数据
我对R有很好的理解，但对JSON文件类型和解析的最佳实践是新手。我在从原始JSON文件构建数据框架时遇到了困难。JSON文件(下面的数据)由重复的测量数据组成，每个用户有多个观察值。
当原始文件被读入r
时jdata<-read_json("./raw.json")
它以"List of 1"该列表为user_ids。在每个user_id中是进一步的列表，如下所示-
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，整个文件中大约有89个用户。
我的目标是最终得到一个矩形数据帧或多个数据帧，我可以像这样合并在一起——我实际上不需要邮政编码。
示例表
我试过jsonlite和tidyverse，我似乎得到的最远的是一个数据框架，在最小的级别上有一个变量-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"]))))
任何帮助/建议更接近上面的表格将不胜感激。我有一种感觉，我不能在不同的关卡中循环它。
下面是一个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
}
}
在rrapply-包中使用rrapply()的另一个(直接的)解决方案:
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
}
}
}
}
}')
我们可以一步一步地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)
给了:
# 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启发的另一种解决方案，我们更改了最后两行:
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)
这给出了几乎相同的结果，除了一个额外的情况，城市是NA:
# 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
tidyverse一个自定义函数unnestable()，用于递归地嵌套写入表像你描述的list的内容。看到有关该列表及其表格格式的详细信息。
解决方案首先确保必要的库存在:
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此表:
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
}
}
}
}
}
则unnestable()将生成如下的tibble
# 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
whichdplyr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
列表格式确切地说，list表示嵌套的分组，由字段{group_1,group_2，…，group_n}，格式必须为:
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(
# .
# .
# .
)
# ...
)
)
# ...
)
)
表格格式给定一个这种形式的list,unnestable()将把它平展成如下形式的表:
# 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 ... ... ... ...
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮