Question

我是r的新手，我正在尝试从给定的表中计算一个索引。我已经计算了下表中coulnmn的最小值，最大值和中值。我在列中有一些NaN。第一列和第二列是纬度和经度，从3开始是土壤湿度。

1            2       3      4        5       6       7        8      9       10       11
28.188  -111.438    NaN     NaN     NaN     NaN      NaN     NaN     NaN     NaN      NaN
28.188  -111.312    362.74  360.723 361.645 356.894 354.188 354.166 355.16  350.148 339.705
28.188  -111.188    189.991 188.153 188.96  184.49  182.028 181.776 182.322 175.402 161.836
28.188  -111.062    312.743 310.278 310.867 306.436 304.178 304.157 306.149 302.238 293.318
28.188  -110.938    146.97  144.582 145.274 141.367 139.66  140.137 142.949 141.15  135.776
28.188  -110.812    149.095 146.86  147.706 143.991 142.258 143.005 146.508 144.388 138.771
28.188  -110.688    147.06  144.825 145.479 141.687 139.826 140.714 144.882 142.481 136.582
28.188  -110.562    145.857 143.75  144.493 140.791 138.866 139.933 144.882 142.148 135.532
28.188  -110.438    135.207 133.254 134.163 130.436 128.345 129.526 135.307 132.01  123.974

我使用以下代码来计算最小值，中位数和最大值;

 data=read.table('NLDAS_MOISTURE_200CM.csv', header=TRUE, sep=',') 
    df=data.frame(data[3:11]) 
    Minimum=apply(X = df, MARGIN=2, FUN = min, na.rm = TRUE) 
    Maximum=apply(X = df, MARGIN=2, FUN = max, na.rm = TRUE) 
    Median=apply(X = df, MARGIN=2, FUN = median, na.rm = TRUE)

现在，我想使用以下条件计算每个单元格列的索引：

1）如果列的单元格值小于列的中值   （单元格值＆lt; Median）Index =（cell   值的中值）/（中值 - 最小值）×100

2）如果列的单元格值大于列的中值   （单元格值＆gt;中位数）索引=（单元格值 - 中位数）/（最大中值）* 100

如果某个机构可以使用if或for循环完成代码，我将不胜感激。感谢

加文

Answer 1

你可以尝试

df[-(1:2)] <-  lapply(df[,-(1:2)], function(x) {med <- median(x, na.rm=TRUE)
                    min1 <- min(x, na.rm=TRUE)
                   max1 <- max(x, na.rm=TRUE)
                  ifelse(x <= med, 100*(x-med)/(med-min1),
                                    100*(x-med)/(max1-med)) })

df[1:5]
#     X1       X2           X3          X4           X5
#1 28.188 -111.438           NA          NA           NA
#2 28.188 -111.312  100.0000000  100.000000  100.0000000
#3 28.188 -111.188   19.5253013   19.690246   19.7010032
#4 28.188 -111.062   76.7090200   76.524161   76.3880913
#5 28.188 -110.938   -8.6049493  -10.013107  -10.6078282
#6 28.188 -110.812    0.4739999    0.473519    0.5177805
#7 28.188 -110.688   -7.9056758   -8.082774   -8.9585261
#8 28.188 -110.562  -17.2526320  -16.622314  -16.8912667
#9 28.188 -110.438 -100.0000000 -100.000000 -100.0000000

或基于已计算的Minimum，Maximum，Median值

df[-(1:2)] <-  Map(function(v, w, x,y) ifelse(v<=w, 100*(v-w)/(w-x),
                  100*(v-w)/(y-w))  ,df[-(1:2)], Median, Minimum, Maximum)

要计算min，max，median，其他选项是使用dplyr

 library(dplyr)
  df %>% 
      mutate_each(funs(min=min(., na.rm=TRUE),
          max=max(., na.rm=TRUE), median=median(., na.rm=TRUE)), 3:11)

数据

df <- structure(list(X1 = c(28.188, 28.188, 28.188, 28.188, 28.188, 
28.188, 28.188, 28.188, 28.188), X2 = c(-111.438, -111.312, -111.188, 
 -111.062, -110.938, -110.812, -110.688, -110.562, -110.438), 
X3 = c(NaN, 362.74, 189.991, 312.743, 146.97, 149.095, 147.06, 
145.857, 135.207), X4 = c(NaN, 360.723, 188.153, 310.278, 
144.582, 146.86, 144.825, 143.75, 133.254), X5 = c(NaN, 361.645, 
188.96, 310.867, 145.274, 147.706, 145.479, 144.493, 134.163
), X6 = c(NaN, 356.894, 184.49, 306.436, 141.367, 143.991, 
141.687, 140.791, 130.436), X7 = c(NaN, 354.188, 182.028, 
304.178, 139.66, 142.258, 139.826, 138.866, 128.345), X8 = c(NaN, 
354.166, 181.776, 304.157, 140.137, 143.005, 140.714, 139.933, 
129.526), X9 = c(NaN, 355.16, 182.322, 306.149, 142.949, 
146.508, 144.882, 144.882, 135.307), X10 = c(NaN, 350.148, 
175.402, 302.238, 141.15, 144.388, 142.481, 142.148, 132.01
), X11 = c(NaN, 339.705, 161.836, 293.318, 135.776, 138.771, 
136.582, 135.532, 123.974)), .Names = c("X1", "X2", "X3", 
"X4", "X5", "X6", "X7", "X8", "X9", "X10", "X11"), class = "data.frame",
 row.names = c(NA, -9L))

Answer 2

以下是您可以立即执行整个操作的方法。这假设您不需要保存中值，最小值和最大值以供日后使用。如果你这样做，请发表评论。

此函数循环向量median的三个函数min，max和x，然后将变量发送到函数环境，使它们可供使用按ifelse中的名称及后续计算。

index <- function(x, ...) {
    funs <- c(med = median, min = min, max = max)
    list2env(lapply(funs, function(f) f(x, ...)), environment())
    ifel <- ifelse(x < med, med - min, max - med)
    (x - med) / ifel * 100
}

然后我们可以使用sapply或lapply或其他来获取索引值。在此之前，我使用cbind来避免新的分配。

indVals <- cbind(df[1:2], sapply(df[-c(1:2)], index, na.rm=TRUE))
indVals[,1:5]
#       X1       X2           X3          X4           X5
# 1 28.188 -111.438          NaN         NaN          NaN
# 2 28.188 -111.312  100.0000000  100.000000  100.0000000
# 3 28.188 -111.188   19.5253013   19.690246   19.7010032
# 4 28.188 -111.062   76.7090200   76.524161   76.3880913
# 5 28.188 -110.938   -8.6049493  -10.013107  -10.6078282
# 6 28.188 -110.812    0.4739999    0.473519    0.5177805
# 7 28.188 -110.688   -7.9056758   -8.082774   -8.9585261
# 8 28.188 -110.562  -17.2526320  -16.622314  -16.8912667
# 9 28.188 -110.438 -100.0000000 -100.000000 -100.0000000

R代码有助于按列计算单元格值的索引

2 个答案:

数据