我会在列表底部放置dput我的列表，以便q可以重现。输入为a而不是x。
我有一个名为x的大型嵌套列表，我正在尝试构建一个数据框，但无法弄明白。
我做了第一部分：
for(i in 1:3){a[[i]]<-x$results[[i]]$experiences
indx <- lengths(a)
zz <- as.data.frame(do.call(rbind,lapply(a, `length<-`, max(indx))))}
为此，我使用了以下答案： Converting nested list (unequal length) to data frame
这给我留下了一个数据框，其中n列为结果，其中n是任何i：
的最大结果 v1 v2 v3
1 NULL NULL NULL
2 * * *
3 NULL NULL NULL
每个*是另一个格式为list(experience = list(duration = ...
例如，第2行第v1列中的第一个*。我不想要总清单。我只想要：
a[[2]][[1]]$experience$start
或原始列表x：
x$results[[2]]$experiences[[1]]$experience$start
我觉得我几乎有一些调整。我试过了：
for(i in 1:3){a[[i]]<-x$results[[i]]$experiences
indx <- lengths(a)
for(y in 1:length(a[[i]])) aa <- rbind(aa,tryCatch(x$results[[i]]$experiences[[y]]$experience$start, error=function(e) print(NA)))
zz <- as.data.frame(do.call(rbind,lapply(aa, `length<-`, max(indx))))}
导致：
v1 v2 v3
1 NA NA NA
2 NA NA NA
3 2014 NA NA
4 2012 NA NA
5 2006 NA NA
6 NA NA NA
7 NA NA NA
在最后一行尝试了cbind而不是rbind，并将所有日期放在第一行。
我也尝试了以下内容：
for(i in 1:3){a[[i]]<-lengths(x$results[[i]]$experiences)
indx <- lengths(a)
for(y in 1:length(indx)){tt[i] <- tryCatch(x$results[[i]]$experiences[[y]]$experience$start, error=function(e) print(""))}
zz <- as.data.frame(do.call(rbind,lapply(tt, `length<-`, max(indx))))}
这很接近，构建正确的格式但只返回第一个结果：
v1 v2 v3
1 NA NA NA
2 2014 NA NA
3 NA NA NA
我想要的格式是：
V1 V2 V3
1 NA NA NA
2 2014 2012 2006
3 NA NA NA
（（样本数据现在在底部））
最新尝试：
执行以下操作但仅返回每个a[[i]]的第一个开始日期，第二个循环我需要使列表aa[i][y]有所不同。
for(i in 1:3){a[[i]]<-x$results[[i]]$experiences
for(y in 1:length(a[[i]])){aa[i][y] = if(is.null(a[[i]][[y]]$experience$start)){"NULL"}else{a[[i]][[y]]$experience$start}}}
因此对于dput2我想要表格：
v1 v2 v3 v4 v5 v6 v7 v8
1 2015
2 2011 2007 null null null null null null
3 2016 2015 2015 2015 2013 2010
我不介意空白是空还是娜
更新
以下答案几乎可行，但是在我的数据中，结构发生了变化，名称的顺序（roleName，持续时间等）发生了变化，因此使用cumsum来确定何时找到新列表。如果您有duration，那么start密钥为9和1，而cumsum部分则标记两个不同的列表。
我写了以下内容：
my.list <- list(structure(
list(
experience = structure(
list(
start = "1",
end = "1",
roleName = "a",
summary = "a",
duration = "a",
current = "a",
org = structure(list(name = "a", url = "a"), .Names = c("name","url")),
location = structure(
list(
displayLocation = NULL,
lat = NULL,
lng = NULL
),
.Names = c("displayLocation",
"lat", "lng")
) ),.Names = c("start", "end", "roleName", "summary", "duration", "current", "org", "location")),
`_meta` = structure(
list(weight = 1L, `_sources` = list(structure(
list(`_origin` = "a"), .Names = "_origin"
))),.Names = c("weight", "_sources"))),.Names = c("experience", "_meta")))
然后：
aa <- lapply(1:length(a), function(y){tryCatch(lapply(1:length(a[[y]]),
function(i){a[[y]][[i]]$experience[names(my.list2[[1]]$experience)]}), error=function(e) print(list()))})
这会改变结构，使key2始终保持正确的顺序。
然而我在这个循环之后找到了另一个问题。
有时我只有经验列表中的roleName。如果连续两次出现，则重复键。 cumsum将它们视为相同的体验，而不是单独的体验。
这意味着我无法创建df3，因为行的标识符重复。即使我可以通过删除麻烦的行，名称也不匹配，因为下面的解决方案中的i匹配使用序列的名称，如果我删除任何更改长度的行。
以下是我提供更多见解的总代码：
for(i in 1:x$count){a[[i]]<-x$results[[i]]$experiences}
aa <- lapply(1:length(a), function(y){tryCatch(lapply(1:length(a[[y]]),
function(i){a[[y]][[i]]$experience[names(my.list2[[1]]$experience)]}), error=function(e) print(list()))})
aaa <- unlist(aa)
dummydf <- data.frame(b=c("start", "end", "roleName", "summary",
"duration", "current", "org.name", "org.url"), key=1:8)
df <- data.frame(a=aaa, b=names(aaa))
df2 <- left_join(df, dummydf)
df2$key2 <- as.factor(cumsum(df2$key < c(0, df2$key[-length(df2$key)])) +1)
df_split <- split(df2, df2$key2)
df3 <- lapply(df_split, function(x){
x %>% select(-c(key, key2)) %>% spread(b, a)
}) %>% data.table::rbindlist(fill=TRUE) %>% t
df3 <- data.frame(df3)
i <- sapply(seq_along(aa), function(y) rep(y, sapply(aa, function(x) length(x))[y])) %>% unlist
names(df3) <- paste0(names(df3), "_", i)
df4 <- data.frame(t(df3))
df4$dates <- as.Date(NA)
df4$dates <- as.Date(df4$start)
df4 <- data.frame(dates = df4$dates)
df4 <- t(df4)
df4 <- data.frame(df4)
names(df4) <- paste0(names(df4), "_", i)
df4[] <- lapply(df4[], as.character)
l1 <- lapply(split(stack(df4), sub('.*_', '', stack(df4)[,2])), '[', 1)
df5 <- t(do.call(cbindPad, l1))
df5 <- data.frame(df5)
cbindpad取自this question
包含问题的新示例代码：
dput3 =
list(list(), list(
structure(list(experience = structure(list(
duration = "1", start = "2014",
end = "3000", roleName = "a",
summary = "aaa",
org = structure(list(name = "a"), .Names = "name"),
location = structure(list(displayLocation = NULL, lat = NULL,
lng = NULL), .Names = c("displayLocation", "lat", "lng"
))), .Names = c("duration", "start", "end", "roleName", "summary",
"org", "location")), `_meta` = structure(list(weight = 1L, `_sources` = list(
structure(list(`_origin` = ""), .Names = "_origin"))), .Names = c("weight",
"_sources"))), .Names = c("experience", "_meta")),
structure(list(
experience = structure(list(end = "3000",
start = "2012", duration = "2",
roleName = "a", summary = "aaa",
org = structure(list(name = "None"), .Names = "name"),
location = structure(list(displayLocation = NULL, lat = NULL, lng = NULL), .Names = c("displayLocation", "lat", "lng"))), .Names = c("duration", "start", "end", "roleName",
"summary", "org", "location")), `_meta` = structure(list(
weight = 1L, `_sources` = list(structure(list(`_origin` = " "), .Names = "_origin"))), .Names = c("weight", "_sources"))), .Names = c("experience", "_meta")),
structure(list(
experience = structure(list(duration = "3",
start = "2006", end = "3000",
roleName = "a", summary = "aaa", org = structure(list(name = " "), .Names = "name"),
location = structure(list(displayLocation = NULL, lat = NULL, lng = NULL), .Names = c("displayLocation", "lat", "lng"))), .Names = c("duration", "start", "end", "roleName",
"summary", "org", "location")), `_meta` = structure(list(weight = 1L, `_sources` = list(structure(list(`_origin` = ""), .Names = "_origin"))), .Names = c("weight",
"_sources"))), .Names = c("experience", "_meta")),
structure(list(
experience = structure(list(roleName = "a",
location = structure(list(displayLocation = NULL, lat = NULL, lng = NULL), .Names = c("displayLocation", "lat", "lng"))), .Names = c("roleName",
"location")), `_meta` = structure(list(
weight = 1L, `_sources` = list(structure(list(`_origin` = " "), .Names = "_origin"))), .Names = c("weight", "_sources"))), .Names = c("experience", "_meta")),
structure(list(
experience = structure(list(roleName = "a",
location = structure(list(displayLocation = NULL, lat = NULL, lng = NULL), .Names = c("displayLocation", "lat", "lng"))), .Names = c("roleName",
"location")), `_meta` = structure(list(
weight = 1L, `_sources` = list(structure(list(`_origin` = " "), .Names = "_origin"))), .Names = c("weight", "_sources"))), .Names = c("experience", "_meta"))
),
list(
structure(list(experience = structure(list(
duration = "1", start = "2014",
end = "3000", roleName = "a",
summary = "aaa",
org = structure(list(name = "a"), .Names = "name"),
location = structure(list(displayLocation = NULL, lat = NULL,
lng = NULL), .Names = c("displayLocation", "lat", "lng"
))), .Names = c("duration", "start", "end", "roleName", "summary",
"org", "location")), `_meta` = structure(list(weight = 1L, `_sources` = list(
structure(list(`_origin` = ""), .Names = "_origin"))), .Names = c("weight",
"_sources"))), .Names = c("experience", "_meta"))))
答案 0 :(得分：1)
也许这可以帮助
library(dplyr)
library(tidyr)
a <- unlist(a)
df <- data.frame(a=a, b=names(a)) %>% mutate(key=cumsum(b=="experience.duration")) %>%
split(.$key) %>% lapply(function(x) x %>% select(-key) %>% spread(b, a)) %>%
do.call(rbind, .) %>% t %>% data.frame
df$key <- rownames(df)
然后您可以过滤感兴趣的行
上述内容相当于
rbind(unlist(a)[1:8], unlist(a)[9:16],unlist(a)[17:24]) %>% t
尝试dput2
a <- unlist(dput2)
library(dplyr)
library(tidyr)
dummydf <- data.frame(b=c("experience.start", "experience.end", "experience.roleName", "experience.summary",
"experience.org", "experience.org.name", "experience.org.url",
"_meta.weight", "_meta._sources._origin", "experience.duration"), key=1:10)
df <- data.frame(a=a, b=names(a))
df2 <- left_join(df, dummydf)
df2$key2 <- as.factor(cumsum(df2$key < c(0, df2$key[-length(df2$key)])) +1)
df_split <- split(df2, df2$key2)
df3 <- lapply(df_split, function(x){
x %>% select(-c(key, key2)) %>% spread(b, a)
}) %>% data.table::rbindlist(fill=TRUE) %>% t
df3 <- data.frame(df3)
i <- sapply(seq_along(dput2), function(y) rep(y, sapply(dput2, function(x) length(x))[y])) %>% unlist
names(df3) <- paste0(names(df3), "_", i)
View(df3)
答案 1 :(得分：0)
使用上面的dput3进行管理以解决问题：
a <- dput3
aa <- lapply(1:length(a), function(y){tryCatch(lapply(1:length(a[[y]]),
function(i){if(is.null(a[[y]][[i]]$experience$start)){"Null"}else{a[[y]][[i]]$experience$start}}),error=function(e) print(list()))})
for(i in 1:length(aa)){for(y in 1:length(aa[[i]])){tryCatch(for(z in length(aa[[i]][[y]]))
{test <- rbind(test, data.frame(key = i, key2= y))},error=function(e) print(0))}}
aaa <- unlist(aa)
df <- data.frame(a=aaa)
df2 <- cbind(df, test)
i <- sapply(seq_along(aa), function(y) rep(y, sapply(aa, function(x) length(x))[y])) %>% unlist
df5 <- data.frame(dates = df2$a)
df5 <- t(df5)
df5 <- data.frame(df5)
names(df5) <- paste0(names(df5), "_", i)
df5[] <- lapply(df5[], as.character)
l1 <- lapply(split(stack(df5), as.numeric(sub('.*_', '', stack(df5)[,2]))), '[', 1)
df6 <- t(do.call(cbindPad, l1))
df6 <- data.frame(df6)
会尝试展开它，以便它可以使用多个垂直（目前在aa我隔离start）