计算值在行中出现的次数答案

【问题标题】：Count how many times value has occured in rows计算值在行中出现的次数
【发布时间】：2021-09-12 16:31:14
【问题描述】：

我有一个这样的数据框：

df5 <- data.frame(
  Ensembl_ID = c("ENSG00000000003", "ENSG00000000005", "ENSG00000000419", 
                 "ENSG00000000457", "ENSG00000000460", "ENSG00000000938", 
                 "ENSG00000000971", "ENSG00000001036", "ENSG00000001084", 
                 "ENSG00000001167" ), 
  `logFC 1` = c(-0.834165161710272, 1.02199443531549, 
                -0.558658947885705, -0.390114219973209, -1.23551839713296, 
                3.11429434221998, 0.283932163407262, -1.16908518620064, 
                -0.597054772455507, -0.593624543273255), 
  `logFC 2` = c(-1.18531035488942, 0.423719727339646, -1.23261719368372, 
                0.0855281133529292, -1.52366830232278, 3.36692586561211, 
                1.00323690950956, -0.000211248816114964, -4.74738483548391, 
                -0.318176231083024), 
  `logFC 3` = c(-0.262659255267546, 1.3962481061442, -0.548673555705647, 
                -0.0149651083306594, -1.45458689193089, 2.54126941463459, 
                1.17711308509307, -1.19425284921181, 1.17788731755683, 
                -0.367897054652365 ), 
  `logFC 4` = c(-0.840752912305256, 0.536548846040064, -0.277409459604357, 
                -0.241073614962264, -0.875313153342293, 1.61789645804321, 
                0.412287101096504, -1.11846661523232, -2.6274528854429, 
                -0.760452698231182), 
  `logFC 5` = c(-0.968784779247286, -0.502809694119192, -0.231526399163731, 
                -0.530038395734114, -0.706006018337411, 3.58264357077653, 
                -0.127521010699219, 0.270523387217103, 1.68335644352003, 
                -0.314902131571829), 
  `logFC 6` = c(-0.481754175843152, -0.440784040523259, -0.532975340622715, 
                -0.182089795101371, -0.564807490336052, 1.74119896504534, 
                -0.96169805631325, -0.721782763145306, -0.433459827401695, 
                -0.727495835245995 ), 
  `logFC 7` = c(-0.889343429110847, 1.07937149728343, -0.215144871523998, 
                -0.92234350748557, -0.832108253417702, 2.02456082994848, 
                -0.0434322861759954, -0.523126561938426, -0.556984056084809, 
                -0.740331742513503), 
  `logFC 8` = c(-0.858141567384178, 1.87728717064375, -0.381047638414538, 
                -0.613568289061259, -1.92838339196505, 2.23393705735665, 
                0.635389543483408, -0.466053620529111, -1.50483745357134, 
                -1.33400859143521), 
  `logFC 9` = c(-0.486388736112514, 0.789390852922639, -0.869434195504952, 
                -0.70405854858187, -1.16488184095428, 2.91497178849082, 
                -2.10331904053714, -0.571130459068143, -0.219526004620518, 
                -0.301435496557957)
)

我想在

当我尝试使用下面的代码找到它时，它返回错误的数字

df5$sumFP <- rowSums(df5[1:10,-1] < -0.2) #for less than -0.2
df5$sumTP <- rowSums(df5[1:10,-1] > 0.2) #for range of -0.2 to 0.2
df5$sumFN <- rowSums(df5[1:10,-1] < 0.2 & df5[1:10,-1] > -0.2) #for range of 0.2 to 1

知道为什么吗？还是有其他方法？

请帮助我。提前致谢

【问题讨论】：

标签： r dataframe

【解决方案1】：

我们可以将 dplyr 与 mutate 结合使用，并为每个条件调用 rowSums：

df5 %>% mutate(lower=rowSums(across(-Ensembl_ID, ~.x<(-0.2))),
               intermediate=rowSums(across(-Ensembl_ID, ~.x>=(-0.2) & .x<=0.2)),
               greater=rowSums(across(-Ensembl_ID, ~.x> 0.2 & .x<1)))

        Ensembl_ID    logFC.1       logFC.2     logFC.3    logFC.4    logFC.5    logFC.6
1  ENSG00000000003 -0.8341652 -1.1853103549 -0.26265926 -0.8407529 -0.9687848 -0.4817542
2  ENSG00000000005  1.0219944  0.4237197273  1.39624811  0.5365488 -0.5028097 -0.4407840
3  ENSG00000000419 -0.5586589 -1.2326171937 -0.54867356 -0.2774095 -0.2315264 -0.5329753
4  ENSG00000000457 -0.3901142  0.0855281134 -0.01496511 -0.2410736 -0.5300384 -0.1820898
5  ENSG00000000460 -1.2355184 -1.5236683023 -1.45458689 -0.8753132 -0.7060060 -0.5648075
6  ENSG00000000938  3.1142943  3.3669258656  2.54126941  1.6178965  3.5826436  1.7411990
7  ENSG00000000971  0.2839322  1.0032369095  1.17711309  0.4122871 -0.1275210 -0.9616981
8  ENSG00000001036 -1.1690852 -0.0002112488 -1.19425285 -1.1184666  0.2705234 -0.7217828
9  ENSG00000001084 -0.5970548 -4.7473848355  1.17788732 -2.6274529  1.6833564 -0.4334598
10 ENSG00000001167 -0.5936245 -0.3181762311 -0.36789705 -0.7604527 -0.3149021 -0.7274958
       logFC.7    logFC.8    logFC.9 lower intermediate greater
1  -0.88934343 -0.8581416 -0.4863887     9            0       0
2   1.07937150  1.8772872  0.7893909     2            0       3
3  -0.21514487 -0.3810476 -0.8694342     9            0       0
4  -0.92234351 -0.6135683 -0.7040585     6            3       0
5  -0.83210825 -1.9283834 -1.1648818     9            0       0
6   2.02456083  2.2339371  2.9149718     0            1       0
7  -0.04343229  0.6353895 -2.1033190     2            2       3
8  -0.52312656 -0.4660536 -0.5711305     7            1       1
9  -0.55698406 -1.5048375 -0.2195260     7            0       0
10 -0.74033174 -1.3340086 -0.3014355     9            0       0

我们也可以使用purrr:map_dfc 和dplyr::bind_cols：

map_dfc(list(lower = \(x) x < (-0.2),
             intermediate = \(x) x >= (-0.2) &  x <= 0.2,
             greater = \(x) x > 0.2 & x <1),
        ~.x(df5) %>% rowSums) %>%
        bind_cols(df5, .)

【讨论】：

谢谢你，亲爱的。你太棒了。谢谢
下次我会接受你的回答。谢谢
没关系，古普塔。您知道您可以随时更改接受的答案，对吗？反正TarJae的回答也很好。

【解决方案2】：

pivot_longer，然后是 summarise，最后是 cbind：

library(dplyr)
library(tidyr)

df5 %>% 
    pivot_longer(
        cols=-Ensembl_ID,
        names_to = "names",
        values_to = "values"
    ) %>% 
    group_by(Ensembl_ID) %>% 
    summarise(lesser =sum(values < -0.2),
              between = sum(values >= -0.2 & values < 0.2),
              greater = sum(values >= 0.2 & values < 1)) %>% 
    cbind(df5[,2:10])

        Ensembl_ID lesser between greater    logFC.1       logFC.2     logFC.3    logFC.4    logFC.5    logFC.6     logFC.7    logFC.8    logFC.9
1  ENSG00000000003      9       0       0 -0.8341652 -1.1853103549 -0.26265926 -0.8407529 -0.9687848 -0.4817542 -0.88934343 -0.8581416 -0.4863887
2  ENSG00000000005      2       0       3  1.0219944  0.4237197273  1.39624811  0.5365488 -0.5028097 -0.4407840  1.07937150  1.8772872  0.7893909
3  ENSG00000000419      9       0       0 -0.5586589 -1.2326171937 -0.54867356 -0.2774095 -0.2315264 -0.5329753 -0.21514487 -0.3810476 -0.8694342
4  ENSG00000000457      6       3       0 -0.3901142  0.0855281134 -0.01496511 -0.2410736 -0.5300384 -0.1820898 -0.92234351 -0.6135683 -0.7040585
5  ENSG00000000460      9       0       0 -1.2355184 -1.5236683023 -1.45458689 -0.8753132 -0.7060060 -0.5648075 -0.83210825 -1.9283834 -1.1648818
6  ENSG00000000938      0       0       0  3.1142943  3.3669258656  2.54126941  1.6178965  3.5826436  1.7411990  2.02456083  2.2339371  2.9149718
7  ENSG00000000971      2       2       3  0.2839322  1.0032369095  1.17711309  0.4122871 -0.1275210 -0.9616981 -0.04343229  0.6353895 -2.1033190
8  ENSG00000001036      7       1       1 -1.1690852 -0.0002112488 -1.19425285 -1.1184666  0.2705234 -0.7217828 -0.52312656 -0.4660536 -0.5711305
9  ENSG00000001084      7       0       0 -0.5970548 -4.7473848355  1.17788732 -2.6274529  1.6833564 -0.4334598 -0.55698406 -1.5048375 -0.2195260
10 ENSG00000001167      9       0       0 -0.5936245 -0.3181762311 -0.36789705 -0.7604527 -0.3149021 -0.7274958 -0.74033174 -1.3340086 -0.3014355

【讨论】：

感谢您的帮助。最优雅和美丽的解决方案。谢谢

【解决方案3】：

这个怎么样。

x <- df5[1:10,-1]

rowSums(x < -0.2)
# 1  2  3  4  5  6  7  8  9 10 
# 9  2  9  6  9  0  2  7  7  9 
rowSums(-0.2 <= x & x < 0.2)
# 1  2  3  4  5  6  7  8  9 10 
# 0  0  0  3  0  0  2  1  0  0 
rowSums(0.2 <= x & x <= 1)
# 1  2  3  4  5  6  7  8  9 10 
# 0  3  0  0  0  0  3  1  0  0

【讨论】：

感谢您的帮助。最优雅和美丽的解决方案。谢谢