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newfinder
Asked: 2024-10-11 18:47:40 +0800 CST

根据特定条件按行查找数值的数量

  • 772

我有这个示例数据框

 df <- data.frame(New=c("X2", "k 5, N 8", "N30","k 6, N 3", "K5", "S12", "K5", "k 1, N 18"),
    K_10=c(NA, NA, 3, 4,0,2,NA, NA), 
    K_11=c(NA, NA, NA, 4,0,3,NA, NA), 
    K_12=c(NA, 2, NA, NA,0,NA,NA,0), 
    K_13=c(0, 3, 5, NA,0,5,NA,NA), 
    K_14=c(NA, 3, 1, 2,10,10,NA,NA),
    K_15=c(NA, 2, 3, 5,15,10,NA,2),
    K_16=c(NA, 10, 1, 6,43,10,NA,56),
    K_17=c(NA, 5, 1, 3,1,10,NA,23),
    K_18=c(NA, 6, 4, 2,0,10,NA,12),
    K_19=c(NA, 3, 8, NA,3,10,NA,90),
    K_20=c(NA, 3, 19, 2,6,10,NA,59),
    K_21=c(NA, 3, 10, 2,8,10,NA,11),
    K_22=c(NA, 3, NA, 2,9,10,NA,10),
    K_23=c(NA, 3, NA, 2,90,10,NA,9))

df
           New K_10 K_11 K_12 K_13 K_14 K_15 K_16 K_17 K_18 K_19 K_20 K_21 K_22 K_23
    1        X2   NA   NA   NA    0   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
    2  k 5, N 8   NA   NA    2    3    3    2   10    5    6    3    3    3    3    3
    3       N30    3   NA   NA    5    1    3    1    1    4    8   19   10   NA   NA
    4  k 6, N 3    4    4   NA   NA    2    5    6    3    2   NA    2    2    2    2
    5        K5    0    0    0    0   10   15   43    1    0    3    6    8    9   90
    6       S12    2    3   NA    5   10   10   10   10   10   10   10   10   10   10
    7        K5   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
     8 k 1, N 18   NA   NA    0   NA   NA    2   56   23   12   90   59   11   10    9

我想为上述数据创建一些附加列。

仅当新列获得固定结构“k，N”时，第一列才会检查每行是否至少有 6 个数值，即“k 5，N 8”，“k 6，N 3”和“k 1，N 18”

当满足第一个附加列的标准时，第二列将计算我们有多少个数值。

仅当新列具有固定结构“k，N”时，第三列才会检查每行是否最多有 5 个数值（1,2,3,4,5），即“k 5，N 8”，“k 6，N 3”和“k 1，N 18”。

第四列将计算当满足第三附加列的标准时我们有多少个数值。

所以我想要类似的东西

  df_N <- data.frame(New=c("X2", "k 5, N 8", "N30","k 6, N 3", "K5", "S12", "K5", "k 1, N 18"),
    K_10=c(NA, NA, 3, 4,0,2,NA, NA), 
    K_11=c(NA, NA, NA, 4,0,3,NA, NA), 
    K_12=c(NA, 2, NA, NA,0,NA,NA,0), 
    K_13=c(0, 3, 5, NA,0,5,NA,NA), 
    K_14=c(NA, 3, 1, 2,10,10,NA,NA),
    K_15=c(NA, 2, 3, 5,15,10,NA,2),
    K_16=c(NA, 10, 1, 6,43,10,NA,56),
    K_17=c(NA, 5, 1, 3,1,10,NA,23),
    K_18=c(NA, 6, 4, 2,0,10,NA,12),
    K_19=c(NA, 3, 8, NA,3,10,NA,90),
    K_20=c(NA, 3, 19, 2,6,10,NA,59),
    K_21=c(NA, 3, 10, 2,8,10,NA,11),
    K_22=c(NA, 3, NA, 2,9,10,NA,10),
    K_23=c(NA, 3, NA, 2,90,10,NA,9),
At_least_6=c("Not Applicable","TRUE", "Not Applicable","TRUE", "Not Applicable", "Not Applicable", "Not Applicable","TRUE"),
Count_at_least_6=c("Not Applicable",8, "Not Applicable",6, "Not Applicable", "Not Applicable", "Not Applicable",5),
At_most_5=c("Not Applicable","FALSE", "Not Applicable","FALSE", "Not Applicable", "Not Applicable", "Not Applicable","FALSE"),
Count_at_most_5=c(0,0,0,0,0,0,0,0)              )


 df_N

    New K_10 K_11 K_12 K_13 K_14 K_15 K_16 K_17 K_18 K_19 K_20 K_21 K_22 K_23     At_least_6 Count_at_least_6      At_most_5 Count_at_most_5
1        X2   NA   NA   NA    0   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA Not Applicable   Not Applicable Not Applicable               0
2  k 5, N 8   NA   NA    2    3    3    2   10    5    6    3    3    3    3    3           TRUE                8          FALSE               0
3       N30    3   NA   NA    5    1    3    1    1    4    8   19   10   NA   NA Not Applicable   Not Applicable Not Applicable               0
4  k 6, N 3    4    4   NA   NA    2    5    6    3    2   NA    2    2    2    2           TRUE                6          FALSE               0
5        K5    0    0    0    0   10   15   43    1    0    3    6    8    9   90 Not Applicable   Not Applicable Not Applicable               0
6       S12    2    3   NA    5   10   10   10   10   10   10   10   10   10   10 Not Applicable   Not Applicable Not Applicable               0
7        K5   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA Not Applicable   Not Applicable Not Applicable               0
8 k 1, N 18   NA   NA    0   NA   NA    2   56   23   12   90   59   11   10    9           TRUE                5          FALSE               0

解释：在“Count_as_least”列的第一行中，我们的值为 7，因为从第六个值开始有 7 个值。换句话说，我们有前 6 个值

8   2    3    3    2   10

从第六个值开始，我们有

10    5    6    3    3    3    3    3

共计 8 个。

最后一列为零，因为倒数第二列的标准不满足。

我知道 dplyr 中的 counts 函数可以完成这个工作但我不确定。

我的真实数据包含数千行这样的行。

1 个回答

  1. Best Answer

    您可以使用 sapply 找出行是否符合条件，然后使用 rowsums 进行计数。在下面的代码中，我使用 df[, -1]，因为我排除了第一列 New，如果您的原始 df 的结构略有不同，您可以调整它。最后，关于您的解释，在列“Count_as_least”的第一行中有 7 个值，因为您添加了示例 df 中没有的初始数字“8”。

    library(tidyverse)
df_N <- df %>% 
  dplyr::mutate(
    # Determine if there are at least 6 numeric values
    At_least_6 = ifelse(grepl("^k \\d+, N \\d+$", New), 
                        rowSums(sapply(df[ , -1], function(x) !is.na(x) & is.numeric(x))) >= 6,
                        "Not Applicable"),
    # Count how many numeric values are at least 6 (subtracting 5)
    Count_at_least_6 = ifelse(At_least_6  == TRUE, 
                              rowSums(sapply(df[ , -1], function(x) !is.na(x) & is.numeric(x))) - 5,
                              "Not Applicable"),
    # Determine if there are at most 5 numeric values
    At_most_5 = ifelse(grepl("^k \\d+, N \\d+$", New), 
                       rowSums(sapply(df[ , -1], function(x) !is.na(x) & is.numeric(x))) <= 5,
                       "Not Applicable"),
    # Count how many numeric values are at most 5
    Count_at_most_5 = ifelse(At_most_5 == TRUE,
                             rowSums(sapply(df[ , -1], function(x) !is.na(x) & is.numeric(x))),
                             0))
df_N
        New K_10 K_11 K_12 K_13 K_14 K_15 K_16 K_17 K_18 K_19 K_20 K_21 K_22 K_23     At_least_6 Count_at_least_6      At_most_5 Count_at_most_5
1        X2   NA   NA   NA    0   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA Not Applicable   Not Applicable Not Applicable               0
2  k 5, N 8   NA   NA    2    3    3    2   10    5    6    3    3    3    3    3           TRUE                7          FALSE               0
3       N30    3   NA   NA    5    1    3    1    1    4    8   19   10   NA   NA Not Applicable   Not Applicable Not Applicable               0
4  k 6, N 3    4    4   NA   NA    2    5    6    3    2   NA    2    2    2    2           TRUE                6          FALSE               0
5        K5    0    0    0    0   10   15   43    1    0    3    6    8    9   90 Not Applicable   Not Applicable Not Applicable               0
6       S12    2    3   NA    5   10   10   10   10   10   10   10   10   10   10 Not Applicable   Not Applicable Not Applicable               0
7        K5   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA Not Applicable   Not Applicable Not Applicable               0
8 k 1, N 18   NA   NA    0   NA   NA    2   56   23   12   90   59   11   10    9           TRUE                5          FALSE               0
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