我有一个数据框中的列表,我想使用 purrr::map()
测试是否有任何 NULL
元素,然后摆脱它们 .
虽然我能够使用sapply来做到这一点,但是 Map 不起作用 . 我读了https://cran.r-project.org/web/packages/purrr/purrr.pdf,但我可以't figure out what'我错过了 .
这是我的sapply代码 - 这很好用:
P_Trans<- P_Trans[!sapply(P_Trans$Group,is.null),]
这是我为 purrr::map
尝试的一些事情,但它们不起作用 .
以下是我尝试过的四件事:
一个)
P_Trans %>% purrr::map(.,~is.null(Group))
b)
P_Trans %>% purrr::map(.,~is.null(.$Group))
C)
P_Trans %>% purrr::map(~is.null(.$Group))
d)
P_Trans %>% purrr::map(~is.null(Group))
有人可以纠正我的错误,让我知道我在做什么,我做错了以上四个选项?
数据:
dput(P_Trans)
structure(list(TransactionID = c("a1", "a1", "a1", "a2", "a2",
"a2", "a3", "a3", "a3", "a3", "a4", "a5", "a5", "a5", "a5", "a5",
"a6", "a6", "a7"), ProductID = c("A", "B", "1", "C", "4", "5",
"D", "C", "7", "8", "H", "1", "2", "3", "3", "1", "H", "15",
"22"), ProductType = c(1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 2,
2, 2, 2, 1, 2, 3), Group = list(structure(list(Group = "Group1"), .Names = "Group", row.names = c(NA,
-1L), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group1"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group1"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = c("Group2", "Group3")), .Names = "Group", row.names = c(NA,
-2L), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group2"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group2"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group3"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = c("Group2", "Group3")), .Names = "Group", row.names = c(NA,
-2L), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group3"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group3"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group5"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group1"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group1"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group1"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group1"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group1"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group5"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), structure(list(
Group = "Group5"), .Names = "Group", row.names = c(NA, -1L
), class = c("tbl_df", "tbl", "data.frame")), NULL)), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -19L), .Names = c("TransactionID",
"ProductID", "ProductType", "Group"))
1 回答
有了所有解决方案:
你循环遍历
P_Trans
的每一列(不是通过项目/行)这些列是原子向量,列表(或data.frame)有名称,原子向量没有名称 .
names(P_Trans[[1]]) # NULL
你打算返回一个列表,而不是data.frame,虽然它之前崩溃了
a与c相同
b与d相同
a)
P_Trans %>% purrr::map(.,~is.null(Group))
:Group
不存在,这里没有任何内容告诉我们我们应该在当前项目中查找它,甚至更少在表中b)
P_Trans %>% purrr::map(.,~is.null(.$Group))
Group
的元素,没有任何元素(即使是第4个),所以$ operator is invalid for atomic vectors
lmap
会帮助你循环遍历列作为P_Trans
的长度一个子列表但是方法也会崩溃,只有最后一个项目会有一个名为Group
(names(P_Trans[4]) # "Group"
)的项目 .由于
map_lgl
被设计为像sapply
那样返回logical
的向量,因此map
等同于你的解决方案P_Trans[!map_lgl(P_Trans$Group,is.null),]
获得你想要的其他方法: