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mr.T
Asked:2020-10-07 20:18:07 +0000 UTC2020-10-07 20:18:07 +0000 UTC

hist() 函数，如何在其中进行等分

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我有一个"Р"带有观察值的矩阵（逐行），对于每一行我都计算分布hist()

破发 - 下注 50

A <- hist(P[n,],breaks = 50,plot = F)

但事实证明，矩阵的每一行 A$breaks都有不同的长度，尽管所有行的长度都"Р"相同，但如何确保 A$breaks 它们的大小始终相同

> head(P)
                P1        P2        P3        P4        P5        P6
 [9802,] -3.264212 -3.008633 -2.739602 -3.035536 -3.129696 -3.210405
 [9803,] -2.922858 -2.660034 -2.949140 -3.041128 -3.119975 -3.014846
 [9804,] -2.582244 -2.864561 -2.954390 -3.031385 -2.928724 -3.031385
 [9805,] -2.780997 -2.868646 -2.943775 -2.843604 -2.943775 -3.094031
 [9806,] -2.794781 -2.868306 -2.770272 -2.868306 -3.015358 -2.917324
 [9807,] -2.801911 -2.705737 -2.801911 -2.946172 -2.849998 -2.753824


> summary(P)
       P1                 P2                 P3                 P4                 P5                 P6          
 Min.   :-5.10760   Min.   :-5.36900   Min.   :-5.74779   Min.   :-6.29262   Min.   :-7.14529   Min.   :-8.36945  
 1st Qu.:-1.16145   1st Qu.:-1.17022   1st Qu.:-1.17861   1st Qu.:-1.18649   1st Qu.:-1.19516   1st Qu.:-1.20249  
 Median :-0.07062   Median :-0.07064   Median :-0.07154   Median :-0.07098   Median :-0.07052   Median :-0.07054  
 Mean   :-0.03485   Mean   :-0.03530   Mean   :-0.03575   Mean   :-0.03623   Mean   :-0.03673   Mean   :-0.03727  
 3rd Qu.: 1.09945   3rd Qu.: 1.10642   3rd Qu.: 1.11345   3rd Qu.: 1.12163   3rd Qu.: 1.12881   3rd Qu.: 1.13619  
 Max.   : 4.72881   Max.   : 5.05836   Max.   : 5.51505   Max.   : 6.10748   Max.   : 6.80844   Max.   : 7.68919

将数据矩阵缩短为 6 列，实际上有 200 列，但我不知道如何制作 200 行的示例

r

1 个回答

  1. Best Answer
    set.seed(1)
P <- matrix(rnorm(200 * 200), 200)
dim(P)
[1] 200 200
A <- hist(P[10, ], breaks = 50, plot = F)
length(A$breaks)
[1] 68
A <- hist(P[100, ], breaks = 50, plot = F)
length(A$breaks)
[1] 62

    显然，设置分割数的选项不起作用。

    但是您可以指定一个划分向量：

    breaks <- seq(range(P)[1], range(P)[2], length.out = 50)
A <- hist(P[10, ], breaks = breaks, plot = F)
length(A$breaks)
[1] 50
A <- hist(P[100, ], breaks = breaks, plot = F)
length(A$breaks)
[1] 50

    划分的范围必须匹配行中的取值范围，所以我取了整个矩阵的取值范围。

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