回归分析未显示摘要 [英] Regression analysis not showing summary
问题描述
以下是我的数据
> x
day sum
1 2015-04-14 129
2 2015-04-15 129
3 2015-04-16 129
4 2015-04-17 899
5 2015-04-18 899
6 2015-04-19 899
7 2015-04-20 899
8 2015-04-21 899
9 2015-04-22 899
10 2015-04-23 899
11 2015-04-24 899
12 2015-04-25 899
13 2015-04-26 899
14 2015-04-27 899
15 2015-04-28 899
16 2015-04-29 899
17 2015-04-30 899
18 2015-05-01 899
19 2015-05-02 899
20 2015-05-03 899
21 2015-05-04 899
22 2015-05-05 899
23 2015-05-06 899
24 2015-05-07 899
25 2015-05-08 899
26 2015-05-09 899
27 2015-05-10 899
28 2015-05-11 899
29 2015-05-12 920
30 2015-05-13 920
31 2015-05-14 920
32 2015-05-15 920
33 2015-05-16 920
34 2015-05-17 920
35 2015-05-18 920
36 2015-05-19 920
37 2015-05-20 920
38 2015-05-21 920
39 2015-05-22 920
40 2015-05-23 920
41 2015-05-24 920
42 2015-05-25 920
43 2015-05-26 920
44 2015-05-27 920
45 2015-05-28 920
46 2015-05-29 920
47 2015-05-30 920
48 2015-05-31 920
49 2015-06-01 1076
我想做一个回归分析，找出总和会变成 6000 的数据.我做了以下，
I wanted to do a regression analysis to find out the data at which the sum would become 6000. I did the following,
> q<-lm(day ~ sum, data=x)
> q
Call:
lm(formula = day ~ sum, data = x)
Coefficients:
(Intercept) sum
1.653e+04 3.584e-02
> as.Date(predict(q, data.frame(sum=6000)))
1
"2015-11-08"
我能够预测日期.但我不确定准确性并想改进它.但我无法查看回归的摘要.我收到以下错误，
I am able to predict the dates. But I am not sure of the accuracy and want to improve it. But I am not able to view the summary of the regression. I get the following error,
> summary(q)
Error in Ops.difftime((f - mean(f)), 2) :
'^' not defined for "difftime" objects
以防万一，变量类型很重要，
Just in case, the variable types matter,
> typeof(x)
[1] "list"
> typeof(x$day)
[1] "double"
> typeof(x$sum)
[1] "integer"
> class(x$day)
[1] "Date"
当我看过以前的论坛时，
When I had a look at a previous forum,
下面给出了解决方案，
我通过将索引从时间值更改为标准整数索引解决了这个问题，一切运行良好.
I solved this by changing the index from time values to a standard integer index, and everything ran fine.
但我不知道该怎么做?
任何人都可以帮我解决这个问题并说出我需要做什么才能在此处获得摘要吗?
Can anybody help me with this and say what I need to do to get the summary here?
谢谢
推荐答案
R 中的日期实际上只是将数值按照一定的规则重新转换为日期格式.例如，Date 格式是自 1970 年 1 月 1 日以来经过的天数，POSIXct 格式是自 1970 年 1 月 1 日以来经过的秒数，参考 UTC 时区.以下是几个例子:
Dates in R are actually just numeric values recast into a date format according to certain rules. For example, Date format is the number of days elapsed since Jan 1, 1970 and POSIXct format is the number of seconds elapsed since Jan 1, 1970 referenced to the UTC time zone. Here are a few examples:
as.numeric(as.Date("1970-01-01", tz="UTC"))
# 0
as.numeric(as.Date("1970-01-05", tz="UTC"))
# 4
as.numeric(as.POSIXct("1970-01-01 00:00:00", tz="UTC"))
# 0
as.numeric(as.POSIXct("1970-01-05 00:00:00", tz="UTC"))
# 345600
解决问题的一种方法是将日期转换为数字格式，对数字数据运行回归，然后在最后转换回日期.
One way to deal with your problem is to convert the dates to a numeric format, run the regression on the numeric data, and then convert back to date at the end.
在下面的代码中，我假设 x$day 以 POSIXct 格式开始.
In the code below, I've assumed x$day starts out in POSIXct format.
# Convert POSIXct date to a numeric value equal to number of days since Jan 1, 1970
x$day.numeric = as.numeric(x$day)/(24*60*60)
# Regression model
q <- lm(day.numeric ~ sum, data=x)
summary(q)
Call:
lm(formula = day.numeric ~ sum, data = x)
Residuals:
Min 1Q Median 3Q Max
-22.253 -10.253 1.747 9.994 20.994
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.653e+04 8.448e+00 1956.996 < 2e-16 ***
sum 3.584e-02 9.550e-03 3.753 0.00048 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 12.67 on 47 degrees of freedom
Multiple R-squared: 0.2306, Adjusted R-squared: 0.2142
F-statistic: 14.09 on 1 and 47 DF, p-value: 0.0004796
# Predict date at which sum = 6000. Use as.Date to convert numeric value
# back to a date based on the number of days since Jan 1, 1970.
as.Date(predict(q, data.frame(sum=6000)), origin="1970-01-01")
1
"2015-11-08"
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