小编Mik*_*kiK的帖子

数据表有向量作为条目 - 如何找出在哪一列中,然后只将向量的第二个条目作为单个整数

我有一个数据表tmp,它看起来像这样(只是一个简短的例子):

dput(tmp)
structure(list(`2020-03-29-00` = list(42.51, 0, 0, 0, 12.32), 
    `2020-03-29-01` = list(46.8, 0, 0, 0, 10.03), `2020-03-29-03` = list(
        c(46.8, 41.87), c(0, 0), c(0, 0), c(0, 0), c(10.03, 10.04
        )), `2020-03-29-04` = list(45.63, 0, 0, 0, 9.24), `2020-03-29-05` = list(
        40.86, 0, 0, 0, 9.06), `2020-03-29-06` = list(45.85, 
        0, 0, 0, 9.19), `2020-03-29-07` = list(43.68, 0, 0, 0, 
        10.39), `2020-03-29-08` = list(47.14, 0, 0, 0, 9.99), 
    `2020-03-29-09` = list(49.06, 0, 0, 0, 11.24)), row.names = c(NA, 
-5L), class …
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r vector data.table

5
推荐指数
1
解决办法
88
查看次数

创建带有副标题的 ggplotly 对象

我正在绘制散点图,ggplot()如下所示:

library(data.table)
library(plotly)
library(ggplot2)
library(lubridate)

dt.allData <- data.table(date = seq(as.Date('2020-01-01'), by = '1 day', length.out = 365),
                         DE = rnorm(365, 4, 1), Austria = rnorm(365, 10, 2), 
                         Czechia = rnorm(365, 1, 2), check.names = FALSE)

## Calculate Pearson correlation coefficient: ##
corrCoeff <- cor(dt.allData$Austria, dt.allData$DE,  method = "pearson", use = "complete.obs")
corrCoeff <- round(corrCoeff, digits = 2)

## Linear regression function extraction by creating linear model: ##
regLine <- lm(DE ~ Austria, data = dt.allData)

## Extract k …
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r ggplot2 ggplotly r-plotly

2
推荐指数
1
解决办法
839
查看次数

标签 统计

r ×2

data.table ×1

ggplot2 ×1

ggplotly ×1

r-plotly ×1

vector ×1