r - dplyr:同一公式中全年每日值的总和和特定每日值的总和

Question

使用df data.frame

date <- rep(as.Date(seq(as.Date("2003-01-01"), 
                        as.Date("2005-12-31"), by = 1), 
                    format="%Y-%m-%d"), 9)
site <- c(rep("Site_1", 3*1096), rep("Site_2", 3*1096), rep("Site_3", 3*1096))
rain <- c(rep(as.numeric(sample(1.1e6:87e6, 1096, replace=T)),3),
               rep(as.numeric(sample(1.3e5:56e6, 1096, replace=T)),3),
               rep(as.numeric(sample(5e5:77e6, 1096, replace=T)),3))
parameter <- rep(c(rep("param_A", 1096), rep("param_B", 1096), rep("param_c", 1096)), 3)
value <- c(runif(1096, 0.005, 2.3)/1e6, 
           runif(1096, 0.5, 3.1)/1e6,
           runif(1096, 0.003, 0.04)/1e6,
           runif(1096, 0.002, 1.7)/1e6, 
           runif(1096, 0.3, 4.5)/1e6,
           runif(1096, 0.001, 0.07)/1e6,
           runif(1096, 0.007, 2.7)/1e6, 
           runif(1096, 0.4, 2.8)/1e6,
           runif(1096, 0.004, 0.09)/1e6)

df <- data.frame( date, site, rain, parameter, value)
df[c(1:4, 8:10, 30:35, 60:65, 90:97, 100:125, 524:645, 
     1000:1100, 1400:1540, 1789:1890, 2100:2250,
     2459:2765, 3942:3987, 4600:4698, 5210:5310, 6081:6154, 7613:7689, 
     8809:8888, 9120:9190, 9600:9650), 5] <- NA

对于每个站点，我想计算每年的每个参数，一个变量让我们将其命名为saturation，它等于

(sum(rain*value)/sum(rain)) for days where value is not NA * sum(rain per year)

我想使用 dplyr 来做到这一点。我尝试了以下代码

library(dplyr)
df1 <- df %>%
  dplyr::mutate(year = factor(format(date, "%Y"))) %>%
  dplyr::arrange(site, year, parameter)  %>%
  dplyr::group_by(site, year, parameter ) %>%
  dplyr::summarise(sum_rain = sum(rain))

df2 <- df %>%
  dplyr::mutate(year = factor(format(date, "%Y"))) %>%
  dplyr::arrange(site, year, parameter)  %>%
  dplyr::group_by(site, year, parameter ) %>%
  dplyr::filter (!is.na(value)) %>%
  dplyr::summarise(specific_days = sum(rain*value)/sum(rain))

saturation <- df1$sum_rain * df2$specific_days

它运行良好，满足了我的需求。但是，我必须创建两个 data.frames df1 和 df2 并将 df1$sum_rain 乘以 df2$specific_days 获取饱和度。无论如何，如果不使用 dplyr 创建两个 data.frames 就可以做到这一点。

Answer 1

我们可以通过将非 NA“值”的 rain 子集化为 is.na

在单个链中执行此操作

res <- df %>%
        mutate(year = factor(format(date, "%Y"))) %>%
        arrange(site, year, parameter)  %>%
        group_by(site, year, parameter ) %>% 
        summarise(sum_rain = sum(rain), 
            specific_days = sum(rain*value, na.rm=TRUE)/sum(rain[!is.na(value)])) %>% 
        mutate(saturation = sum_rain * specific_days)
res %>%
    as.data.frame()
#      site year parameter    sum_rain    specific_days saturation
#1  Site_1 2003   param_A 15988875602 0.00000123589041 19760.4980
#2  Site_1 2003   param_B 15988875602 0.00000172552158 27589.1499
#3  Site_1 2003   param_c 15988875602 0.00000002161544   345.6067
#4  Site_1 2004   param_A 15180127505 0.00000116507160 17685.9355
#5  Site_1 2004   param_B 15180127505 0.00000181695952 27581.6772
#6  Site_1 2004   param_c 15180127505 0.00000002185010   331.6873
#7  Site_1 2005   param_A 16058234005 0.00000120130563 19290.8469
#8  Site_1 2005   param_B 16058234005 0.00000186185975 29898.1795
#9  Site_1 2005   param_c 16058234005 0.00000002049335   329.0870
#10 Site_2 2003   param_A  9930134442 0.00000079639249  7908.2845
#11 Site_2 2003   param_B  9930134442 0.00000246576645 24485.3923
#12 Site_2 2003   param_c  9930134442 0.00000003348046   332.4655
#13 Site_2 2004   param_A 10926778631 0.00000088141235  9630.9976
#14 Site_2 2004   param_B 10926778631 0.00000244015257 26663.0070
#15 Site_2 2004   param_c 10926778631 0.00000003448817   376.8447
#16 Site_2 2005   param_A  9599581600 0.00000089477811  8589.4955
#17 Site_2 2005   param_B  9599581600 0.00000238522373 22897.1498
#18 Site_2 2005   param_c  9599581600 0.00000003442887   330.5027
#19 Site_3 2003   param_A 13711985538 0.00000142896664 19593.9700
#20 Site_3 2003   param_B 13711985538 0.00000157700917 21623.9270
#21 Site_3 2003   param_c 13711985538 0.00000004665944   639.7935
#22 Site_3 2004   param_A 14371047715 0.00000134324260 19303.8035
#23 Site_3 2004   param_B 14371047715 0.00000156583784 22502.7303
#24 Site_3 2004   param_c 14371047715 0.00000004859102   698.3039
#25 Site_3 2005   param_A 13729491381 0.00000131305086 18027.5205
#26 Site_3 2005   param_B 13729491381 0.00000159005889 21830.6999
#27 Site_3 2005   param_c 13729491381 0.00000004616979   633.8878





identical(df1['sum_rain'], res['sum_rain'])
#[1] TRUE

identical(df2['specific_days'], res['specific_days'])
#[1] TRUE

无需执行另一个加入。这给出了 OP 帖子中的预期输出，并且没有提供任何不正确的输出。

---

或者这也可以用 data.table

library(data.table)
setDT(df)[, .(sum_rain = sum(rain), 
              specific_days = sum(rain*value, na.rm=TRUE)/sum(rain[!is.na(value)])),
            by =  .(site, year= factor(format(date, "%Y")), parameter)
      ][, saturation := sum_rain * specific_days][]
#      site year parameter    sum_rain    specific_days saturation
# 1: Site_1 2003   param_A 15988875602 0.00000123589041 19760.4980
# 2: Site_1 2004   param_A 15180127505 0.00000116507160 17685.9355
# 3: Site_1 2005   param_A 16058234005 0.00000120130563 19290.8469
# 4: Site_1 2003   param_B 15988875602 0.00000172552158 27589.1499
# 5: Site_1 2004   param_B 15180127505 0.00000181695952 27581.6772
# 6: Site_1 2005   param_B 16058234005 0.00000186185975 29898.1795
# 7: Site_1 2003   param_c 15988875602 0.00000002161544   345.6067
# 8: Site_1 2004   param_c 15180127505 0.00000002185010   331.6873
# 9: Site_1 2005   param_c 16058234005 0.00000002049335   329.0870
#10: Site_2 2003   param_A  9930134442 0.00000079639249  7908.2845
#11: Site_2 2004   param_A 10926778631 0.00000088141235  9630.9976
#12: Site_2 2005   param_A  9599581600 0.00000089477811  8589.4955
#13: Site_2 2003   param_B  9930134442 0.00000246576645 24485.3923
#14: Site_2 2004   param_B 10926778631 0.00000244015257 26663.0070
#15: Site_2 2005   param_B  9599581600 0.00000238522373 22897.1498
#16: Site_2 2003   param_c  9930134442 0.00000003348046   332.4655
#17: Site_2 2004   param_c 10926778631 0.00000003448817   376.8447
#18: Site_2 2005   param_c  9599581600 0.00000003442887   330.5027
#19: Site_3 2003   param_A 13711985538 0.00000142896664 19593.9700
#20: Site_3 2004   param_A 14371047715 0.00000134324260 19303.8035
#21: Site_3 2005   param_A 13729491381 0.00000131305086 18027.5205
#22: Site_3 2003   param_B 13711985538 0.00000157700917 21623.9270
#23: Site_3 2004   param_B 14371047715 0.00000156583784 22502.7303
#24: Site_3 2005   param_B 13729491381 0.00000159005889 21830.6999
#25: Site_3 2003   param_c 13711985538 0.00000004665944   639.7935
#26: Site_3 2004   param_c 14371047715 0.00000004859102   698.3039
#27: Site_3 2005   param_c 13729491381 0.00000004616979   633.8878