查找每日数据与 R 中每月平均值的百分比差异
Finding percentage difference of daily data compared to monthly means in R
我想知道不同地点的每日温度与指定的月平均温度有多少差异。我正在考虑让这个百分比差异值。例如，指定的月平均值为 20，而我有一些天数为 15(-25%)、25(+25%) 和 10 (-50%)。
我能想到的唯一方法是在每个位置创建一个具有月均值的重复列，然后使用 diff 函数或百分比差异公式计算列之间的差异。我想知道是否有更优雅、更简单的方法适合大数据？
然后我想使用这个每日趋势或差异并将其应用于一组不同的月度平均值以将其分解为每日数据。例如，假设每月平均值为 10，我有几天趋势为 +25% (12.5)、-50% (5) 和 -25% (7.5)。同样，是否有更优雅或更简单的方法？
如有任何帮助，我们将不胜感激。我对 R 还是很陌生！
这是一些示例数据：
示例数据
date <- c("2009-01-01", "2009-01-02", "2009-01-03", "2009-01-04","2009-01-05",
"2009-01-01", "2009-01-02", "2009-01-03", "2009-01-04","2009-01-05",
"2009-01-01", "2009-01-02", "2009-01-03", "2009-01-04","2009-01-05")
location <- c("A", "A", "A", "A", "A",
"B", "B", "B", "B", "B",
"C", "C", "C", "C", "C")
daily_temp <- c(10, 12, 12, 9, 8,
13, 14, 18, 8, 11,
14, 18, 20, 16, 17)
data_daily <- cbind(date, location, daily_temp)
mean_monthly <- c(12, 14, 16)
location_monthly <- c("A", "B", "C")
data_monthly <- cbind(mean_monthly, location_monthly)
以正确的格式制作源数据以供分析
df.daily <- as.data.frame( data_daily, stringsAsFactors = F)
df.monthly <- as.data.frame( data_monthly, stringsAsFactors = F)
整洁宇宙
library( tidyverse )
df.daily <- as.data.frame( data_daily, stringsAsFactors = FALSE )
df.monthly <- as.data.frame( data_monthly, stringsAsFactors = FALSE )
df.daily %>%
left_join( df.monthly, by = c( "location" = "location_monthly" ) ) %>%
mutate( daily_temp = as.numeric( daily_temp ) ) %>%
mutate( mean_monthly = as.numeric( mean_monthly ) ) %>%
mutate( delta_temp = ( daily_temp - mean_monthly ) / mean_monthly )
# date location daily_temp mean_monthly delta_temp
# 1 2009-01-01 A 10 12 -0.16666667
# 2 2009-01-02 A 12 12 0.00000000
# 3 2009-01-03 A 12 12 0.00000000
# 4 2009-01-04 A 9 12 -0.25000000
# 5 2009-01-05 A 8 12 -0.33333333
# 6 2009-01-01 B 13 14 -0.07142857
# 7 2009-01-02 B 14 14 0.00000000
# 8 2009-01-03 B 18 14 0.28571429
# 9 2009-01-04 B 8 14 -0.42857143
# 10 2009-01-05 B 11 14 -0.21428571
# 11 2009-01-01 C 14 16 -0.12500000
# 12 2009-01-02 C 18 16 0.12500000
# 13 2009-01-03 C 20 16 0.25000000
# 14 2009-01-04 C 16 16 0.00000000
# 15 2009-01-05 C 17 16 0.06250000
data.table
#less readable but usually faster , especially on larger datasets
library( data.table )
setDT( df.monthly )[, mean_monthly := as.numeric( mean_monthly )][setDT( df.daily )[, daily_temp := as.numeric( daily_temp )], on = c( "location_monthly==location" )][, delta_temp := ( daily_temp - mean_monthly ) / mean_monthly ][]
基准
data.table稍有优势
microbenchmark::microbenchmark( tidyverse = {df.daily %>%
left_join( df.monthly, by = c( "location" = "location_monthly" ) ) %>%
mutate( daily_temp = as.numeric( daily_temp ) ) %>%
mutate( mean_monthly = as.numeric( mean_monthly ) ) %>%
mutate( delta_temp = ( daily_temp - mean_monthly ) / mean_monthly )},
data.table = {setDT(df.monthly)[, mean_monthly := as.numeric( mean_monthly )][setDT(df.daily)[, daily_temp := as.numeric( daily_temp )], on = c( "location_monthly==location" )][, delta_temp := ( daily_temp - mean_monthly ) / mean_monthly ][]},
times = 100)
# Unit: milliseconds
# expr min lq mean median uq max neval
# tidyverse 2.318527 2.408303 2.579056 2.454999 2.513293 13.104373 100
# data.table 1.515959 1.590221 1.669511 1.643545 1.702141 2.345037 100
根据@Wimpel 的回复，这里有一些总结位置差异的方法。
df.combo <-
df.daily%>%
left_join( df.monthly, by = c( "location" = "location_monthly" ) ) %>%
mutate( daily_temp = as.numeric( daily_temp ) ) %>%
mutate( mean_monthly = as.numeric( mean_monthly ) ) %>%
mutate( delta_temp = ( daily_temp - mean_monthly ) / mean_monthly ) %>%
# Here I add the difference in degrees between daily temp and monthly avg temp
mutate( temp_dif = daily_temp - mean_monthly)
# For each location, what are some stats about those temp_dif values?
df.loc.stats <-
df.combo %>%
group_by(location) %>%
summarize(mean_dif = mean(temp_dif),
mean_abs_dif = mean(abs(temp_dif)),
SD_dif = sd(temp_dif))
df.loc.stats table 显示位置 B 的温度变化最大（例如，使用平均绝对差或标准差测量），而 A 在平均温度下最低，C 是最高：
df.loc.stats
# A tibble: 3 x 4
location mean_dif mean_abs_dif SD_dif
<chr> <dbl> <dbl> <dbl>
1 A -1.8 1.8 1.79
2 B -1.2 2.8 3.70
3 C 1 1.8 2.24
我想知道不同地点的每日温度与指定的月平均温度有多少差异。我正在考虑让这个百分比差异值。例如，指定的月平均值为 20，而我有一些天数为 15(-25%)、25(+25%) 和 10 (-50%)。
我能想到的唯一方法是在每个位置创建一个具有月均值的重复列，然后使用 diff 函数或百分比差异公式计算列之间的差异。我想知道是否有更优雅、更简单的方法适合大数据？
然后我想使用这个每日趋势或差异并将其应用于一组不同的月度平均值以将其分解为每日数据。例如，假设每月平均值为 10，我有几天趋势为 +25% (12.5)、-50% (5) 和 -25% (7.5)。同样，是否有更优雅或更简单的方法？
如有任何帮助，我们将不胜感激。我对 R 还是很陌生！
这是一些示例数据：
示例数据
date <- c("2009-01-01", "2009-01-02", "2009-01-03", "2009-01-04","2009-01-05",
"2009-01-01", "2009-01-02", "2009-01-03", "2009-01-04","2009-01-05",
"2009-01-01", "2009-01-02", "2009-01-03", "2009-01-04","2009-01-05")
location <- c("A", "A", "A", "A", "A",
"B", "B", "B", "B", "B",
"C", "C", "C", "C", "C")
daily_temp <- c(10, 12, 12, 9, 8,
13, 14, 18, 8, 11,
14, 18, 20, 16, 17)
data_daily <- cbind(date, location, daily_temp)
mean_monthly <- c(12, 14, 16)
location_monthly <- c("A", "B", "C")
data_monthly <- cbind(mean_monthly, location_monthly)
以正确的格式制作源数据以供分析
df.daily <- as.data.frame( data_daily, stringsAsFactors = F)
df.monthly <- as.data.frame( data_monthly, stringsAsFactors = F)
整洁宇宙
library( tidyverse )
df.daily <- as.data.frame( data_daily, stringsAsFactors = FALSE )
df.monthly <- as.data.frame( data_monthly, stringsAsFactors = FALSE )
df.daily %>%
left_join( df.monthly, by = c( "location" = "location_monthly" ) ) %>%
mutate( daily_temp = as.numeric( daily_temp ) ) %>%
mutate( mean_monthly = as.numeric( mean_monthly ) ) %>%
mutate( delta_temp = ( daily_temp - mean_monthly ) / mean_monthly )
# date location daily_temp mean_monthly delta_temp
# 1 2009-01-01 A 10 12 -0.16666667
# 2 2009-01-02 A 12 12 0.00000000
# 3 2009-01-03 A 12 12 0.00000000
# 4 2009-01-04 A 9 12 -0.25000000
# 5 2009-01-05 A 8 12 -0.33333333
# 6 2009-01-01 B 13 14 -0.07142857
# 7 2009-01-02 B 14 14 0.00000000
# 8 2009-01-03 B 18 14 0.28571429
# 9 2009-01-04 B 8 14 -0.42857143
# 10 2009-01-05 B 11 14 -0.21428571
# 11 2009-01-01 C 14 16 -0.12500000
# 12 2009-01-02 C 18 16 0.12500000
# 13 2009-01-03 C 20 16 0.25000000
# 14 2009-01-04 C 16 16 0.00000000
# 15 2009-01-05 C 17 16 0.06250000
data.table
#less readable but usually faster , especially on larger datasets
library( data.table )
setDT( df.monthly )[, mean_monthly := as.numeric( mean_monthly )][setDT( df.daily )[, daily_temp := as.numeric( daily_temp )], on = c( "location_monthly==location" )][, delta_temp := ( daily_temp - mean_monthly ) / mean_monthly ][]
基准
data.table稍有优势
microbenchmark::microbenchmark( tidyverse = {df.daily %>%
left_join( df.monthly, by = c( "location" = "location_monthly" ) ) %>%
mutate( daily_temp = as.numeric( daily_temp ) ) %>%
mutate( mean_monthly = as.numeric( mean_monthly ) ) %>%
mutate( delta_temp = ( daily_temp - mean_monthly ) / mean_monthly )},
data.table = {setDT(df.monthly)[, mean_monthly := as.numeric( mean_monthly )][setDT(df.daily)[, daily_temp := as.numeric( daily_temp )], on = c( "location_monthly==location" )][, delta_temp := ( daily_temp - mean_monthly ) / mean_monthly ][]},
times = 100)
# Unit: milliseconds
# expr min lq mean median uq max neval
# tidyverse 2.318527 2.408303 2.579056 2.454999 2.513293 13.104373 100
# data.table 1.515959 1.590221 1.669511 1.643545 1.702141 2.345037 100
根据@Wimpel 的回复，这里有一些总结位置差异的方法。
df.combo <-
df.daily%>%
left_join( df.monthly, by = c( "location" = "location_monthly" ) ) %>%
mutate( daily_temp = as.numeric( daily_temp ) ) %>%
mutate( mean_monthly = as.numeric( mean_monthly ) ) %>%
mutate( delta_temp = ( daily_temp - mean_monthly ) / mean_monthly ) %>%
# Here I add the difference in degrees between daily temp and monthly avg temp
mutate( temp_dif = daily_temp - mean_monthly)
# For each location, what are some stats about those temp_dif values?
df.loc.stats <-
df.combo %>%
group_by(location) %>%
summarize(mean_dif = mean(temp_dif),
mean_abs_dif = mean(abs(temp_dif)),
SD_dif = sd(temp_dif))
df.loc.stats table 显示位置 B 的温度变化最大（例如，使用平均绝对差或标准差测量），而 A 在平均温度下最低，C 是最高：
df.loc.stats
# A tibble: 3 x 4
location mean_dif mean_abs_dif SD_dif
<chr> <dbl> <dbl> <dbl>
1 A -1.8 1.8 1.79
2 B -1.2 2.8 3.70
3 C 1 1.8 2.24