在R中,合并2个数据帧,同时保持第一个数据帧

时间:2017-03-17 06:52:54

标签: r data-manipulation

当我了解在merge()中默认为TRUE的附加参数'sort'时,我真的认为我有解决这个问题的方法。但是,将此设置为false并没有帮助。下面是我的代码演示,包括我得到的结果和我想要的结果:

df2 = structure(list(player = c("Marvin Williams", "Spencer Hawes", 
"Jeremy Lin", "Kemba Walker", "P.J. Hairston", "Rudy Gay", "Rajon Rondo", 
"DeMarcus Cousins", "Ben McLemore", "Willie Cauley-Stein"), global.player.id = c(263884L, 
329824L, 340730L, 462980L, 609567L, 266358L, 262882L, 509450L, 
604898L, 699950L), team.name = c("Hornets", "Hornets", "Hornets", 
"Hornets", "Grizzlies", "Kings", "Kings", "Kings", "Kings", "Kings"
)), .Names = c("player", "global.player.id", "team.name"), class = "data.frame", row.names = c(47L, 
48L, 52L, 53L, 225L, 389L, 390L, 395L, 398L, 401L))

df1 = structure(list(global.player.id = c(-1L, 262882L, 266358L, 509450L, 
604898L, 699950L, 263884L, 329824L, 340730L, 462980L, 609567L, 
-1L, 262882L, 266358L, 509450L, 604898L, 699950L, 263884L, 329824L, 
340730L, 462980L, 609567L, -1L, 262882L, 266358L), x_loc = c(47.17753, 
13.57165, 46.45843, 26.68803, 52.16717, 47.20201, 60.097, 47.20201, 
52.16717, 65.1302, 46.45843, 47.19141, 13.61702, 46.5355, 26.71856, 
52.25433, 47.27324, 60.08215, 47.27324, 52.25433, 65.11267, 46.5355, 
46.82163, 13.66478, 46.57545), y_loc = c(26.44326, 25.18298, 
18.46573, 25.48557, 33.09177, 31.09372, 22.79717, 31.09372, 33.09177, 
26.39671, 18.46573, 26.5187, 25.17431, 18.42014, 25.53807, 33.11185, 
31.01197, 22.76307, 31.01197, 33.11185, 26.40227, 18.42014, 26.72834, 
25.17784, 18.35961), order = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25)), .Names = c("global.player.id", 
"x_loc", "y_loc", "order"), row.names = c("1", "2", "3", "4", 
"5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", 
"16", "17", "18", "19", "20", "21", "22", "23", "24", "25"), class = "data.frame")

以上是我正在使用的数据框架。当我将df2合并到它时,我想保持df1的顺序。我在这里处理时间序列数据,因此数据帧的顺序很重要。 df1中的order列只是为了测试df1是否正在被洗牌(我不想在合并后使用额外的代码对订单进行排序)。

这是我尝试过的:

merge(df1, df2, by = 'global.player.id', all.x = TRUE)

   global.player.id    x_loc    y_loc order              player team.name
1                -1 47.17753 26.44326     1                <NA>      <NA>
2                -1 46.82163 26.72834    23                <NA>      <NA>
3                -1 47.19141 26.51870    12                <NA>      <NA>
4            262882 13.57165 25.18298     2         Rajon Rondo     Kings
5            262882 13.61702 25.17431    13         Rajon Rondo     Kings
6            262882 13.66478 25.17784    24         Rajon Rondo     Kings
7            263884 60.08215 22.76307    18     Marvin Williams   Hornets
8            263884 60.09700 22.79717     7     Marvin Williams   Hornets
9            266358 46.53550 18.42014    14            Rudy Gay     Kings
10           266358 46.45843 18.46573     3            Rudy Gay     Kings
11           266358 46.57545 18.35961    25            Rudy Gay     Kings
12           329824 47.27324 31.01197    19       Spencer Hawes   Hornets
13           329824 47.20201 31.09372     8       Spencer Hawes   Hornets
14           340730 52.16717 33.09177     9          Jeremy Lin   Hornets
15           340730 52.25433 33.11185    20          Jeremy Lin   Hornets
16           462980 65.13020 26.39671    10        Kemba Walker   Hornets
17           462980 65.11267 26.40227    21        Kemba Walker   Hornets
18           509450 26.71856 25.53807    15    DeMarcus Cousins     Kings
19           509450 26.68803 25.48557     4    DeMarcus Cousins     Kings
20           604898 52.16717 33.09177     5        Ben McLemore     Kings
21           604898 52.25433 33.11185    16        Ben McLemore     Kings
22           609567 46.53550 18.42014    22       P.J. Hairston Grizzlies
23           609567 46.45843 18.46573    11       P.J. Hairston Grizzlies
24           699950 47.20201 31.09372     6 Willie Cauley-Stein     Kings
25           699950 47.27324 31.01197    17 Willie Cauley-Stein     Kings

最初在df1中,订单按1-25排序,现在全部乱序。很明显,df1以我不想要的方式被洗牌。这是我将sort = FALSE传递给合并函数时的输出:

merge(df1, df2, by = 'global.player.id', all.x = TRUE, sort = FALSE)

global.player.id x_loc y_loc order player team.name

1            262882 13.57165 25.18298     2         Rajon Rondo     Kings
2            262882 13.61702 25.17431    13         Rajon Rondo     Kings
3            262882 13.66478 25.17784    24         Rajon Rondo     Kings
4            266358 46.53550 18.42014    14            Rudy Gay     Kings
5            266358 46.45843 18.46573     3            Rudy Gay     Kings
6            266358 46.57545 18.35961    25            Rudy Gay     Kings
7            509450 26.71856 25.53807    15    DeMarcus Cousins     Kings
8            509450 26.68803 25.48557     4    DeMarcus Cousins     Kings
9            604898 52.16717 33.09177     5        Ben McLemore     Kings
10           604898 52.25433 33.11185    16        Ben McLemore     Kings
11           699950 47.20201 31.09372     6 Willie Cauley-Stein     Kings
12           699950 47.27324 31.01197    17 Willie Cauley-Stein     Kings
13           263884 60.08215 22.76307    18     Marvin Williams   Hornets
14           263884 60.09700 22.79717     7     Marvin Williams   Hornets
15           329824 47.27324 31.01197    19       Spencer Hawes   Hornets
16           329824 47.20201 31.09372     8       Spencer Hawes   Hornets
17           340730 52.16717 33.09177     9          Jeremy Lin   Hornets
18           340730 52.25433 33.11185    20          Jeremy Lin   Hornets
19           462980 65.13020 26.39671    10        Kemba Walker   Hornets
20           462980 65.11267 26.40227    21        Kemba Walker   Hornets
21           609567 46.53550 18.42014    22       P.J. Hairston Grizzlies
22           609567 46.45843 18.46573    11       P.J. Hairston Grizzlies
23               -1 47.17753 26.44326     1                <NA>      <NA>
24               -1 46.82163 26.72834    23                <NA>      <NA>
25               -1 47.19141 26.51870    12                <NA>      <NA>

也不是我想要的,因为订单再次出现故障。

无论如何调用合并函数而不完全改组传递的第一个数据帧参数,或者我完全没有运气。如果是这样,这似乎是merge()函数的主要缺点。谢谢!

1 个答案:

答案 0 :(得分:2)

您可以使用join

中的plyr
library(plyr)
plyr::join(df1,df2, by='global.player.id')

结果未排序。