这个问题在这里已有答案：
我有以下向量，包括几秒钟的时间：
times <- c(93.813, 89.662, 89.246, 88.641, 88.759, 89.034, 88.783, 88.651, 88.664, 89.028, 88.784, 88.633, 88.616, 88.300, 88.217, 87.954, 88.377, 88.203,
87.836, 87.832, 87.815, 88.514, 88.637, 88.496, 87.338, 87.450, 87.638, 87.511, 87.547, 87.240, 87.235, 87.266, 87.362, 87.457, 87.376, 87.465,
87.108, 87.358, 87.229, 87.273, 87.043, 86.739, 87.397, 87.540, 87.435, 87.448, 87.098, 87.184, 87.071, 87.238, 86.890, 87.898, 86.983, 86.540,
86.830, 86.610, 85.580, 86.627)
而且我希望返回另一个向量，该向量将是该元素的“最佳最小值”。结果将是：
c(93.813, 89.662, 89.246, 88.641, 88.641, 88.641, 88.641, ...)
等等，直到更短的时间发生。我一直在玩lag，就像这样：
ifelse(times < lag(times), times, lag(times))
但那没用。
你是在追求cummin（即vector的累积最小值）吗？
cummin(times)
# [1] 93.813 89.662 89.246 88.641 88.641 88.641 88.641 88.641 88.641 88.641
#[11] 88.641 88.633 88.616 88.300 88.217 87.954 87.954 87.954 87.836 87.832
#[21] 87.815 87.815 87.815 87.815 87.338 87.338 87.338 87.338 87.338 87.240
#[31] 87.235 87.235 87.235 87.235 87.235 87.235 87.108 87.108 87.108 87.108
#[41] 87.043 86.739 86.739 86.739 86.739 86.739 86.739 86.739 86.739 86.739
#[51] 86.739 86.739 86.739 86.540 86.540 86.540 85.580 85.580
也许有人可以提出更优雅的解决方案，但有效的解决方案如下：
current = NULL
min.vec = NULL
for(i in 1:length(times)){
current = c(current, times[i])
min.vec = c(min.vec, min(current))
}
> min.vec
[1] 93.813 89.662 89.246 88.641 88.641 88.641 88.641 88.641 88.641
[10] 88.641 88.641 88.633 88.616 88.300 88.217 87.954 87.954 87.954
[19] 87.836 87.832 87.815 87.815 87.815 87.815 87.338 87.338 87.338
[28] 87.338 87.338 87.240 87.235 87.235 87.235 87.235 87.235 87.235
[37] 87.108 87.108 87.108 87.108 87.043 86.739 86.739 86.739 86.739
[46] 86.739 86.739 86.739 86.739 86.739 86.739 86.739 86.739 86.540
[55] 86.540 86.540 85.580 85.580