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问创建循环/函数，将缺失的行添加到dataframe并填充0
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Stack Overflow用户

提问于 2014-08-06 16:17:54

回答 1查看 90关注 0票数 0

我总结了一个数据框架，并希望添加缺少的年份，并在适当的情况下填写这些丢失的行w/ 0。以下是我的起始数据：

start_data <- structure(list(park = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("apis", 
"grpo", "isro", "miss", "piro", "sacn", "slbe", "voya"), class = "factor"), 
    loc_01 = structure(c(1L, 1L, 1L, 1L, 4L, 4L, 4L, 4L, 4L, 
    4L, 2L, 2L, 2L, 2L, 6L, 6L, 6L, 6L, 6L, 6L), .Label = c("apis", 
    "isro", "miss", "non_apis", "non_grpo", "non_isro", "non_miss", 
    "non_piro", "non_sacn", "non_slbe", "non_voya", "piro", "sacn", 
    "slbe", "voya"), class = "factor"), year = c(2005L, 2006L, 
    2007L, 2008L, 2004L, 2005L, 2006L, 2007L, 2008L, 2009L, 2003L, 
    2005L, 2006L, 2007L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L
    ), agriculture = c(0, 0, 0, 0, 2.83549420428, 0, 26.41126099384, 
    9.07370206906, 11.00043405833, 1.440345049, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0), beaver = c(0, 0.29706355242, 1.25997210478, 
    1.5123175298, 9.14483092902, 0.70214089206, 2.78157443836, 
    4.42825988163, 0.9900762968, 2.3401234612, 2.8808849429, 
    1.2604019414, 0.54011663526, 0.729245712, 5.45502002852, 
    2.7912116718, 3.0604650244, 1.51253347654, 0.9002514858, 
    2.7548091776), blowdown = c(0, 0, 0, 0, 0, 0, 2.23207970694, 
    0, 0, 0, 0, 0, 0, 0.81011784036, 0, 0, 0, 0, 0, 0)), class = "data.frame", row.names = c(NA, 
-20L), .Names = c("park", "loc_01", "year", "agriculture", "beaver", 
"blowdown"))

看起来是这样的：

   park   loc_01 year agriculture    beaver  blowdown
1  apis     apis 2005    0.000000 0.0000000 0.0000000
2  apis     apis 2006    0.000000 0.2970636 0.0000000
3  apis     apis 2007    0.000000 1.2599721 0.0000000
4  apis     apis 2008    0.000000 1.5123175 0.0000000
5  apis non_apis 2004    2.835494 9.1448309 0.0000000
6  apis non_apis 2005    0.000000 0.7021409 0.0000000
7  apis non_apis 2006   26.411261 2.7815744 2.2320797
8  apis non_apis 2007    9.073702 4.4282599 0.0000000
9  apis non_apis 2008   11.000434 0.9900763 0.0000000
10 apis non_apis 2009    1.440345 2.3401235 0.0000000
11 isro     isro 2003    0.000000 2.8808849 0.0000000
12 isro     isro 2005    0.000000 1.2604019 0.0000000
13 isro     isro 2006    0.000000 0.5401166 0.0000000
14 isro     isro 2007    0.000000 0.7292457 0.8101178
15 isro non_isro 2003    0.000000 5.4550200 0.0000000
16 isro non_isro 2004    0.000000 2.7912117 0.0000000
17 isro non_isro 2005    0.000000 3.0604650 0.0000000
18 isro non_isro 2006    0.000000 1.5125335 0.0000000
19 isro non_isro 2007    0.000000 0.9002515 0.0000000
20 isro non_isro 2008    0.000000 2.7548092 0.0000000

根据公园的标识，我想在缺少的年份里加上行。例如，在park == apis中，apis和non_apis都应该存在2005:2009年。在这里，我可以引用一个列表，比如apis.yrs <- 2004:2009。park == isro，2003:2008应该再次出现，无论是isro还是non_isro。再一次，可以为这些年创建另一个列表，isro.yrs <- 2003:2008。添加这些新行时，agriculture、beaver和blowdown都应该填充为0。

这是我的最终目标：

goal <- structure(list(park = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L), .Label = c("", "apis", "isro"), class = "factor"), loc_01 = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 4L, 4L, 4L, 4L, 4L, 4L, 2L, 2L, 2L, 2L, 2L, 
2L, 5L, 5L, 5L, 5L, 5L, 5L), .Label = c("apis", "isro", "miss", 
"non_apis", "non_isro", "non_miss", "non_piro", "non_sacn", "non_slbe", 
"non_voya", "piro", "sacn", "slbe", "voya"), class = "factor"), 
    year = c(2004L, 2005L, 2006L, 2007L, 2008L, 2009L, 2004L, 
    2005L, 2006L, 2007L, 2008L, 2009L, 2003L, 2004L, 2005L, 2006L, 
    2007L, 2008L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L), 
    agriculture = c(0, 0, 0, 0, 0, 0, 2.835494204, 0, 26.41126099, 
    9.073702069, 11.00043406, 1.440345049, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0), beaver = c(0, 0, 0.297063552, 1.259972105, 
    1.51231753, 0, 9.144830929, 0.702140892, 2.781574438, 4.428259882, 
    0.990076297, 2.340123461, 2.880884943, 0, 1.260401941, 0.540116635, 
    0.729245712, 0, 5.455020029, 2.791211672, 3.060465024, 1.512533477, 
    0.900251486, 2.754809178), blowdown = c(0, 0, 0, 0, 0, 0, 
    0, 0, 2.232079707, 0, 0, 0, 0, 0, 0, 0, 0.81011784, 0, 0, 
    0, 0, 0, 0, 0)), class = "data.frame", row.names = c(NA, 
-24L), .Names = c("park", "loc_01", "year", "agriculture", "beaver", 
"blowdown"))

看起来像这样..。

   park   loc_01 year agriculture    beaver  blowdown
1  apis     apis 2004    0.000000 0.0000000 0.0000000
2  apis     apis 2005    0.000000 0.0000000 0.0000000
3  apis     apis 2006    0.000000 0.2970636 0.0000000
4  apis     apis 2007    0.000000 1.2599721 0.0000000
5  apis     apis 2008    0.000000 1.5123175 0.0000000
6  apis     apis 2009    0.000000 0.0000000 0.0000000
7  apis non_apis 2004    2.835494 9.1448309 0.0000000
8  apis non_apis 2005    0.000000 0.7021409 0.0000000
9  apis non_apis 2006   26.411261 2.7815744 2.2320797
10 apis non_apis 2007    9.073702 4.4282599 0.0000000
11 apis non_apis 2008   11.000434 0.9900763 0.0000000
12 apis non_apis 2009    1.440345 2.3401235 0.0000000
13 isro     isro 2003    0.000000 2.8808849 0.0000000
14 isro     isro 2004    0.000000 0.0000000 0.0000000
15 isro     isro 2005    0.000000 1.2604019 0.0000000
16 isro     isro 2006    0.000000 0.5401166 0.0000000
17 isro     isro 2007    0.000000 0.7292457 0.8101178
18 isro     isro 2008    0.000000 0.0000000 0.0000000
19 isro non_isro 2003    0.000000 5.4550200 0.0000000
20 isro non_isro 2004    0.000000 2.7912117 0.0000000
21 isro non_isro 2005    0.000000 3.0604650 0.0000000
22 isro non_isro 2006    0.000000 1.5125335 0.0000000
23 isro non_isro 2007    0.000000 0.9002515 0.0000000
24 isro non_isro 2008    0.000000 2.7548092 0.0000000

其中，2004年、2009年添加了loc_01==apis，而2004年、2008年添加了loc_01==isro。

-al

loops

回答 1

Stack Overflow用户

回答已采纳

发布于 2014-08-06 16:33:50

对于这类问题，expand.grid + merge是有用的。

你可以试试这样的东西：

步骤1:使用您希望包含的所有组合创建一个data.frame。

toMerge <- rbind(
  expand.grid(park = "apis", 
              loc_01 = c("apis", "non_apis"), 
              year = 2004:2009),
  expand.grid(park = "isro",
              loc_01 = c("isro", "non_isro"),
              year = 2003:2008)
)

步骤2:使用原始的merge进行data.frame。

merge(start_data, toMerge, all = TRUE)
#    park   loc_01 year agriculture    beaver  blowdown
# 1  apis     apis 2004          NA        NA        NA
# 2  apis     apis 2005    0.000000 0.0000000 0.0000000
# 3  apis     apis 2006    0.000000 0.2970636 0.0000000
# 4  apis     apis 2007    0.000000 1.2599721 0.0000000
# 5  apis     apis 2008    0.000000 1.5123175 0.0000000
# 6  apis     apis 2009          NA        NA        NA
# 7  apis non_apis 2004    2.835494 9.1448309 0.0000000
# 8  apis non_apis 2005    0.000000 0.7021409 0.0000000
# 9  apis non_apis 2006   26.411261 2.7815744 2.2320797
# 10 apis non_apis 2007    9.073702 4.4282599 0.0000000
# 11 apis non_apis 2008   11.000434 0.9900763 0.0000000
# 12 apis non_apis 2009    1.440345 2.3401235 0.0000000
# 13 isro     isro 2003    0.000000 2.8808849 0.0000000
# 14 isro     isro 2004          NA        NA        NA
# 15 isro     isro 2005    0.000000 1.2604019 0.0000000
# 16 isro     isro 2006    0.000000 0.5401166 0.0000000
# 17 isro     isro 2007    0.000000 0.7292457 0.8101178
# 18 isro     isro 2008          NA        NA        NA
# 19 isro non_isro 2003    0.000000 5.4550200 0.0000000
# 20 isro non_isro 2004    0.000000 2.7912117 0.0000000
# 21 isro non_isro 2005    0.000000 3.0604650 0.0000000
# 22 isro non_isro 2006    0.000000 1.5125335 0.0000000
# 23 isro non_isro 2007    0.000000 0.9002515 0.0000000
# 24 isro non_isro 2008    0.000000 2.7548092 0.0000000

从这里开始，用0替换NA并不困难。

票数 2

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/25165299

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