【发布时间】:2017-12-06 00:00:10
【问题描述】:
我有一个包含加利福尼亚州公立学校关闭和开放日期的数据集。可在问题底部找到here 或dput()。数据还列出了它是什么类型的学校以及它在哪里。我正在尝试创建一个运行总计列,该列还考虑到学校关闭和学校类型。
这是我想出的解决方案,基本上需要我根据使用ifelse的条件编码许多不同的1和0:
# open charter schools
pubschls$open_chart <- ifelse(pubschls$Charter=="Y" & is.na(pubschls$ClosedDate)==TRUE, 1, 0)
# open public schools
pubschls$open_pub <- ifelse(pubschls$Charter=="N" & is.na(pubschls$ClosedDate)==TRUE, 1, 0)
# closed charters
pubschls$closed_chart <- ifelse(pubschls$Charter=="Y" & is.na(pubschls$ClosedDate)==FALSE, 1, 0)
# closed public schools
pubschls$closed_pub <- ifelse(pubschls$Charter=="N" & is.na(pubschls$ClosedDate)==FALSE, 1, 0)
lausd <- filter(pubschls, NCESDist=="0622710")
# count number open during each year
然后我将各列相互减去以获得总数。
la_schools_count <- aggregate(lausd[c('open_chart','closed_chart','open_pub','closed_pub')],
by=list(year(lausd$OpenDate)), sum)
# find net charters by subtracting closed from open
la_schools_count$net_chart <- la_schools_count$open_chart - la_schools_count$closed_chart
# find net public schools by subtracting closed from open
la_schools_count$net_pub <- la_schools_count$open_pub - la_schools_count$closed_pub
# add running totals
la_schools_count$cum_chart <- cumsum(la_schools_count$net_chart)
la_schools_count$cum_pub <- cumsum(la_schools_count$net_pub)
# total totals
la_schools_count$total <- la_schools_count$cum_chart + la_schools_count$cum_pub
我的输出如下所示:
la_schools_count <- select(la_schools_count, "year", "cum_chart", "cum_pub", "pen_rate", "total")
year cum_chart cum_pub pen_rate total
1 1952 1 0 100.00000 1
2 1956 1 1 50.00000 2
3 1969 1 2 33.33333 3
4 1980 55 469 10.49618 524
5 1989 55 470 10.47619 525
6 1990 55 470 10.47619 525
7 1991 55 473 10.41667 528
8 1992 55 476 10.35782 531
9 1993 55 477 10.33835 532
10 1994 56 478 10.48689 534
11 1995 57 478 10.65421 535
12 1996 57 479 10.63433 536
13 1997 58 481 10.76067 539
14 1998 59 480 10.94620 539
15 1999 61 480 11.27542 541
16 2000 61 481 11.25461 542
17 2001 62 482 11.39706 544
18 2002 64 484 11.67883 548
19 2003 73 485 13.08244 558
20 2004 83 496 14.33506 579
21 2005 90 524 14.65798 614
22 2006 96 532 15.28662 628
23 2007 90 534 14.42308 624
24 2008 97 539 15.25157 636
25 2009 108 546 16.51376 654
26 2010 124 566 17.97101 690
27 2011 140 580 19.44444 720
28 2012 144 605 19.22563 749
29 2013 162 609 21.01167 771
30 2014 179 611 22.65823 790
31 2015 195 611 24.19355 806
32 2016 203 614 24.84700 817
33 2017 211 619 25.42169 830
我只是想知道这是否可以以更好的方式完成。像 apply 根据条件对所有行的语句?
dput:
structure(list(CDSCode = c("19647330100289", "19647330100297",
"19647330100669", "19647330100677", "19647330100743", "19647330100750"
), OpenDate = structure(c(12324, 12297, 12240, 12299, 12634,
12310), class = "Date"), ClosedDate = structure(c(NA, 15176,
NA, NA, NA, NA), class = "Date"), Charter = c("Y", "Y", "Y",
"Y", "Y", "Y")), .Names = c("CDSCode", "OpenDate", "ClosedDate",
"Charter"), row.names = c(NA, -6L), class = c("tbl_df", "tbl",
"data.frame"))
【问题讨论】:
-
什么是
pen_rate?你还没解释呢。 -
是的。 pen_rate 不是很重要。就是渗透率。基本上我把包机的数量除以总数。看起来很有用,但我相信有更好的方法来显示比例。
标签: r if-statement cumsum