我有一个数据框，我希望运行一些复杂的'if'语句。
提供一些背景信息 - 这是欧盟不同地区的数据。我想在“结果”中查看国家/地区的变化。在欧盟，我们有不同级别的行政边界（表中称为“区域”） - 1和2（表中的“级别”）。我希望探索级别'1'的变异性，但是一些国家的人口较少，因此我必须考虑第2级的变异性。我创建了一个缩短的数据集来说明我的观点。
df >- structure(list(region = structure(c(2L, 4L, 6L, 8L, 9L, 13L,
16L, 17L, 18L, 21L, 23L, 26L, 27L, 33L, 34L, 41L, 37L, 43L, 48L,
50L, 56L, 54L, 59L, 3L, 1L, 7L, 5L, 10L, 11L, 12L, 14L, 15L,
20L, 19L, 22L, 24L, 25L, 28L, 29L, 30L, 31L, 32L, 35L, 36L, 42L,
38L, 40L, 39L, 45L, 44L, 46L, 49L, 51L, 47L, 52L, 53L, 57L, 58L,
55L, 61L, 60L), .Label = c("AT13", "AT2", "AT22", "AT3", "BE10",
"BE2", "BE21", "BE3", "CH0", "CH01", "CH02", "CH03", "CZ0", "CZ01",
"CZ02", "DE4", "DEC", "DK0", "DK02", "DK04", "EE0", "EE00", "FI1",
"FI19", "FI1D", "FI2", "FR1", "FR10", "FR21", "FR22", "FR23",
"FR24", "FR8", "IE0", "IE01", "IE02", "NL1", "NL23", "NL31",
"NL33", "NL4", "NL41", "NO0", "NO01", "NO03", "NO07", "PL11",
"PL2", "PL21", "PL3", "PL41", "PL52", "PL62", "SE1", "SE11",
"SE2", "SE22", "SE32", "SI0", "SI01", "SI02"), class = "factor"),
result = c(24.43, 20.37, 23.53, 25.51, 17.73, 30.61, 46.2,
43.75, 25.32, 53.32, 34.25, 46.15, 34.59, 38.06, 18.6, 32.29,
28.57, 22.36, 40.98, 34.53, 21.09, 23.89, 43.15, 25.73, 30.06,
26.64, 16.78, 18.75, 17.51, 19.58, 28.63, 25.44, 24.29, 30.73,
53.32, 37.31, 38.19, 34.59, 53.33, 39.02, 22.92, 35.44, 19.07,
18.38, 29.26, 38.6, 31.51, 21.54, 22.93, 17.86, 30.77, 39.87,
44.7, 36.8, 37.5, 33.33, 19.05, 30.77, 17.98, 39.71, 45.66
), level = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L), country = structure(c(1L, 1L, 2L, 2L,
3L, 4L, 5L, 5L, 6L, 7L, 8L, 8L, 9L, 9L, 10L, 11L, 11L, 12L,
13L, 13L, 14L, 14L, 15L, 1L, 1L, 2L, 2L, 3L, 3L, 3L, 4L,
4L, 6L, 6L, 7L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 11L,
11L, 11L, 11L, 12L, 12L, 12L, 13L, 13L, 13L, 13L, 13L, 14L,
14L, 14L, 15L, 15L), .Label = c("AT", "BE", "CH", "CZ", "DE",
"DK", "EE", "FI", "FR", "IE", "NL", "NO", "PL", "SE", "SI"
), class = "factor")), .Names = c("region", "result", "level",
"country"), class = "data.frame", row.names = c(NA, -61L))
所以，
我想要if语句删除行IF级别= 1并且只有一个国家/地区的级别为1的记录（即它将删除国家CH，CZ，DK，EE，IE，NO和SI ）。 （因此将删除第5,6,9,10,15,18和23行） 然后：
我想要第二个if语句删除行IF级别== 2并且国家IS在级别1的数据框中表示（已应用语句1）。
如果声明我可以做简单的调节：
df<-nuts1_2[!(df$level==1),]
......但我在第二部分上挣扎（即，坚果1 $ $等级== 1＆amp; df $ country'只出现一次'??）
我应该留下一个数据框，详情如下：
region result level country
1 AT2 24.43 1 AT
2 AT3 20.37 1 AT
3 BE2 23.53 1 BE
4 BE3 25.51 1 BE
5 DE4 46.20 1 DE
6 DEC 43.75 1 DE
7 FI1 34.25 1 FI
8 FI2 46.15 1 FI
9 FR1 34.59 1 FR
10 FR8 38.06 1 FR
11 NL4 32.29 1 NL
12 NL1 28.57 1 NL
13 PL2 40.98 1 PL
14 PL3 34.53 1 PL
15 SE2 21.09 1 SE
16 SE1 23.89 1 SE
17 CH01 18.75 2 CH
18 CH02 17.51 2 CH
19 CH03 19.58 2 CH
20 CZ01 28.63 2 CZ
21 CZ02 25.44 2 CZ
22 DK04 24.29 2 DK
23 DK02 30.73 2 DK
24 EE00 53.32 2 EE
25 IE01 19.07 2 IE
26 IE02 18.38 2 IE
27 NO03 22.93 2 NO
28 NO01 17.86 2 NO
29 NO07 30.77 2 NO
30 SI02 39.71 2 SI
31 SI01 45.66 2 SI
在一个理想的世界中，如果这些行可以成为另一个名为“已删除”的数据框的子集 - 这将有助于我的记录保存。
感谢上述任何帮助。
答案 0 :(得分：3)
使用data.table包可以执行以下操作：
library(data.table)
setDT(df1)[, if(!(level==1 & .N==1)) .SD, by = .(country,level)
][, unlev := uniqueN(level), by = country
][!(unlev==2 & level==2)][,unlev:=NULL][]
给出：
country level region result 1: AT 1 AT2 24.43 2: AT 1 AT3 20.37 3: BE 1 BE2 23.53 4: BE 1 BE3 25.51 5: DE 1 DE4 46.20 6: DE 1 DEC 43.75 7: FI 1 FI1 34.25 8: FI 1 FI2 46.15 9: FR 1 FR1 34.59 10: FR 1 FR8 38.06 11: NL 1 NL4 32.29 12: NL 1 NL1 28.57 13: PL 1 PL2 40.98 14: PL 1 PL3 34.53 15: SE 1 SE2 21.09 16: SE 1 SE1 23.89 17: CH 2 CH01 18.75 18: CH 2 CH02 17.51 19: CH 2 CH03 19.58 20: CZ 2 CZ01 28.63 21: CZ 2 CZ02 25.44 22: DK 2 DK04 24.29 23: DK 2 DK02 30.73 24: EE 2 EE00 53.32 25: IE 2 IE01 19.07 26: IE 2 IE02 18.38 27: NO 2 NO03 22.93 28: NO 2 NO01 17.86 29: NO 2 NO07 30.77 30: SI 2 SI02 39.71 31: SI 2 SI01 45.66
您可以使用dplyr：
应用相同的逻辑library(dplyr)
df1 %>%
group_by(country,level) %>%
mutate(n = n()) %>%
filter(!(level==1 & n==1)) %>%
group_by(country) %>%
mutate(unlev = length(unique(level))) %>%
filter(!(unlev==2 & level==2)) %>%
select(-n, -unlev)
答案 1 :(得分：2)
我们可以使用dplyr
library(dplyr)
df %>%
group_by(country,level) %>%
filter(!(level==1 & n()==1)) %>%
group_by(country) %>%
filter(!(n_distinct(level)==2 & level==2)) %>%
as.data.frame()
# region result level country
#1 AT2 24.43 1 AT
#2 AT3 20.37 1 AT
#3 BE2 23.53 1 BE
#4 BE3 25.51 1 BE
#5 DE4 46.20 1 DE
#6 DEC 43.75 1 DE
#7 FI1 34.25 1 FI
#8 FI2 46.15 1 FI
#9 FR1 34.59 1 FR
#10 FR8 38.06 1 FR
#11 NL4 32.29 1 NL
#12 NL1 28.57 1 NL
#13 PL2 40.98 1 PL
#14 PL3 34.53 1 PL
#15 SE2 21.09 1 SE
#16 SE1 23.89 1 SE
#17 CH01 18.75 2 CH
#18 CH02 17.51 2 CH
#19 CH03 19.58 2 CH
#20 CZ01 28.63 2 CZ
#21 CZ02 25.44 2 CZ
#22 DK04 24.29 2 DK
#23 DK02 30.73 2 DK
#24 EE00 53.32 2 EE
#25 IE01 19.07 2 IE
#26 IE02 18.38 2 IE
#27 NO03 22.93 2 NO
#28 NO01 17.86 2 NO
#29 NO07 30.77 2 NO
#30 SI02 39.71 2 SI
#31 SI01 45.66 2 SI