dplyr：全年日值总和和同一公式中特定日值的总和答案

【问题标题】：dplyr: sum of daily values for whole year and sum of specific daily values in the same formuladplyr：全年日值总和和同一公式中特定日值的总和
【发布时间】：2016-06-01 04:42:39
【问题描述】：

使用dfdata.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，我想计算每一年，对于每个参数，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 来执行此操作。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

它工作得很好，给了我想要的东西。但是，我必须创建两个 data.frames df1 和 df2 并将 df1$sum_rain 乘以 df2$specific_days 以获得 saturation。在不使用 dplyr 创建两个 data.frames 的情况下是否可以做到这一点。

【问题讨论】：

标签： r dplyr

【解决方案1】：

我们可以通过用is.na 子集rain 为非NA '值' 来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

无需再做一次join。这给出了 OP 帖子中的预期输出，并且不会提供任何不正确的输出。

或者这也可以通过data.table来完成

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

【讨论】：

感谢您的宝贵时间和帮助

【解决方案2】：

您可以使用mutate 添加重复值的列，而不会像summarise 那样折叠data.frame：

df %>% tbl_df() %>%    # for printing
    group_by(site, year = lubridate::year(date), parameter) %>%    # add variables inline
    mutate(sum_rain = sum(rain)) %>%    # add column but don't collapse df
    filter(!is.na(value)) %>% 
    mutate(specific_days = sum(rain*value)/sum(rain), 
           saturation = sum_rain * specific_days) %>%    # add vars dependent on previous
    arrange(site, year, parameter) %>% 
    full_join(df)    # reinsert NA rows

# Source: local data frame [9,864 x 9]
# Groups: site, year, parameter [?]
# 
#          date   site     rain parameter        value  year    sum_rain specific_days saturation
#        (date) (fctr)    (dbl)    (fctr)        (dbl) (int)       (dbl)         (dbl)      (dbl)
# 1  2003-01-05 Site_1 32113070   param_A 1.050169e-07  2003 16033161650  1.225226e-06   19644.24
# 2  2003-01-06 Site_1 37830442   param_A 1.854250e-06  2003 16033161650  1.225226e-06   19644.24
# 3  2003-01-07 Site_1 76670445   param_A 1.386651e-06  2003 16033161650  1.225226e-06   19644.24
# 4  2003-01-11 Site_1 80337620   param_A 4.348852e-07  2003 16033161650  1.225226e-06   19644.24
# 5  2003-01-12 Site_1 77468528   param_A 1.118393e-06  2003 16033161650  1.225226e-06   19644.24
# 6  2003-01-13 Site_1 12609166   param_A 5.386190e-08  2003 16033161650  1.225226e-06   19644.24
# 7  2003-01-14 Site_1 80655681   param_A 1.881504e-06  2003 16033161650  1.225226e-06   19644.24
# 8  2003-01-15 Site_1 73617496   param_A 1.558818e-06  2003 16033161650  1.225226e-06   19644.24
# 9  2003-01-16 Site_1 30367141   param_A 2.068242e-06  2003 16033161650  1.225226e-06   19644.24
# 10 2003-01-17 Site_1 16743355   param_A 1.551760e-06  2003 16033161650  1.225226e-06   19644.24
# ..        ...    ...      ...       ...          ...   ...         ...           ...        ...

或者如果您只想要分组变量和饱和度，

df %>% tbl_df() %>%
    group_by(site, year = lubridate::year(date), parameter) %>%
    mutate(sum_rain = sum(rain), add = TRUE) %>%
    filter(!is.na(value)) %>% 
    summarise(saturation = unique(sum_rain * sum(rain*value)/sum(rain)))

# Source: local data frame [27 x 4]
# Groups: site, year [?]
# 
#      site  year parameter saturation
#    (fctr) (dbl)    (fctr)      (dbl)
# 1  Site_1  2003   param_A 19644.2422
# 2  Site_1  2003   param_B 28599.3730
# 3  Site_1  2003   param_c   320.6451
# 4  Site_1  2004   param_A 18333.6141
# 5  Site_1  2004   param_B 28856.5608
# 6  Site_1  2004   param_c   357.9545
# 7  Site_1  2005   param_A 17621.4250
# 8  Site_1  2005   param_B 27565.1503
# 9  Site_1  2005   param_c   338.8673
# 10 Site_2  2003   param_A  8584.3319
# ..    ...   ...       ...        ...

【讨论】：

感谢您的宝贵时间和帮助