list[lengths(problemtype_list) == 0] <- NA
do.call(rbind.data.frame, Map(data.frame, A=vendor_list, B=problemtype_list))
#           A             B
# 1       ibm NVD-CWE-Other
# 2    apache NVD-CWE-Other
# 3 canonical NVD-CWE-Other
# 4     apple NVD-CWE-Other
# 5    novell NVD-CWE-Other
# 6      <NA>       CWE-824

删除：

keepthese <- (lengths(vendor_list) > 0) & (lengths(problemtype_list) > 0)
keepthese
# [1]  TRUE  TRUE FALSE
vendor_list <- vendor_list[keepthese]
problemtype_list <- problemtype_list[keepthese]
do.call(rbind.data.frame, Map(data.frame, A=vendor_list, B=problemtype_list))
#           A             B
# 1       ibm NVD-CWE-Other
# 2    apache NVD-CWE-Other
# 3 canonical NVD-CWE-Other
# 4     apple NVD-CWE-Other
# 5    novell NVD-CWE-Other

Answer 3

你可以用mapply解决问题

df=mapply(c,vendor_list,problemtype_list) 这给你一个 data.frame 包含列表中的值

在 R 中组合两个具有不同结构的列表

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list

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