我有一个名为df
的数据框:
df <- structure(list(opty_id = c(123L, 147L, 165L, 198L, 203L, 209L, 219L, 271L, 284L),
amount_opty = c(855252.21, 590259.86, 742818.76, 742818.76, 211760, 149190.3, 118924.22, 414663.687, 403313.319 ),
percent = c(NA, 0.309841175, NA, 0, 0.714923732, 0.295474594, 0.202868953, NA, 0.027372467)),
.Names = c("opty_id", "amount_opty", "percent"),
class = "data.frame", row.names = c(NA, -9L))
percent
列是与前一行相比的百分比增长。
我需要找出percent
小于10%的行对。
所需的输出为: id 8和id 9 (基本上是结果id和-1)
答案 0 :(得分:2)
你可以这样做。
将之前的diff
除以当前value
并找到分数(百分比):
c(df$diff[-1], NA) / df$value
# [1] 0.3333333 0.0000000 0.0400000 0.1666667 NA
检查分数差异是否小于0.1
(10%)
c(df$diff[-1], NA) / df$value < 0.1
# [1] FALSE TRUE TRUE FALSE NA
在您的数据中查找ID:
which(c(df$diff[-1], NA) / df$value < 0.1) + 1
# [1] 3 4
数据:
df <- structure(list(id = 1:5, value = c(60L, 40L, 50L, 48L, 40L),
diff = c(0L, 20L, 0L, 2L, 8L)), .Names = c("id", "value", "diff"), row.names = c(NA, -5L), class = "data.frame")
答案 1 :(得分:0)
您可以使用which()
查找df$percent
中小于0.1 且也大于0的元素:
main_id <- which(df$percent < 0.1 & df$percent > 0.0) # gives 9
target <- c(main_id - 1, main_id)
target
# [1] 8 9
如果您不想要行号,但想要opty_id
,请使用:
df$opty_id[target]
# [1] 271 284