Question

我怎样才能以

的形式转换时间序列矩阵

      dDomingo  dSegunda     dTerça   dQuarta    dQuinta    dSexta    dSábado
[1,] -1.783558 0.5123224  0.5997068 0.4680476  0.6521374 0.6276698 -1.0763264
[2,] -1.730086 0.8552685  0.6426731 0.6377859  0.3995325 0.2944567 -1.0996309
[3,] -1.503333 0.9450167 -0.4125656 0.8446433  0.6839187 0.5397113 -1.0973919
[4,] -1.720261 0.3044134  0.6722075 0.5313041  0.6411365 0.7105043 -1.1393050
[5,] -1.234701 0.9189446  0.9670409 0.8713827 -1.1208727 0.3465987 -0.7483936

进入这个

           V1     V2
[1,] -1.783558 dDomingo        
[2,]  0.5123224 dSegunda 
[3,]  0.5997068 dTerça 
[4,]  0.4680476 dQuarta
[5,]  0.6521374 dQuinta 
[6,]  0.6276698 dSexta 
[7,] -1.0763264 dSábado
[8,] -1.730086  dDomingo

......等等，直到

[35,] -0.7483936  dSábado

请注意，我想要的新数据格式只有两列。第一个将包含最后一个矩阵的变量值。第二列将包含变量值为的相应列。

时间序列矩阵数据如下：

structure(c(-1.78355763661757, -1.73008589314791, -1.50333251627705, 
-1.72026078275706, -1.23470060199216, 0.512322374645035, 0.855268450002498, 
0.945016691336727, 0.304413384239457, 0.91894460806156, 0.599706795271625, 
0.642673149573306, -0.412565627467277, 0.672207513790216, 0.967040903075167, 
0.468047601527563, 0.637785901287577, 0.844643267672228, 0.531304065337568, 
0.871382716325882, 0.652137447647579, 0.39953254735831, 0.683918734842367, 
0.64113649695194, -1.12087270379329, 0.627669809872133, 0.294456709215146, 
0.539711347678816, 0.710504348497859, 0.346598697399641, -1.07632639234637, 
-1.09963086428893, -1.09739189778582, -1.13930502605998, -0.748393619076803
), .Dim = c(5L, 7L), .Dimnames = list(NULL, c("dDomingo", "dSegunda", 
"dTerça", "dQuarta", "dQuinta", "dSexta", "dSábado")))

Answer 1

在此基本代码末尾的注释中使用

alert('<?php echo $cat_name; ?>');

alert('<?php echo $cat_name; ?>');

或者

将创建一个包含35行的双列数据框;但请注意，m将使用ggplot2绘制多变量动物园时间序列，而不需要任何显式变换。如果您希望每个系列都在一个单独的面板中，请删除stack(as.data.frame(m))。有关详细信息和示例，请参阅library(zoo) fortify.zoo(zoo(m), melt = TRUE)和autoplot.zoo。

facet = NULL

注意

我们假设输入?autoplot.zoo是：

?fortify.zoo

Answer 2

您需要先将matrix转换为data frame。然后使用tidyr::gather转换为长格式

mat1 <- structure(c(-1.78355763661757, -1.73008589314791, -1.50333251627705, 
  -1.72026078275706, -1.23470060199216, 0.512322374645035, 0.855268450002498, 
  0.945016691336727, 0.304413384239457, 0.91894460806156, 0.599706795271625, 
  0.642673149573306, -0.412565627467277, 0.672207513790216, 0.967040903075167, 
  0.468047601527563, 0.637785901287577, 0.844643267672228, 0.531304065337568, 
  0.871382716325882, 0.652137447647579, 0.39953254735831, 0.683918734842367, 
  0.64113649695194, -1.12087270379329, 0.627669809872133, 0.294456709215146, 
  0.539711347678816, 0.710504348497859, 0.346598697399641, -1.07632639234637, 
  -1.09963086428893, -1.09739189778582, -1.13930502605998, -0.748393619076803
  ), .Dim = c(5L, 7L), .Dimnames = list(NULL, c("dDomingo", "dSegunda", 
  "dTerça", "dQuarta", "dQuinta", "dSexta", "dSábado")))

df <- data.frame(mat1)

library(tidyr)
df %>% 
  gather(key, value)

#>         key      value
#> 1  dDomingo -1.7835576
#> 2  dDomingo -1.7300859
#> 3  dDomingo -1.5033325
#> 4  dDomingo -1.7202608
#> 5  dDomingo -1.2347006
#> 6  dSegunda  0.5123224
#> 7  dSegunda  0.8552685
#> 8  dSegunda  0.9450167
#> 35  dSábado -0.7483936

# rename to whatever you want
df %>% 
  gather(key = "V1", value = "V2")

#>          V1         V2
#> 1  dDomingo -1.7835576
#> 2  dDomingo -1.7300859
#> 3  dDomingo -1.5033325
#> 4  dDomingo -1.7202608
#> 5  dDomingo -1.2347006
#> 6  dSegunda  0.5123224
#> 7  dSegunda  0.8552685
#> 8  dSegunda  0.9450167
#> 35  dSábado -0.7483936

修改：要返回wide格式，请使用tidyr::spread

library(dplyr)

df_long %>% 
  group_by(key) %>% 
  mutate(rowid = row_number()) %>% 
  spread(key, value) %>% 
  select(-rowid)

  dDomingo dSegunda dTerça dQuarta dQuinta dSexta dSábado
     <dbl>    <dbl>  <dbl>   <dbl>   <dbl>  <dbl>   <dbl>
1    -1.78    0.512  0.600   0.468   0.652  0.628  -1.08 
2    -1.73    0.855  0.643   0.638   0.400  0.294  -1.10 
3    -1.50    0.945 -0.413   0.845   0.684  0.540  -1.10 
4    -1.72    0.304  0.672   0.531   0.641  0.711  -1.14 
5    -1.23    0.919  0.967   0.871  -1.12   0.347  -0.748

由reprex package（v0.2.0）创建于2018-04-15。

Answer 3

t（你的结构在这里）

t(structure(c(-1.78355763661757, -1.73008589314791, -1.50333251627705, 
-1.72026078275706, -1.23470060199216, 0.512322374645035, 0.855268450002498, 
0.945016691336727, 0.304413384239457, 0.91894460806156, 0.599706795271625, 
0.642673149573306, -0.412565627467277, 0.672207513790216, 0.967040903075167, 
0.468047601527563, 0.637785901287577, 0.844643267672228, 0.531304065337568, 
0.871382716325882, 0.652137447647579, 0.39953254735831, 0.683918734842367, 
0.64113649695194, -1.12087270379329, 0.627669809872133, 0.294456709215146, 
0.539711347678816, 0.710504348497859, 0.346598697399641, -1.07632639234637, 
-1.09963086428893, -1.09739189778582, -1.13930502605998, -0.748393619076803
), .Dim = c(5L, 7L), .Dimnames = list(NULL, c("dDomingo", "dSegunda", 
"dTerça", "dQuarta", "dQuinta", "dSexta", "dSábado"))))

输出

              [,1]       [,2]       [,3]       [,4]       [,5]
dDomingo -1.7835576 -1.7300859 -1.5033325 -1.7202608 -1.2347006
dSegunda  0.5123224  0.8552685  0.9450167  0.3044134  0.9189446
dTerça    0.5997068  0.6426731 -0.4125656  0.6722075  0.9670409
dQuarta   0.4680476  0.6377859  0.8446433  0.5313041  0.8713827
dQuinta   0.6521374  0.3995325  0.6839187  0.6411365 -1.1208727
dSexta    0.6276698  0.2944567  0.5397113  0.7105043  0.3465987
dSábado  -1.0763264 -1.0996309 -1.0973919 -1.1393050 -0.7483936

将时间序列矩阵转换为ggplot格式

3 个答案:

注意