I am new to loops in R and have a relatively simple dataset to process. My sample dataset consists of timestamps - time, cell phone id: id, and cell phone battery level: level My objective is to produce an output which takes the rate of battery decline over time, taking account recharge cycles. The beginning of a cycle can be identified where the following record's level is greater than the previous. In other words, while level <= lag(level), the cycle should continue, but as soon as level > lag(level), the cycle should restart
我是R中的循环新手,并且有一个相对简单的数据集来处理。我的样本数据集包括时间戳 - 时间,手机ID:id和手机电池电量:级别我的目标是产生一个输出,该输出随着时间的推移考虑电池电量下降的速率,考虑到充电周期。可以识别循环的开始,其中后续记录的级别大于前一级别。换句话说,当水平<=滞后(水平)时,循环应该继续,但是一旦水平>滞后(水平),循环应该重新开始
> test
time id level
1: 2017-12-25 14:10:03 1 81
2: 2017-12-25 14:20:03 1 81
3: 2017-12-25 14:30:04 1 81
4: 2017-12-25 14:40:04 1 73
5: 2017-12-25 14:50:04 1 70
6: 2017-12-25 15:00:03 1 70
7: 2017-12-25 15:10:04 1 65
8: 2017-12-25 15:20:04 1 62
9: 2017-12-25 15:30:04 1 61
10: 2017-12-25 15:40:04 1 60
11: 2017-12-25 15:50:03 1 60
12: 2017-12-25 16:00:04 1 60
13: 2017-12-25 16:10:04 1 95
14: 2017-12-25 16:20:03 1 95
15: 2017-12-25 16:30:04 1 95
16: 2017-12-25 16:40:04 1 94
17: 2017-12-25 16:50:04 1 92
18: 2017-12-25 17:00:03 1 90
19: 2017-12-25 17:10:04 1 81
20: 2017-12-25 17:20:03 1 79
21: 2017-12-25 17:30:03 2 100
22: 2017-12-25 17:40:03 2 100
23: 2017-12-25 17:50:03 2 100
24: 2017-12-25 18:00:03 2 90
25: 2017-12-25 18:10:03 2 85
26: 2017-12-25 18:20:03 2 75
27: 2017-12-25 18:30:04 2 65
28: 2017-12-25 18:40:03 2 54
29: 2017-12-25 18:50:03 2 32
30: 2017-12-25 19:00:03 2 11
31: 2017-12-25 19:10:04 2 92
32: 2017-12-25 19:20:04 2 92
33: 2017-12-25 19:30:03 2 91
34: 2017-12-25 19:40:04 2 90
35: 2017-12-25 19:50:04 2 90
36: 2017-12-25 20:00:03 2 81
37: 2017-12-25 20:10:03 2 79
38: 2017-12-25 20:20:04 2 99
39: 2017-12-25 20:30:04 2 96
40: 2017-12-25 20:40:03 2 96
In the sample dataset above, the intended output would look like this, where difftime = the difference in time between where the cycle started and stopped, diffcharge = the difference in battery level between where the cycle started and stopped, and rate = diffcharge/difftime
在上面的示例数据集中,预期输出将如下所示,其中difftime =循环开始和停止之间的时间差,diffcharge =循环开始和停止之间的电池电量差异,以及rate = diffcharge / difftime
> outcome
id start recharge difftime diffcharge rate
1 1 2017-12-25 14:10:03 2017-12-25 16:00:04 110.0167 21 0.1908801
2 1 2017-12-25 16:10:04 2017-12-25 17:20:03 69.98333 16 0.2286259
3 2 2017-12-25 17:30:03 2017-12-25 19:00:03 90 89 0.9888889
4 2 2017-12-25 19:10:04 2017-12-25 20:10:03 59.98333 13 0.2167269
5 2 2017-12-25 20:20:04 2017-12-25 20:40:03 19.98333 3 0.1501251
I have tried so far simply to create a while loop that concatenates the levels from each cycle, after which I can take the min, max, etc. with the following code but this does not produce the intended output.
到目前为止,我已经尝试过简单地创建一个while循环来连接每个循环的级别,之后我可以使用以下代码获取min,max等,但这不会产生预期的输出。
raw_data <- test
unique_id = unique(test$id)
for (id in unique_id)
{
onePhone <- raw_data[ which(raw_data$id == id), ]
onePhone <- onePhone[order(onePhone$time, decreasing = FALSE),]
cycle <- NULL
if(nrow(onePhone) >=2 ){
for(i in 2:nrow(onePhone)) {
while(onePhone[i-1,"level"] >= onePhone[i,"level"])
{
i = i+1
cycle <- c(z, onePhone[i,"level"])
}
print(cycle)
}
}
}
Any advice on how to use data.table, dplyr, or a simple while loop would be appreciated. Here is the sample data:
有关如何使用data.table,dplyr或简单while循环的任何建议将不胜感激。以下是示例数据:
> dput(test)
structure(list(time = structure(c(1514229003.91212, 1514229603.61297,
1514230204.14629, 1514230804.81938, 1514231404.36784, 1514232003.73393,
1514232604.17933, 1514233204.00143, 1514233804.68755, 1514234404.15599,
1514235003.99419, 1514235604.68204, 1514236204.18828, 1514236803.66526,
1514237404.0434, 1514238004.40609, 1514238604.02003, 1514239203.42238,
1514239804.19495, 1514240403.15927, 1514241003.87092, 1514241603.93167,
1514242203.77223, 1514242803.66758, 1514243403.33705, 1514244003.25017,
1514244604.05367, 1514245203.7921, 1514245803.2651, 1514246403.63888,
1514247004.02684, 1514247604.04009, 1514248203.99929, 1514248804.07401,
1514249404.11004, 1514250003.74613, 1514250603.88962, 1514251204.19115,
1514251804.06932, 1514252403.94181), class = c("POSIXct", "POSIXt"
), tzone = "EST"), id = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2), level = c(81, 81, 81, 73, 70, 70, 65, 62,
61, 60, 60, 60, 95, 95, 95, 94, 92, 90, 81, 79, 100, 100, 100,
90, 85, 75, 65, 54, 32, 11, 92, 92, 91, 90, 90, 81, 79, 99, 96,
96)), .Names = c("time", "id", "level"), class = c("data.table",
"data.frame"), row.names = c(NA, -40L), .internal.selfref = <pointer: 0x102010778>)
3 个解决方案
#1
2
Using @Hugh approach in first step and then getting the end result:
在第一步中使用@Hugh方法,然后获得最终结果:
test[, cycle := cumsum(level > shift(level, fill = first(level))), by = "id"]
x <- test[, .(start = min(time),
recharge = max(time),
diffcharge = max(level) - min(level)),
by = .(id, cycle)]
x[, difftime := as.numeric(recharge - start)]
x[, rate := diffcharge / difftime]
x
# id cycle start recharge diffcharge difftime rate
# 1: 1 0 2017-12-25 14:10:03 2017-12-25 16:00:04 21 110.01283 0.1908868
# 2: 1 1 2017-12-25 16:10:04 2017-12-25 17:20:03 16 69.98285 0.2286274
# 3: 2 0 2017-12-25 17:30:03 2017-12-25 19:00:03 89 89.99613 0.9889314
# 4: 2 1 2017-12-25 19:10:04 2017-12-25 20:10:03 13 59.99771 0.2166749
# 5: 2 2 2017-12-25 20:20:04 2017-12-25 20:40:03 3 19.99584 0.1500312
#2
1
If test is a data.table, you can use cumsum with shift. (shift is a function from data.table; it's the same as lag.)
如果test是data.table,你可以使用带有shift的cumsum。 (shift是data.table中的函数;它与lag相同。)
test[, cycle := cumsum(level > shift(level, fill = first(level))), by = "id"]
#3
1
Assuming you read test from a csv file:
假设您从csv文件中读取测试:
test<-read.csv("test.csv",stringsAsFactors = F)
test$DateTime<-paste(test$Date,test$time,by=" ")
test$Charge<-FALSE
test$Charge[1:((nrow(test)-1))]<-diff(test$level)>0
start=test[which(test$Charge)+1,]$DateTime
start<-c(test$DateTime[1],start)
start<-dmy_hms(start)
recharge<-filter(test,Charge)$DateTime
recharge<-c(recharge,tail(test$DateTime,1))
recharge<-dmy_hms(recharge)
difftime=recharge-start
startLevel=test[which(test$Charge)+1,]$level
startLevel=c(test$level[1],startLevel)
endLevel=filter(test,Charge)$level
endLevel=c(endLevel,tail(test$level,1))
diffcharge=startLevel-endLevel
rate=diffcharge/as.numeric(difftime)
id=filter(test,Charge)$id
id=c(id,tail(test$id,1))
outcome=data.frame(id=id,start=start,recharge=recharge,difftime=difftime,diffcharge=diffcharge,rate=rate)
#1
2
Using @Hugh approach in first step and then getting the end result:
在第一步中使用@Hugh方法,然后获得最终结果:
test[, cycle := cumsum(level > shift(level, fill = first(level))), by = "id"]
x <- test[, .(start = min(time),
recharge = max(time),
diffcharge = max(level) - min(level)),
by = .(id, cycle)]
x[, difftime := as.numeric(recharge - start)]
x[, rate := diffcharge / difftime]
x
# id cycle start recharge diffcharge difftime rate
# 1: 1 0 2017-12-25 14:10:03 2017-12-25 16:00:04 21 110.01283 0.1908868
# 2: 1 1 2017-12-25 16:10:04 2017-12-25 17:20:03 16 69.98285 0.2286274
# 3: 2 0 2017-12-25 17:30:03 2017-12-25 19:00:03 89 89.99613 0.9889314
# 4: 2 1 2017-12-25 19:10:04 2017-12-25 20:10:03 13 59.99771 0.2166749
# 5: 2 2 2017-12-25 20:20:04 2017-12-25 20:40:03 3 19.99584 0.1500312
#2
1
If test is a data.table, you can use cumsum with shift. (shift is a function from data.table; it's the same as lag.)
如果test是data.table,你可以使用带有shift的cumsum。 (shift是data.table中的函数;它与lag相同。)
test[, cycle := cumsum(level > shift(level, fill = first(level))), by = "id"]
#3
1
Assuming you read test from a csv file:
假设您从csv文件中读取测试:
test<-read.csv("test.csv",stringsAsFactors = F)
test$DateTime<-paste(test$Date,test$time,by=" ")
test$Charge<-FALSE
test$Charge[1:((nrow(test)-1))]<-diff(test$level)>0
start=test[which(test$Charge)+1,]$DateTime
start<-c(test$DateTime[1],start)
start<-dmy_hms(start)
recharge<-filter(test,Charge)$DateTime
recharge<-c(recharge,tail(test$DateTime,1))
recharge<-dmy_hms(recharge)
difftime=recharge-start
startLevel=test[which(test$Charge)+1,]$level
startLevel=c(test$level[1],startLevel)
endLevel=filter(test,Charge)$level
endLevel=c(endLevel,tail(test$level,1))
diffcharge=startLevel-endLevel
rate=diffcharge/as.numeric(difftime)
id=filter(test,Charge)$id
id=c(id,tail(test$id,1))
outcome=data.frame(id=id,start=start,recharge=recharge,difftime=difftime,diffcharge=diffcharge,rate=rate)