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R language objects are stored in JSON
2022-06-27 14:03:00 【Analysis of breeding data】
Reference resources
Getting started with JSON and jsonlite
0. Installation package
install.packages("rjson")
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1. vector
library(rjson)
vector = c(1:10)
vector
toJSON(vector)
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> ### vector
> vector = c(1:10) # vector
> vector
[1] 1 2 3 4 5 6 7 8 9 10
> vector_j = toJSON(vector)
> vector_j # json In the form of
[1] "[1,2,3,4,5,6,7,8,9,10]"
> fromJSON(vector_j) # Turn into R
[1] 1 2 3 4 5 6 7 8 9 10
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2. Data frame
### data.frame
data("npk")
npk # R Data frame
npk_j = toJSON(npk)
npk_j # Json Data frame
fromJSON(npk_j) # Turn into R list
as.data.frame(fromJSON(npk_j)) # Change to data frame
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> ### data.frame
> data("npk")
> npk # R Data frame
block N P K yield
1 1 0 1 1 49.5
2 1 1 1 0 62.8
3 1 0 0 0 46.8
4 1 1 0 1 57.0
5 2 1 0 0 59.8
6 2 1 1 1 58.5
7 2 0 0 1 55.5
8 2 0 1 0 56.0
9 3 0 1 0 62.8
10 3 1 1 1 55.8
11 3 1 0 0 69.5
12 3 0 0 1 55.0
13 4 1 0 0 62.0
14 4 1 1 1 48.8
15 4 0 0 1 45.5
16 4 0 1 0 44.2
17 5 1 1 0 52.0
18 5 0 0 0 51.5
19 5 1 0 1 49.8
20 5 0 1 1 48.8
21 6 1 0 1 57.2
22 6 1 1 0 59.0
23 6 0 1 1 53.2
24 6 0 0 0 56.0
> npk_j = toJSON(npk)
> npk_j # Json Data frame
[1] "{\"block\":[\"1\",\"1\",\"1\",\"1\",\"2\",\"2\",\"2\",\"2\",\"3\",\"3\",\"3\",\"3\",\"4\",\"4\",\"4\",\"4\",\"5\",\"5\",\"5\",\"5\",\"6\",\"6\",\"6\",\"6\"],\"N\":[\"0\",\"1\",\"0\",\"1\",\"1\",\"1\",\"0\",\"0\",\"0\",\"1\",\"1\",\"0\",\"1\",\"1\",\"0\",\"0\",\"1\",\"0\",\"1\",\"0\",\"1\",\"1\",\"0\",\"0\"],\"P\":[\"1\",\"1\",\"0\",\"0\",\"0\",\"1\",\"0\",\"1\",\"1\",\"1\",\"0\",\"0\",\"0\",\"1\",\"0\",\"1\",\"1\",\"0\",\"0\",\"1\",\"0\",\"1\",\"1\",\"0\"],\"K\":[\"1\",\"0\",\"0\",\"1\",\"0\",\"1\",\"1\",\"0\",\"0\",\"1\",\"0\",\"1\",\"0\",\"1\",\"1\",\"0\",\"0\",\"0\",\"1\",\"1\",\"1\",\"0\",\"1\",\"0\"],\"yield\":[49.5,62.8,46.8,57,59.8,58.5,55.5,56,62.8,55.8,69.5,55,62,48.8,45.5,44.2,52,51.5,49.8,48.8,57.2,59,53.2,56]}"
>
> fromJSON(npk_j) # Turn into R list
$`block`
[1] "1" "1" "1" "1" "2" "2" "2" "2" "3" "3" "3" "3" "4" "4" "4" "4" "5"
[18] "5" "5" "5" "6" "6" "6" "6"
$N
[1] "0" "1" "0" "1" "1" "1" "0" "0" "0" "1" "1" "0" "1" "1" "0" "0" "1"
[18] "0" "1" "0" "1" "1" "0" "0"
$P
[1] "1" "1" "0" "0" "0" "1" "0" "1" "1" "1" "0" "0" "0" "1" "0" "1" "1"
[18] "0" "0" "1" "0" "1" "1" "0"
$K
[1] "1" "0" "0" "1" "0" "1" "1" "0" "0" "1" "0" "1" "0" "1" "1" "0" "0"
[18] "0" "1" "1" "1" "0" "1" "0"
$yield
[1] 49.5 62.8 46.8 57.0 59.8 58.5 55.5 56.0 62.8 55.8 69.5 55.0 62.0 48.8
[15] 45.5 44.2 52.0 51.5 49.8 48.8 57.2 59.0 53.2 56.0
> as.data.frame(fromJSON(npk_j)) # Change to data frame
block N P K yield
1 1 0 1 1 49.5
2 1 1 1 0 62.8
3 1 0 0 0 46.8
4 1 1 0 1 57.0
5 2 1 0 0 59.8
6 2 1 1 1 58.5
7 2 0 0 1 55.5
8 2 0 1 0 56.0
9 3 0 1 0 62.8
10 3 1 1 1 55.8
11 3 1 0 0 69.5
12 3 0 0 1 55.0
13 4 1 0 0 62.0
14 4 1 1 1 48.8
15 4 0 0 1 45.5
16 4 0 1 0 44.2
17 5 1 1 0 52.0
18 5 0 0 0 51.5
19 5 1 0 1 49.8
20 5 0 1 1 48.8
21 6 1 0 1 57.2
22 6 1 1 0 59.0
23 6 0 1 1 53.2
24 6 0 0 0 56.0
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3. matrix
### matrix
mat = matrix(1:9,3,3)
mat
mat_j = toJSON(mat,indent = "0")
mat_j
fromJSON(mat_j)
matrix(fromJSON(mat_j),3,3)
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> ### matrix
> mat = matrix(1:9,3,3)
> mat
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 5 8
[3,] 3 6 9
>
> mat_j = toJSON(mat,indent = "0")
> mat_j
[1] "[1,2,3,4,5,6,7,8,9]"
> fromJSON(mat_j)
[1] 1 2 3 4 5 6 7 8 9
> matrix(fromJSON(mat_j),3,3)
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 5 8
[3,] 3 6 9
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4. Analysis of variance
### Analysis of variance
data(npk)
anova = summary(aov(yield~ N*P*K + block, data=npk))
anova
a_j = toJSON(anova)
a_j
as.data.frame(fromJSON(a_j))
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> ### Analysis of variance
> data(npk)
> anova = summary(aov(yield~ N*P*K + block, data=npk))
> anova
Df Sum Sq Mean Sq F value Pr(>F)
N 1 189.3 189.28 12.259 0.00437 **
P 1 8.4 8.40 0.544 0.47490
K 1 95.2 95.20 6.166 0.02880 *
block 5 343.3 68.66 4.447 0.01594 *
N:P 1 21.3 21.28 1.378 0.26317
N:K 1 33.1 33.14 2.146 0.16865
P:K 1 0.5 0.48 0.031 0.86275
Residuals 12 185.3 15.44
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
> a_j = toJSON(anova)
> a_j
[1] "[{\"Df\":[1,1,1,5,1,1,1,12],\"Sum Sq\":[189.281666666667,8.40166666666664,95.2016666666665,343.295,21.2816666666668,33.1350000000001,0.48166666666667,185.286666666667],\"Mean Sq\":[189.281666666667,8.40166666666664,95.2016666666665,68.659,21.2816666666668,33.1350000000001,0.48166666666667,15.4405555555556],\"F value\":[12.2587342136509,0.544129816860359,6.16568920231712,4.44666642679811,1.37829669341202,2.14597200733998,0.0311949051919551,\"NA\"],\"Pr(>F)\":[0.00437181182579937,0.474904092674435,0.0287950535002327,0.0159387902081939,0.263165282877167,0.168647878500492,0.862752085685407,\"NA\"]}]"
>
> as.data.frame(fromJSON(a_j))
Df Sum.Sq Mean.Sq F.value.12.2587342136509
1 1 189.2816667 189.2816667 12.25873
2 1 8.4016667 8.4016667 12.25873
3 1 95.2016667 95.2016667 12.25873
4 5 343.2950000 68.6590000 12.25873
5 1 21.2816667 21.2816667 12.25873
6 1 33.1350000 33.1350000 12.25873
7 1 0.4816667 0.4816667 12.25873
8 12 185.2866667 15.4405556 12.25873
F.value.0.544129816860359 F.value.6.16568920231712
1 0.5441298 6.165689
2 0.5441298 6.165689
3 0.5441298 6.165689
4 0.5441298 6.165689
5 0.5441298 6.165689
6 0.5441298 6.165689
7 0.5441298 6.165689
8 0.5441298 6.165689
F.value.4.44666642679811 F.value.1.37829669341202
1 4.446666 1.378297
2 4.446666 1.378297
3 4.446666 1.378297
4 4.446666 1.378297
5 4.446666 1.378297
6 4.446666 1.378297
7 4.446666 1.378297
8 4.446666 1.378297
F.value.2.14597200733998 F.value.0.0311949051919551 F.value..NA.
1 2.145972 0.03119491 NA
2 2.145972 0.03119491 NA
3 2.145972 0.03119491 NA
4 2.145972 0.03119491 NA
5 2.145972 0.03119491 NA
6 2.145972 0.03119491 NA
7 2.145972 0.03119491 NA
8 2.145972 0.03119491 NA
Pr..F..0.00437181182579937 Pr..F..0.474904092674435
1 0.004371812 0.4749041
2 0.004371812 0.4749041
3 0.004371812 0.4749041
4 0.004371812 0.4749041
5 0.004371812 0.4749041
6 0.004371812 0.4749041
7 0.004371812 0.4749041
8 0.004371812 0.4749041
Pr..F..0.0287950535002327 Pr..F..0.0159387902081939
1 0.02879505 0.01593879
2 0.02879505 0.01593879
3 0.02879505 0.01593879
4 0.02879505 0.01593879
5 0.02879505 0.01593879
6 0.02879505 0.01593879
7 0.02879505 0.01593879
8 0.02879505 0.01593879
Pr..F..0.263165282877167 Pr..F..0.168647878500492
1 0.2631653 0.1686479
2 0.2631653 0.1686479
3 0.2631653 0.1686479
4 0.2631653 0.1686479
5 0.2631653 0.1686479
6 0.2631653 0.1686479
7 0.2631653 0.1686479
8 0.2631653 0.1686479
Pr..F..0.862752085685407 Pr..F...NA.
1 0.8627521 NA
2 0.8627521 NA
3 0.8627521 NA
4 0.8627521 NA
5 0.8627521 NA
6 0.8627521 NA
7 0.8627521 NA
8 0.8627521 NA
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anova Of Json Format , Turn into R When the format , Use data.frame The effect is not very good .
If there's a good solution , Update again !
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