R 根据另一列计算两天和五天的平均值
我希望根据另一个变量（停留）的数量计算2天和5天的平均温度和湿度。如果住院时间为0（患者今天入院），则应根据同一日期（第1天）和前一天（第2天）的值计算平均值。同样，对于停留1，平均值是根据日期前2和3的值计算得出的。对于停留时间为8天及以上的所有值，从当前日期前的第9天和第10天开始计算2天的平均值。停留0天的5天平均值基于第1、2、3、4和5天的值。下表显示了计算的推导方式 需要示例输出R 根据另一列计算两天和五天的平均值,r,average,R,Average,我希望根据另一个变量（停留）的数量计算2天和5天的平均温度和湿度。如果住院时间为0（患者今天入院），则应根据同一日期（第1天）和前一天（第2天）的值计算平均值。同样，对于停留1，平均值是根据日期前2和3的值计算得出的。对于停留时间为8天及以上的所有值，从当前日期前的第9天和第10天开始计算2天的平均值。停留0天的5天平均值基于第1、2、3、4和5天的值。下表显示了计算的推导方式 需要示例输出 Date stay Temperature Humidity 2d temp 5d t
Date stay Temperature Humidity 2d temp 5d temp
9-Mar-98 6 4.23 74.32 na na
10-Mar-98 1 5.16 70.33 2.12 na
11-Mar-98 8 7.39 65.77 na na
14-Mar-98 3 6.63 66.46 6.27 3.35
23-Mar-98 2 11.03 62.94 11.13 13.97
24-Mar-98 10 10.87 57.35 10.09 8.78
4-Apr-98 0 9.64 59.21 8.68 9.51
5-Apr-98 5 9.70 88.30 16.14 13.81
stay 2d average 5d averages
0 1,2 1,2,3,4,5
1 2,3 2,3,4,5,6
2 3,4 3,4,5,6,7
3 4,5 4,5,6,7,8
4 5,6 5,6,7,8,9
5 6,7 6,7,8,9,10
6 7,8 7,8,9,10,11
7 8,9 8,9,10,11,12
8 9,10 9,10,11,12,13
> dput(mydata)
structure(list(date = structure(c(10294, 10295, 10296, 10297,
10298, 10299, 10300, 10301, 10302, 10303, 10304, 10305, 10306,
10307, 10308, 10309, 10310, 10311, 10312, 10313, 10314, 10315,
10316, 10317, 10318, 10319, 10320, 10321, 10322, 10323, 10324
), class = "Date"), stay = c(6, 1, 8, 11, 27, 3, 4, 5, 11, 13,
2, 17, 26, 6, 2, 10, 5, 2, 11, 24, 8, 11, 2, 8, 7, 30, 0, 5,
1, 2, 2), temperature = c(4.23000001907349, 5.15541648864746,
7.38499999046326, 9.47041666507721, 7.61999988555908, 6.62625002861023,
8.71875, 11.4608333110809, 11.2570832967758, 14.5691666603088,
10.3120833337307, 11.1216666698456, 11.1420832872391, 11.241666674614,
11.03125, 10.8691666722298, 12.4862499237061, 13.9341666698456,
11.8995833396912, 12.3716666698456, 12.5091667175293, 16.3833332061768,
15.8945832252502, 7.26666665077209, 7.0091667175293, 7.73125004768372,
9.63833332061768, 9.7045833170414, 11.4941666126251, 11.1304166316986,
11.3908333778381), humid = c(74.3199996948242, 70.3308334350586,
65.7658309936523, 69.2799987792969, 83.1170806884766, 66.4599990844727,
67.4225006103516, 85.7504196166992, 89.9520797729492, 65.2566680908203,
43.3604164123535, 51.7508316040039, 54.6866683959961, 68.2958297729492,
62.9420852661133, 57.3504180908203, 66.4137496948242, 57.6333351135254,
78.9029159545898, 84.5666656494141, 84.2004165649414, 71.2779159545898,
74.0320816040039, 65.2512512207031, 58.8224983215332, 62.4949989318848,
59.2054176330566, 88.2983322143555, 71.2545852661133, 78.0783309936523,
51.9004173278809)), datalabel = "", time.stamp = " 3 Apr 2015 22:09", .Names = c("date",
"stay", "temperature", "humid"), formats = c("%dD_m_Y", "%9.0g",
"%9.0g", "%9.0g"), types = c(255L, 255L, 255L, 255L), val.labels = c("",
"", "", ""), var.labels = c(" ", "", "temp", "rh"), expansion.fields = list(
c("_dta", "_lang_list", "default"), c("_dta", "_lang_c",
"default")), row.names = c("1", "2", "3", "4", "5", "6",
"7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17",
"18", "19", "20", "21", "22", "23", "24", "25", "26", "27", "28",
"29", "30", "31"), version = 12L, class = "data.frame")
你能举例说明你的预期产出吗？我不确定这是否正确 这是一个简单的循环。对于每一行，取停留值，加1并增加2（或5），然后提取与这两个指数对应的
温度平均值mydata <- structure(list(date=structure(c(10294,10295,10296,10297,10298,10299,10300,10301,10302,10303,10304,10305,10306,10307,10308,10309,10310,10311,10312,10313,10314,10315,10316,10317,10318,10319,10320,10321,10322,10323,10324),class="Date"),stay=c(6,1,8,11,27,3,4,5,11,13,2,17,26,6,2,10,5,2,11,24,8,11,2,8,7,30,0,5,1,2,2),temperature=c(4.23000001907349,5.15541648864746,7.38499999046326,9.47041666507721,7.61999988555908,6.62625002861023,8.71875,11.4608333110809,11.2570832967758,14.5691666603088,10.3120833337307,11.1216666698456,11.1420832872391,11.241666674614,11.03125,10.8691666722298,12.4862499237061,13.9341666698456,11.8995833396912,12.3716666698456,12.5091667175293,16.3833332061768,15.8945832252502,7.26666665077209,7.0091667175293,7.73125004768372,9.63833332061768,9.7045833170414,11.4941666126251,11.1304166316986,11.3908333778381),humid=c(74.3199996948242,70.3308334350586,65.7658309936523,69.2799987792969,83.1170806884766,66.4599990844727,67.4225006103516,85.7504196166992,89.9520797729492,65.2566680908203,43.3604164123535,51.7508316040039,54.6866683959961,68.2958297729492,62.9420852661133,57.3504180908203,66.4137496948242,57.6333351135254,78.9029159545898,84.5666656494141,84.2004165649414,71.2779159545898,74.0320816040039,65.2512512207031,58.8224983215332,62.4949989318848,59.2054176330566,88.2983322143555,71.2545852661133,78.0783309936523,51.9004173278809)),datalabel="",time.stamp="3Apr201522:09",.Names=c("date","stay","temperature","humid"),formats=c("%dD_m_Y","%9.0g","%9.0g","%9.0g"),types=c(255L,255L,255L,255L),val.labels=c("","","",""),var.labels=c("","","temp","rh"),expansion.fields=list(c("_dta","_lang_list","default"),c("_dta","_lang_c","default")),row.names=c("1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31"),version=12L,class="data.frame")
res <- lapply(mydata[, 'stay'], function(x)
c('two' = mean(mydata[(x + 1):(x + 2), 'temperature'], na.rm = TRUE),
'five' = mean(mydata[(x + 1):(x + 5), 'temperature'], na.rm = TRUE)))
cbind(mydata, do.call('rbind', res))
# date stay temperature humid two five
# 1 1998-03-09 6 4.230000 74.32000 10.089792 11.263583
# 2 1998-03-10 1 5.155416 70.33083 6.270208 7.251417
# 3 1998-03-11 8 7.385000 65.76583 12.913125 11.680417
# 4 1998-03-12 11 9.470417 69.28000 11.131875 11.081167
# 5 1998-03-13 27 7.620000 83.11708 10.599375 10.930000
# 6 1998-03-14 3 6.626250 66.46000 8.545208 8.779250
# 7 1998-03-15 4 8.718750 67.42250 7.123125 9.136583
# 8 1998-03-16 5 11.460833 85.75042 7.672500 10.526417
# 9 1998-03-17 11 11.257083 89.95208 11.131875 11.081167
# 10 1998-03-18 13 14.569167 65.25667 11.136458 11.912500
# 11 1998-03-19 2 10.312083 43.36042 8.427708 7.964083
# 12 1998-03-20 17 11.121667 51.75083 12.916875 13.419583
# 13 1998-03-21 26 11.142083 54.68667 9.671458 10.671667
# 14 1998-03-22 6 11.241667 68.29583 10.089792 11.263583
# 15 1998-03-23 2 11.031250 62.94209 8.427708 7.964083
# 16 1998-03-24 10 10.869167 57.35042 10.716875 10.969750
# 17 1998-03-25 5 12.486250 66.41375 7.672500 10.526417
# 18 1998-03-26 2 13.934167 57.63334 8.427708 7.964083
# 19 1998-03-27 11 11.899583 78.90292 11.131875 11.081167
# 20 1998-03-28 24 12.371667 84.56667 7.370208 9.115500
# 21 1998-03-29 8 12.509167 84.20042 12.913125 11.680417
# 22 1998-03-30 11 16.383333 71.27792 11.131875 11.081167
# 23 1998-03-31 2 15.894583 74.03208 8.427708 7.964083
# 24 1998-04-01 8 7.266667 65.25125 12.913125 11.680417
# 25 1998-04-02 7 7.009167 58.82250 11.358958 11.744167
# 26 1998-04-03 30 7.731250 62.49500 11.390833 11.390833
# 27 1998-04-04 0 9.638333 59.20542 4.692708 6.772167
# 28 1998-04-05 5 9.704583 88.29833 7.672500 10.526417
# 29 1998-04-06 1 11.494167 71.25459 6.270208 7.251417
# 30 1998-04-07 2 11.130417 78.07833 8.427708 7.964083
# 31 1998-04-08 2 11.390833 51.90042 8.427708 7.964083
mydata欢迎来到SO！请注意，此站点不是代码分发器。这里通常不会收到为您编写代码的请求。请花点时间查看常见问题解答，看看如何提问，以获得满意的答案。我会使用dplyr windows函数lead
/lag
为最后一天、最后一天的温度添加额外的列数。。。并使用ifelse
进行案例区分，例如mydata@ckluss我是R新手，链接示例有点难以理解。当我尝试您提供的代码时，我收到错误消息“找不到函数”%I已编辑问题并添加所需输出。你的剧本似乎不考虑停留。例如，对于停留时间为2的最后一行（4月8日），2天的平均值是11,49和9.7的总和（即当前日期的3天和4天）