Python 在r中,输出绘图正在剪切或不在正确的帧中

Python 在r中,输出绘图正在剪切或不在正确的帧中,python,r,data-visualization,data-science,Python,R,Data Visualization,Data Science,在r中绘制用于eda分析的hist+密度图时,输出图未正确拟合到帧中 hist(Absenteeism_Data$Absenteeism.time.in.hours, col = "peachpuff", border = "black", prob = TRUE, xlab = "Absenteeism.time.in.hour", main = "Absenteeism_Data") lines(density(Absenteeism_Data$Absenteeism.time.

在r中绘制用于eda分析的hist+密度图时,输出图未正确拟合到帧中

  hist(Absenteeism_Data$Absenteeism.time.in.hours, col = "peachpuff", border = "black", prob = TRUE, xlab = "Absenteeism.time.in.hour", main = "Absenteeism_Data")
    lines(density(Absenteeism_Data$Absenteeism.time.in.hours, na.rm = TRUE), lwd = 2, col = "chocolate3")

由于您没有提供示例供我们使用,我将使用一些内置数据进行演示。我可以使用iris数据复制您遇到的问题类型

hist(iris$Sepal.Width, col = "peachpuff", border = "black", 
    prob = TRUE, xlab = "Sepal.Width", main = "Iris Data", breaks=5)
lines(density(iris$Sepal.Width, na.rm = TRUE), lwd = 2, col = "chocolate3")

问题是R在生成直方图时决定y值的范围。它不知道你的计划密度图。所以你必须告诉它要留足够的空间。您可以通过先运行“密度”并找到所需的最大值来完成此操作

DENS = density(iris$Sepal.Width, na.rm = TRUE)
YMax = max(DENS$y)
hist(iris$Sepal.Width, col = "peachpuff", border = "black", ylim=c(0,YMax),
    prob = TRUE, xlab = "Sepal.Width", main = "Iris Data", breaks=5)
lines(DENS, lwd = 2, col = "chocolate3")

欢迎来到SO。请花时间阅读stackoverflow.com/help/how-to-ask。它将帮助你设计出可靠的问题,希望能得到有用的答案。请不要截图。您能否通过共享您的数据样本使您的问题重现,以便其他人能够提供帮助(请不要使用
str()
head()
或截图)?您可以使用和包来帮助您实现这一点。另见&