Bogaso's questions

这是我上一篇文章《在单个图的顶部和底部绘制 2 个直方图》中的后续问题

我画了以下一对直方图

library(ggplot2)
set.seed(1)
dat = rbind(data.frame('val' = rnorm(100), 'met' = 'Metric1'), data.frame('val' = rt(100, 2), 'met' = 'Metric12'))

ggplot(dat, aes(x = val, fill = met)) + 
  geom_histogram(data    = dat[dat$met == 'Metric1',], 
                 breaks  = seq(-10, 10, 0.5),
                 mapping = aes(y = after_stat(density)),
                 colour  = "black",
                 alpha   = 0.3) +
  stat_function(fun = dnorm, args = list(mean = 0, sd = 1)) +
  geom_histogram(data    = dat[dat$met == 'Metric12', ],
                 breaks  = seq(-10, 10, 0.5),
                 mapping = aes(y = -after_stat(density)),
                 colour  = "black",
                 alpha   = 0.3)+
  coord_cartesian(xlim = c(-10, 10))

用这个我可以绘制normal density plot第一个直方图，但无法叠加第二个直方图的正常密度曲线。

关于如何绘制normal density plot第二个直方图的任何建议都很棒。

感谢您的时间。

我想在同一个绘图窗口上绘制两个直方图，但垂直放置。可以在https://blogs.sas.com/content/graphicallyspeaking/files/2013/11/MirrorHistogramVert.png找到一个这样的示例

下面是我的代码

library(ggplot2)
set.seed(1)
dat = rbind(data.frame('val' = rnorm(100), 'met' = 'Metric1'), data.frame('val' = rt(100, 2), 'met' = 'Metric12'))

ggplot(dat, aes(x = val, y = met, color = met)) + geom_histogram()

但是我遇到了以下错误

Error in `geom_histogram()`:
! Problem while computing stat.
ℹ Error occurred in the 1st layer.
Caused by error in `setup_params()`:
! `stat_bin()` must only have an x or y aesthetic.
Run `rlang::last_trace()` to see where the error occurred.

我还想为Normal distribution两个直方图添加概率曲线。

任何建议什么可能是正确的方法都将非常有帮助

我有以下 histogra 沉思 ggplot

library(ggplot2)
n = 1000
mean = 0
sd = 1
binwidth = 0.3 # passed to geom_histogram and stat_function
set.seed(1)
df <- data.frame(x = rnorm(n, mean, sd))

myplot = ggplot(df, aes(x = x, mean = mean, sd = sd, binwidth = binwidth, n = n)) +
    theme_bw() +
    geom_histogram(binwidth = binwidth, 
        colour = "white", fill = "cornflowerblue", size = 0.1)

现在我想在顶部中间区域添加一个自定义标签，但标签顶部应该留出一些边距，比如说 10 点。我尝试了以下方法

myplot+
geom_label(aes(x = mean((df$x)), y = 100), label = 'Plot with y', label.padding = unit(.7, 'lines'), fill = '#000000', color = '#ffffff', alpha = 0.2)

有没有什么方法可以动态地（仅基于数据而不进行任何硬编码）设置y = 100将标签保持在顶部并以 10 点的边距？

我有以下ggplot

library(ggplot2)
col_define = c('red', 'orange', 'blue', 'lightblue', 'green')
names(col_define) = as.vector(unique(diamonds$cut))

ggplot(diamonds, aes(color)) +
  geom_bar(aes(fill = cut)) +
  scale_fill_manual(values = col_define, breaks = names(col_define), aesthetics = c("fill")) + 
    theme(panel.grid.major = element_blank(),
        panel.grid.minor = element_blank(),
        legend.position=c(0,1),
        legend.justification = c(0,1),
        legend.box.margin = margin(5, l = 5, unit = 'mm'),
        legend.box = 'horizontal',
        legend.box.background = element_rect( fill = alpha('#7f7f7f', .2), size = 0.1, linetype = "solid", color = "#595959"),
        axis.text.x = element_text(angle = 30, hjust = 1)
        ) 

如您所见，图例似乎不在正确的位置。我想将图例放在绘图区域的左上角，但上边距和左边距会有 5 个像素。

此外，边距颜色和alpha 值也似乎无法正确显示。另外，图例项应水平放置

您能帮忙纠正一下吗？

我有下面的代码

import pandas as pd
from dateutil.relativedelta import relativedelta
date = pd.to_datetime(['2001-01-01', '2003-01-01'])
date + relativedelta(months = +6)

基本上，我想从日期数据框中计算出提前 6 个月的日期。

上述代码失败。

TypeError: unsupported operand type(s) for +: 'DatetimeArray' and 'relativedelta'

您能帮忙修正这个代码吗？

我有下面的代码

import pandas as pd
pd.to_datetime(['8/23/1999']).strftime("%m/%d/%Y").astype('str')

这会产生08/23/1999

但是我想得到8/23/1999

这个案例需要使用什么特定的格式吗？

我有以下自定义功能

import pandas as pd
def MyFn(DF : pd.DataFrame) -> float :
  return DF['Col_A'].values[1] - DF['Col_B'].values[1]

但是我想强制用户提供一个包含 2 列且列名称为的数据框'Col_A' and 'Col_B'

任何关于我如何做到这一点的见解将不胜感激。

我有一个自定义函数，如下所示

Function abc(Seed As Long)
Static iset As Integer
Static gset As Double
MsgBox Rnd(Seed)
MsgBox Rnd(Seed)
abc = 0
End Function

但是我Rnd(Seed)从 中得到了两个不同的值MsgBox。然而，自从我修复了种子后，我期望得到相同的值。

你能帮我了解一下种子是如何运作的吗？rnd()

我有以下数字序列

dat = c(-0.958694836980719, -0.105869068515775, -0.980669802804642, 
0.028491800336812, -0.635061963060457, 0, -0.930716072199491, 
0, 0.588929981219519, 0, 0.282785133581983, 0.611611557545295, 
0.0378059192851003, -0.797706616778708, 0, 0, 0, 0, 0.517465164205096, 
0, -0.488017358301909, 0, 0, -0.25055111810459, -0.649502253262175, 
-0.665111940088685, 0, 0.0833606598934977, -0.514514991719384, 
0.317596020495366, 0.794602807208168, 0, 0, -0.694999957450694, 
0.68257063515541, 0, -0.624026837516857, 0, 0, 0.450339396971535, 
-0.0302201203415504, -0.579393349186543, -0.844405771995823, 
0.315863139331068, -0.171564746000156, -0.0996391017024767, 0, 
0.0838315913186335, -0.36374768393003, 0, -0.572951822576261, 
-0.352439656458088, -0.637019777744324, 0, 0, -0.0952080968332089, 
0.617610001126072, -0.0816285346831291, 0.365239637846338, -0.0470848081799582, 
-0.925681187001364, 0, 0, 0.516154738675926, 0, 0.335416263139046, 
0.532290710398372, 0.18945326903775, 0.288998846320578, 0.125846440933334, 
-0.279555383136218, -0.456389602581116, -0.716237311784933, -0.0920396169199712, 
-0.2813560662731, 0.345024808219092, 0.338383493565635, 0.0058064242368383, 
0, 0.967537135446715, 0, 0.875822485251258, -0.431060076692186, 
-0.822882194591966, -0.62446221874739, -0.348475036137595, 0, 
0, 0.560600291351039, -0.855141781405395, 0, -0.706490388562219, 
0, -0.0451735735541755, 0.113810585454296, 0, -0.283307362362865, 
0.557656832607336, 0, 0, -0.909282421745824, -0.638976539326668, 
0.393719257131686, 0.301397306195678, 0, -0.74000532620085, 0.831188707386854, 
0.786577908437602, 0.296505948686095, 0.139539200765132, 0.88548929301196, 
0, 0.416614048955629, -0.316088049464881, 0, -0.323222691008726, 
-0.227387382853164, -0.562929988503375, 0, 0.283457267127375, 
0.713770547038207, 0.390959387881678, 0, 0, 0, 0.130514217066274, 
0.511687471126713, 0, 0.259730193040464, 0.741689274343481, -0.775924686373506, 
-0.495098678357968, 0.284476197633141, -0.900591805602638, -0.276707687274933, 
-0.191991699142624, 0, -0.916979262244761, 0.769473198941637, 
-0.241554713076157, 0, 0, 0, -0.231727168460227, -0.761155897450598, 
-0.678432614215555, -0.934782559884297, 0, 0, -0.314088267640064, 
0.186322473577494, 0, -0.235062452954516, 0, -0.314446614967701, 
-0.290302655565895, 0, 0, 0.144997859475891, 0, -0.840827052729484, 
0.88274732032249, 0.228399769981503, 0, 0.109512538112691, 0.671159365334607, 
0, 0, 0.0383683666391103, -0.798745428998881, 0, 0, 0, 0.244742671600118, 
-0.567358245295884, 0.509559617984882, 0.909915275452086, 0, 
0.904785111614818, 0, 0.207396095012435, 0, -0.156956691582582, 
0.776618542355675, -0.555791786131894, 0, 0.932355178469579, 
0.429624993163275, 0.0220608551322564, 0.146385826283492, 0, 
0.111119149224947, 0, 0.200025553735095, -0.429542452648371, 
-0.0528214778849886, 0, 0.353971870563417, 0, 0.768878060423797
)

我正在计算，但是'n' consecutive zeros上述顺序发生了很多次。

R中有没有直接的函数可以实现同样的目的？

0/1我有一个值列表，即

MyList = [0,0,0,0,1,0,1,0,0,0,1,1,0]

我的目标是 2 计算像 那样的对的数量(0,0) and (0,1)。

在上面的例子中，有5对(0,0)（即第一、第二、第三、第八、第九位置），并且(0,1)有3次这样的出现。

有什么函数/方法可以实现这一点吗？