Q：特殊条件，R的变化率；ForLoop/应用/滞后
我开始被R打湿了脚，我对时间序列的概念是全新的。谁能告诉我正确的方向来计算每月%的变化 我有不同年份、不同月份、不同城镇的数据&价格和变化率如是 在第二步中，我想在上表中取从9月开始到8月的所有可能对的平均比率->也就是说，9比10，9比11，…，9比8，10比11，…，10比8。。。7_8 计算： 变化率函数：a-b/b*100，其中a表示新月份，b表示上月 平均_比率：所有年份和城镇中各月份的平均值 %变化：log1+平均/x*100， 其中x是开始月份和结束月份之间的距离Q：特殊条件，R的变化率；ForLoop/应用/滞后,r,for-loop,if-statement,rows,R,For Loop,If Statement,Rows,我开始被R打湿了脚，我对时间序列的概念是全新的。谁能告诉我正确的方向来计算每月%的变化 我有不同年份、不同月份、不同城镇的数据&价格和变化率如是 在第二步中，我想在上表中取从9月开始到8月的所有可能对的平均比率->也就是说，9比10，9比11，…，9比8，10比11，…，10比8。。。7_8 计算： 变化率函数：a-b/b*100，其中a表示新月份，b表示上月 平均_比率：所有年份和城镇中各月份的平均值 %变化：log1+平均/x*100， 其中x是开始月份和结束月份之间的距离 struct
structure(list(hrvyear = c(1270, 1270, 1272, 1272, 1275, 1275
), m = c(5, 12, 2, 4, 2, 3), town = c("Chesterford", "Chesterford",
"Lopham", "Lopham", "Lopham", "Lopham"), `mean(price)` = c(80,
64, 74, 78, 59, 61)), row.names = c(NA, -6L), groups = structure(list(
hrvyear = c(1270, 1270, 1272, 1272, 1275, 1275), m = c(5,
12, 2, 4, 2, 3), .rows = structure(list(1L, 2L, 3L, 4L, 5L,
6L), ptype = integer(0), class = c("vctrs_list_of", "vctrs_vctr",
"list"))), row.names = c(NA, 6L), class = c("tbl_df", "tbl",
"data.frame"), .drop = TRUE), class = c("grouped_df", "tbl_df",
"tbl", "data.frame"))
我希望问题是清楚的。非常感谢您的建议。到目前为止，我在第一步中使用了此代码。但是，很明显，我不喜欢在每个月组重复多次该函数
may_july <- complete_mc %>%
filter(
m %in% c(5,7)
) %>%
arrange(town, hrvyear, m)
# create new column, to check whether the previous month is from the same year and the same town (e.g. we start with may to july comparison)
roc <- c()
for (i in 1:nrow(may_july)) {
if(may_july$hrvyear[i+1] == may_july$hrvyear[i] & may_july$town[i+1] == may_july$town[i]) {
roc <- c(roc, TRUE)
} else {
roc <- c(roc, FALSE)
}
}
# add FALSE for the first row of the roc column, as no previous row exists,
# and in order to combine matrix with vector
roc <- c(FALSE, roc)
tm <- cbind(may_july, roc)
# if previous month is from the same year and the same town, calculate the ratio,
# if not, add NA
roc2 <- c()
for(i in 1:nrow(may_july)) {
if(roc[i]==TRUE) {
roc2 <- c(roc2, (may_july$mean_price[i+1] - may_july$mean_price[i]) / (may_july$mean_price[i]))
} else {
roc2 <- c(roc2, NA)
}
}
# combine matrix with the final ratios
tt <- cbind(may_july, roc2)
roc3 <- na.omit(roc2)
# calculate the rate of change with the average ratio
may_to_july <- (log(1+mean(roc3))/2)*100
mean(roc3)
´´´
到目前为止，我在第一步中使用了这段代码。但是，很明显，我不喜欢在每个月组重复多次该函数
may_july <- complete_mc %>%
filter(
m %in% c(5,7)
) %>%
arrange(town, hrvyear, m)
# create new column, to check whether the previous month is from the same year and the same town (e.g. we start with may to july comparison)
roc <- c()
for (i in 1:nrow(may_july)) {
if(may_july$hrvyear[i+1] == may_july$hrvyear[i] & may_july$town[i+1] == may_july$town[i]) {
roc <- c(roc, TRUE)
} else {
roc <- c(roc, FALSE)
}
}
# add FALSE for the first row of the roc column, as no previous row exists,
# and in order to combine matrix with vector
roc <- c(FALSE, roc)
tm <- cbind(may_july, roc)
# if previous month is from the same year and the same town, calculate the ratio,
# if not, add NA
roc2 <- c()
for(i in 1:nrow(may_july)) {
if(roc[i]==TRUE) {
roc2 <- c(roc2, (may_july$mean_price[i+1] - may_july$mean_price[i]) / (may_july$mean_price[i]))
} else {
roc2 <- c(roc2, NA)
}
}
# combine matrix with the final ratios
tt <- cbind(may_july, roc2)
roc3 <- na.omit(roc2)
# calculate the rate of change with the average ratio
may_to_july <- (log(1+mean(roc3))/2)*100
mean(roc3)
´´´
您为此编写的函数几乎可以工作，但不要忘记将am$`meanprice`[i]-am$`meanprice`[i-1]放在括号中，这样在进行减法之前就不会进行除法 一个更简单的答案是在data.tables中使用shift函数，它类似于dplyr中的超前滞后函数。它们根据传递的参数选择前后行
library(data.table)
dt <- as.data.table(structure(list(hrvyear = c(1270, 1270, 1272, 1272, 1275, 1275
), m = c(5, 12, 2, 4, 2, 3), town = c("Chesterford", "Chesterford",
"Lopham", "Lopham", "Lopham", "Lopham"), `mean(price)` = c(80,
64, 74, 78, 59, 61)), row.names = c(NA, -6L), groups = structure(list(
hrvyear = c(1270, 1270, 1272, 1272, 1275, 1275), m = c(5,
12, 2, 4, 2, 3), .rows = structure(list(1L, 2L, 3L, 4L, 5L,
6L), ptype = integer(0), class = c("vctrs_list_of", "vctrs_vctr",
"list"))), row.names = c(NA, 6L), class = c("tbl_df", "tbl",
"data.frame"), .drop = TRUE), class = c("grouped_df", "tbl_df",
"tbl", "data.frame")))
# this changes the name of your mean(price)
colnames(dt)[4] <- 'price'
dt[, rate := (price - shift(price))/price * 100]
dt
hrvyear m town price rate
1: 1270 5 Chesterford 80 NA
2: 1270 12 Chesterford 64 -25.000000
3: 1272 2 Lopham 74 13.513514
4: 1272 4 Lopham 78 5.128205
5: 1275 2 Lopham 59 -32.203390
6: 1275 3 Lopham 61 3.278689
您为此编写的函数几乎可以工作，但不要忘记将am$`meanprice`[i]-am$`meanprice`[i-1]放在括号中，这样在进行减法之前就不会进行除法 一个更简单的答案是在data.tables中使用shift函数，它类似于dplyr中的超前滞后函数。它们根据传递的参数选择前后行
library(data.table)
dt <- as.data.table(structure(list(hrvyear = c(1270, 1270, 1272, 1272, 1275, 1275
), m = c(5, 12, 2, 4, 2, 3), town = c("Chesterford", "Chesterford",
"Lopham", "Lopham", "Lopham", "Lopham"), `mean(price)` = c(80,
64, 74, 78, 59, 61)), row.names = c(NA, -6L), groups = structure(list(
hrvyear = c(1270, 1270, 1272, 1272, 1275, 1275), m = c(5,
12, 2, 4, 2, 3), .rows = structure(list(1L, 2L, 3L, 4L, 5L,
6L), ptype = integer(0), class = c("vctrs_list_of", "vctrs_vctr",
"list"))), row.names = c(NA, 6L), class = c("tbl_df", "tbl",
"data.frame"), .drop = TRUE), class = c("grouped_df", "tbl_df",
"tbl", "data.frame")))
# this changes the name of your mean(price)
colnames(dt)[4] <- 'price'
dt[, rate := (price - shift(price))/price * 100]
dt
hrvyear m town price rate
1: 1270 5 Chesterford 80 NA
2: 1270 12 Chesterford 64 -25.000000
3: 1272 2 Lopham 74 13.513514
4: 1272 4 Lopham 78 5.128205
5: 1275 2 Lopham 59 -32.203390
6: 1275 3 Lopham 61 3.278689