解析 r 中的多级 json 文件
Parse multi-level json file in r
我对 R 有很好的理解,但对 JSON 文件类型和解析的最佳实践不熟悉。我在从原始 JSON 文件构建数据框时遇到困难。 JSON 文件(下面的数据)由每个用户有多个观察值的重复测量数据组成。
当raw文件读入r时
jdata<-read_json("./raw.json")
它以“列表 1”的形式出现,该列表为 user_ids。在每个 user_id 中还有更多列表,例如 -
jdata$user_id$`sjohnson`$date$`2020-09-25`$city
最后一个位置实际上分为两个选项 - $city 或 $zip。在最高级别,完整文件中大约有 89 个用户。
我的目标是最终得到一个矩形数据框或多个数据框,我可以像这样合并在一起 - 我实际上不需要邮政编码。
example table
我试过 jsonlite 和 tidyverse,我似乎得到的最远的是一个数据框,在最小级别有一个变量 - 城市和邮政编码交替行
使用这个
df <- as.data.frame(matrix(unlist(jdata), nrow=length(unlist(jdata["users"]))))
任何 help/suggestions 接近上述 table 的人将不胜感激。我有一种感觉,我无法通过不同的级别将其循环回来。
这是原始 json 文件结构的示例:
{
"user_id": {
"sjohnson": {
"date": {
"2020-09-25": {
"city": "Denver",
"zip": "80014"
},
"2020-10-01": {
"city": "Atlanta",
"zip": "30301"
},
"2020-11-04": {
"city": "Jacksonville",
"zip": "14001"
}
},
"asmith: {
"date": {
"2020-10-16": {
"city": "Cleavland",
"zip": "34321"
},
"2020-11-10": {
"City": "Elmhurst",
"zip": "00013
},
"2020-11-10 08:49:36": {
"location": null,
"timestamp": 1605016176013
}
}
这是 tidyverse
: a custom function unnestable()
designed to recursively unnest into a table the contents of a list
中的解决方案,就像您描述的那样。有关此类列表格式及其 table 的详细信息,请参阅 详细信息 。
解决方案
首先确保存在必要的库:
library(jsonlite)
library(tidyverse)
然后定义unnestable()
函数如下:
unnestable <- function(v) {
# If we've reached the bottommost list, simply treat it as a table...
if(all(sapply(
X = v,
# Check that each element is a single value (or NULL).
FUN = function(x) {
is.null(x) || purrr::is_scalar_atomic(x)
},
simplify = TRUE
))) {
v %>%
# Replace any NULLs with NAs to preserve blank fields...
sapply(
FUN = function(x) {
if(is.null(x))
NA
else
x
},
simplify = FALSE
) %>%
# ...and convert this bottommost list into a table.
tidyr::as_tibble()
}
# ...but if this list contains another nested list, then recursively unnest its
# contents and combine their tabular results.
else if(purrr::is_scalar_list(v)) {
# Take the contents within the nested list...
v[[1]] %>%
# ...apply this 'unnestable()' function to them recursively...
sapply(
FUN = unnestable,
simplify = FALSE,
USE.NAMES = TRUE
) %>%
# ...and stack their results.
dplyr::bind_rows(.id = names(v)[1])
}
# Otherwise, the format is unrecognized and yields no results.
else {
NULL
}
}
最后对JSON数据进行如下处理:
# Read the JSON file into an R list.
jdata <- jsonlite::read_json("./raw.json")
# Flatten the R list into a table, via 'unnestable()'
flat_data <- unnestable(jdata)
# View the raw table.
flat_data
当然,您可以根据需要重新格式化 table:
library(lubridate)
flat_data <- flat_data %>%
dplyr::transmute(
user_id = as.character(user_id),
date = lubridate::as_datetime(date),
city = as.character(city)
) %>%
dplyr::distinct()
# View the reformatted table.
flat_data
结果
给定一个 raw.json
文件,就像这里采样的那样
{
"user_id": {
"sjohnson": {
"date": {
"2020-09-25": {
"city": "Denver",
"zip": "80014"
},
"2020-10-01": {
"city": "Atlanta",
"zip": "30301"
},
"2020-11-04": {
"city": "Jacksonville",
"zip": "14001"
}
}
},
"asmith": {
"date": {
"2020-10-16": {
"city": "Cleavland",
"zip": "34321"
},
"2020-11-10": {
"city": "Elmhurst",
"zip": "00013"
},
"2020-11-10 08:49:36": {
"location": null,
"timestamp": 1605016176013
}
}
}
}
}
然后 unnestable()
会产生一个 tibble
这样的
# A tibble: 6 x 6
user_id date city zip location timestamp
<chr> <chr> <chr> <chr> <lgl> <dbl>
1 sjohnson 2020-09-25 Denver 80014 NA NA
2 sjohnson 2020-10-01 Atlanta 30301 NA NA
3 sjohnson 2020-11-04 Jacksonville 14001 NA NA
4 asmith 2020-10-16 Cleavland 34321 NA NA
5 asmith 2020-11-10 Elmhurst 00013 NA NA
6 asmith 2020-11-10 08:49:36 NA NA NA 1605016176013
其中 dplyr
将格式化为以下结果:
# A tibble: 6 x 3
user_id date city
<chr> <dttm> <chr>
1 sjohnson 2020-09-25 00:00:00 Denver
2 sjohnson 2020-10-01 00:00:00 Atlanta
3 sjohnson 2020-11-04 00:00:00 Jacksonville
4 asmith 2020-10-16 00:00:00 Cleavland
5 asmith 2020-11-10 00:00:00 Elmhurst
6 asmith 2020-11-10 08:49:36 NA
详情
列表格式
准确的说,list
代表字段{group_1
, group_2
, ..., group_n
}的嵌套分组,它必须是表格:
list(
group_1 = list(
"value_1" = list(
group_2 = list(
"value_1.1" = list(
# .
# .
# .
group_n = list(
"value_1.1.….n.1" = list(
field_a = 1,
field_b = TRUE
),
"value_1.1.….n.2" = list(
field_a = 2,
field_c = "2"
)
# ...
)
),
"value_1.2" = list(
# .
# .
# .
)
# ...
)
),
"value_2" = list(
group_2 = list(
"value_2.1" = list(
# .
# .
# .
group_n = list(
"value_2.1.….n.1" = list(
field_a = 3,
field_d = 3.0
)
# ...
)
),
"value_2.2" = list(
# .
# .
# .
)
# ...
)
)
# ...
)
)
Table格式
给定这种形式的 list
,unnestable()
会将其展平为以下形式的 table:
# A tibble: … x …
group_1 group_2 ... group_n field_a field_b field_c field_d
<chr> <chr> ... <chr> <dbl> <lgl> <chr> <dbl>
1 value_1 value_1.1 ... value_1.1.….n.1 1 TRUE NA NA
2 value_1 value_1.1 ... value_1.1.….n.2 2 NA 2 NA
3 value_1 value_1.2 ... value_1.2.….n.1 ... ... ... ...
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
j value_2 value_2.1 ... value_2.1.….n.1 3 NA NA 3
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
k value_2 value_2.2 ... value_2.2.….n.1 ... ... ... ...
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
我们可以一步一步构建我们想要的结构:
library(jsonlite)
library(tidyverse)
df <- fromJSON('{
"user_id": {
"sjohnson": {
"date": {
"2020-09-25": {
"city": "Denver",
"zip": "80014"
},
"2020-10-01": {
"city": "Atlanta",
"zip": "30301"
},
"2020-11-04": {
"city": "Jacksonville",
"zip": "14001"
}
}
},
"asmith": {
"date": {
"2020-10-16": {
"city": "Cleavland",
"zip": "34321"
},
"2020-11-10": {
"city": "Elmhurst",
"zip": "00013"
},
"2020-11-10 08:49:36": {
"location": null,
"timestamp": 1605016176013
}
}
}
}
}')
df %>%
bind_rows() %>%
pivot_longer(everything(), names_to = 'user_id') %>%
unnest_longer(value, indices_to = 'date') %>%
unnest_longer(value, indices_to = 'var') %>%
mutate(city = unlist(value)) %>%
filter(var == 'city') %>%
select(-var, -value)
给出:
# A tibble: 5 x 3
user_id date city
<chr> <chr> <chr>
1 sjohnson 2020-09-25 Denver
2 sjohnson 2020-10-01 Atlanta
3 sjohnson 2020-11-04 Jacksonville
4 asmith 2020-10-16 Cleavland
5 asmith 2020-11-10 Elmhurst
受@Greg 启发的替代解决方案,我们更改最后两行:
df %>%
bind_rows() %>%
pivot_longer(everything(), names_to = 'user_id') %>%
unnest_longer(value, indices_to = 'date') %>%
unnest_longer(value, indices_to = 'var') %>%
mutate(value = unlist(value)) %>%
pivot_wider(names_from = "var") %>%
select(user_id, date, city)
除了城市为 NA 的另一种情况外,这给出了几乎相同的结果:
# A tibble: 6 x 3
user_id date city
<chr> <chr> <chr>
1 sjohnson 2020-09-25 Denver
2 sjohnson 2020-10-01 Atlanta
3 sjohnson 2020-11-04 Jacksonville
4 asmith 2020-10-16 Cleavland
5 asmith 2020-11-10 Elmhurst
6 asmith 2020-11-10 08:49:36 NA
另一个(直接的)解决方案使用 rrapply
-package 中的 rrapply()
进行繁重的工作:
library(rrapply)
library(dplyr)
rrapply(jdata, how = "melt") %>%
filter(L5 == "city") %>%
select(user_id = L2, date = L4, city = value)
#> user_id date city
#> 1 sjohnson 2020-09-25 Denver
#> 2 sjohnson 2020-10-01 Atlanta
#> 3 sjohnson 2020-11-04 Jacksonville
#> 4 asmith 2020-10-16 Cleavland
#> 5 asmith 2020-11-10 Elmhurst
数据
jdata <- jsonlite::fromJSON('{
"user_id": {
"sjohnson": {
"date": {
"2020-09-25": {
"city": "Denver",
"zip": "80014"
},
"2020-10-01": {
"city": "Atlanta",
"zip": "30301"
},
"2020-11-04": {
"city": "Jacksonville",
"zip": "14001"
}
}
},
"asmith": {
"date": {
"2020-10-16": {
"city": "Cleavland",
"zip": "34321"
},
"2020-11-10": {
"city": "Elmhurst",
"zip": "00013"
},
"2020-11-10 08:49:36": {
"location": null,
"timestamp": 1605016176013
}
}
}
}
}')
我对 R 有很好的理解,但对 JSON 文件类型和解析的最佳实践不熟悉。我在从原始 JSON 文件构建数据框时遇到困难。 JSON 文件(下面的数据)由每个用户有多个观察值的重复测量数据组成。
当raw文件读入r时
jdata<-read_json("./raw.json")
它以“列表 1”的形式出现,该列表为 user_ids。在每个 user_id 中还有更多列表,例如 -
jdata$user_id$`sjohnson`$date$`2020-09-25`$city
最后一个位置实际上分为两个选项 - $city 或 $zip。在最高级别,完整文件中大约有 89 个用户。
我的目标是最终得到一个矩形数据框或多个数据框,我可以像这样合并在一起 - 我实际上不需要邮政编码。
example table
我试过 jsonlite 和 tidyverse,我似乎得到的最远的是一个数据框,在最小级别有一个变量 - 城市和邮政编码交替行 使用这个
df <- as.data.frame(matrix(unlist(jdata), nrow=length(unlist(jdata["users"]))))
任何 help/suggestions 接近上述 table 的人将不胜感激。我有一种感觉,我无法通过不同的级别将其循环回来。
这是原始 json 文件结构的示例:
{
"user_id": {
"sjohnson": {
"date": {
"2020-09-25": {
"city": "Denver",
"zip": "80014"
},
"2020-10-01": {
"city": "Atlanta",
"zip": "30301"
},
"2020-11-04": {
"city": "Jacksonville",
"zip": "14001"
}
},
"asmith: {
"date": {
"2020-10-16": {
"city": "Cleavland",
"zip": "34321"
},
"2020-11-10": {
"City": "Elmhurst",
"zip": "00013
},
"2020-11-10 08:49:36": {
"location": null,
"timestamp": 1605016176013
}
}
这是 tidyverse
: a custom function unnestable()
designed to recursively unnest into a table the contents of a list
中的解决方案,就像您描述的那样。有关此类列表格式及其 table 的详细信息,请参阅 详细信息 。
解决方案
首先确保存在必要的库:
library(jsonlite)
library(tidyverse)
然后定义unnestable()
函数如下:
unnestable <- function(v) {
# If we've reached the bottommost list, simply treat it as a table...
if(all(sapply(
X = v,
# Check that each element is a single value (or NULL).
FUN = function(x) {
is.null(x) || purrr::is_scalar_atomic(x)
},
simplify = TRUE
))) {
v %>%
# Replace any NULLs with NAs to preserve blank fields...
sapply(
FUN = function(x) {
if(is.null(x))
NA
else
x
},
simplify = FALSE
) %>%
# ...and convert this bottommost list into a table.
tidyr::as_tibble()
}
# ...but if this list contains another nested list, then recursively unnest its
# contents and combine their tabular results.
else if(purrr::is_scalar_list(v)) {
# Take the contents within the nested list...
v[[1]] %>%
# ...apply this 'unnestable()' function to them recursively...
sapply(
FUN = unnestable,
simplify = FALSE,
USE.NAMES = TRUE
) %>%
# ...and stack their results.
dplyr::bind_rows(.id = names(v)[1])
}
# Otherwise, the format is unrecognized and yields no results.
else {
NULL
}
}
最后对JSON数据进行如下处理:
# Read the JSON file into an R list.
jdata <- jsonlite::read_json("./raw.json")
# Flatten the R list into a table, via 'unnestable()'
flat_data <- unnestable(jdata)
# View the raw table.
flat_data
当然,您可以根据需要重新格式化 table:
library(lubridate)
flat_data <- flat_data %>%
dplyr::transmute(
user_id = as.character(user_id),
date = lubridate::as_datetime(date),
city = as.character(city)
) %>%
dplyr::distinct()
# View the reformatted table.
flat_data
结果
给定一个 raw.json
文件,就像这里采样的那样
{
"user_id": {
"sjohnson": {
"date": {
"2020-09-25": {
"city": "Denver",
"zip": "80014"
},
"2020-10-01": {
"city": "Atlanta",
"zip": "30301"
},
"2020-11-04": {
"city": "Jacksonville",
"zip": "14001"
}
}
},
"asmith": {
"date": {
"2020-10-16": {
"city": "Cleavland",
"zip": "34321"
},
"2020-11-10": {
"city": "Elmhurst",
"zip": "00013"
},
"2020-11-10 08:49:36": {
"location": null,
"timestamp": 1605016176013
}
}
}
}
}
然后 unnestable()
会产生一个 tibble
这样的
# A tibble: 6 x 6
user_id date city zip location timestamp
<chr> <chr> <chr> <chr> <lgl> <dbl>
1 sjohnson 2020-09-25 Denver 80014 NA NA
2 sjohnson 2020-10-01 Atlanta 30301 NA NA
3 sjohnson 2020-11-04 Jacksonville 14001 NA NA
4 asmith 2020-10-16 Cleavland 34321 NA NA
5 asmith 2020-11-10 Elmhurst 00013 NA NA
6 asmith 2020-11-10 08:49:36 NA NA NA 1605016176013
其中 dplyr
将格式化为以下结果:
# A tibble: 6 x 3
user_id date city
<chr> <dttm> <chr>
1 sjohnson 2020-09-25 00:00:00 Denver
2 sjohnson 2020-10-01 00:00:00 Atlanta
3 sjohnson 2020-11-04 00:00:00 Jacksonville
4 asmith 2020-10-16 00:00:00 Cleavland
5 asmith 2020-11-10 00:00:00 Elmhurst
6 asmith 2020-11-10 08:49:36 NA
详情
列表格式
准确的说,list
代表字段{group_1
, group_2
, ..., group_n
}的嵌套分组,它必须是表格:
list(
group_1 = list(
"value_1" = list(
group_2 = list(
"value_1.1" = list(
# .
# .
# .
group_n = list(
"value_1.1.….n.1" = list(
field_a = 1,
field_b = TRUE
),
"value_1.1.….n.2" = list(
field_a = 2,
field_c = "2"
)
# ...
)
),
"value_1.2" = list(
# .
# .
# .
)
# ...
)
),
"value_2" = list(
group_2 = list(
"value_2.1" = list(
# .
# .
# .
group_n = list(
"value_2.1.….n.1" = list(
field_a = 3,
field_d = 3.0
)
# ...
)
),
"value_2.2" = list(
# .
# .
# .
)
# ...
)
)
# ...
)
)
Table格式
给定这种形式的 list
,unnestable()
会将其展平为以下形式的 table:
# A tibble: … x …
group_1 group_2 ... group_n field_a field_b field_c field_d
<chr> <chr> ... <chr> <dbl> <lgl> <chr> <dbl>
1 value_1 value_1.1 ... value_1.1.….n.1 1 TRUE NA NA
2 value_1 value_1.1 ... value_1.1.….n.2 2 NA 2 NA
3 value_1 value_1.2 ... value_1.2.….n.1 ... ... ... ...
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
j value_2 value_2.1 ... value_2.1.….n.1 3 NA NA 3
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
k value_2 value_2.2 ... value_2.2.….n.1 ... ... ... ...
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
我们可以一步一步构建我们想要的结构:
library(jsonlite)
library(tidyverse)
df <- fromJSON('{
"user_id": {
"sjohnson": {
"date": {
"2020-09-25": {
"city": "Denver",
"zip": "80014"
},
"2020-10-01": {
"city": "Atlanta",
"zip": "30301"
},
"2020-11-04": {
"city": "Jacksonville",
"zip": "14001"
}
}
},
"asmith": {
"date": {
"2020-10-16": {
"city": "Cleavland",
"zip": "34321"
},
"2020-11-10": {
"city": "Elmhurst",
"zip": "00013"
},
"2020-11-10 08:49:36": {
"location": null,
"timestamp": 1605016176013
}
}
}
}
}')
df %>%
bind_rows() %>%
pivot_longer(everything(), names_to = 'user_id') %>%
unnest_longer(value, indices_to = 'date') %>%
unnest_longer(value, indices_to = 'var') %>%
mutate(city = unlist(value)) %>%
filter(var == 'city') %>%
select(-var, -value)
给出:
# A tibble: 5 x 3
user_id date city
<chr> <chr> <chr>
1 sjohnson 2020-09-25 Denver
2 sjohnson 2020-10-01 Atlanta
3 sjohnson 2020-11-04 Jacksonville
4 asmith 2020-10-16 Cleavland
5 asmith 2020-11-10 Elmhurst
受@Greg 启发的替代解决方案,我们更改最后两行:
df %>%
bind_rows() %>%
pivot_longer(everything(), names_to = 'user_id') %>%
unnest_longer(value, indices_to = 'date') %>%
unnest_longer(value, indices_to = 'var') %>%
mutate(value = unlist(value)) %>%
pivot_wider(names_from = "var") %>%
select(user_id, date, city)
除了城市为 NA 的另一种情况外,这给出了几乎相同的结果:
# A tibble: 6 x 3
user_id date city
<chr> <chr> <chr>
1 sjohnson 2020-09-25 Denver
2 sjohnson 2020-10-01 Atlanta
3 sjohnson 2020-11-04 Jacksonville
4 asmith 2020-10-16 Cleavland
5 asmith 2020-11-10 Elmhurst
6 asmith 2020-11-10 08:49:36 NA
另一个(直接的)解决方案使用 rrapply
-package 中的 rrapply()
进行繁重的工作:
library(rrapply)
library(dplyr)
rrapply(jdata, how = "melt") %>%
filter(L5 == "city") %>%
select(user_id = L2, date = L4, city = value)
#> user_id date city
#> 1 sjohnson 2020-09-25 Denver
#> 2 sjohnson 2020-10-01 Atlanta
#> 3 sjohnson 2020-11-04 Jacksonville
#> 4 asmith 2020-10-16 Cleavland
#> 5 asmith 2020-11-10 Elmhurst
数据
jdata <- jsonlite::fromJSON('{
"user_id": {
"sjohnson": {
"date": {
"2020-09-25": {
"city": "Denver",
"zip": "80014"
},
"2020-10-01": {
"city": "Atlanta",
"zip": "30301"
},
"2020-11-04": {
"city": "Jacksonville",
"zip": "14001"
}
}
},
"asmith": {
"date": {
"2020-10-16": {
"city": "Cleavland",
"zip": "34321"
},
"2020-11-10": {
"city": "Elmhurst",
"zip": "00013"
},
"2020-11-10 08:49:36": {
"location": null,
"timestamp": 1605016176013
}
}
}
}
}')