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,我想计算每年,对于每个参数,变量让它命名saturation为等于它的位置
(sum(rain*value)/sum(rain)) for days where value is not NA * sum(rain per year)
我想用这个来做dplyr.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
它运作良好,给了我想要的东西.但是,我不得不创建两个data.frames df1和df2繁衍df1$sum_rain通过df2$specific_days获得saturation.不管怎么说,如果没有使用dplyr创建两个data.frames就可以做到这一点.
我们可以通过rain对非NA'值'进行子集化来在单个链中完成此操作is.na
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
不需要做另一件事join.这给出了OP的帖子中的预期输出,并且没有提供任何不正确的输出.
或者也可以这样做 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