这是我的数据框：
structure(list(Year = c(1975L, 1975L, 1975L, 1975L, 1975L, 1975L,
1975L, 1975L, 1975L, 1975L, 1975L, 1975L, 1976L, 1976L), Month = c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 1L, 2L), A = c(0.419727177,
0.411522634, 0.415627598, 0.425350915, 0.431778929, 0.455892409,
0.464252553, 0.473933649, 0.48947626, 0.481231954, 0.495049505,
0.49419323, 0.4927322, 0.493778392), log.S.grb = c(-0.86815035734372,
-0.887891256732457, -0.877965616764487, -0.854840768392932, -0.839841560146749,
-0.785498442482632, -0.767326579572197, -0.746687948097975, -0.714419316655485,
-0.731405892265036, -0.703097511313113, -0.704828684428783, -0.707789457396815,
-0.705668461635217), D = c(10.2641, 9.7704, 9.3694, 9.2403, 9.4459,
9.4826, 10.4272, 10.3805, 10.4835, 11.4103, 10.988, 10.7708,
9.2987, 8.6161), E = c(5.68, 5.4, 5.53, 5.5, 5.2, 5.86, 6.25,
6.36, 6.58, 5.51, 5.54, 5.2, 4.73, 5), J = c(1.15663289, 1.05923536,
0.938740721, 0.890710069, 1.012043355, 0.843618397, 0.850583558,
0.957856493, 0.888553262, 1.391339534, 1.309574432, 1.322714922,
1.247746749, 0.894350421), st3_st.grb = c(NA, NA, NA, 1.33095889507878,
4.80496965857083, 9.24671742818556, 8.75141888207357, 9.31536120487736,
7.10791258271466, 3.59206873071611, 4.35904367848617, 0.959063222670167,
2.36164348682203, -0.257095032210358), idiff_3m.grb = c(1.08206512614859,
1.03439401057093, 0.909556730440464, 0.886569121738501, 1.00676358283349,
0.856864819463321, 0.983497311932147, 0.946390153374566, 0.917819573656709,
1.38813518035337, 1.28343598984861, 1.31491300418167, 1.08566149047631,
0.860507535403032), EXCESS = c("NA", "NA", "NA", "NA", "NA",
"NA", "NA", "NA", "NA", "NA", "NA", "NA", "NA", "NA")), row.names = c(NA,
14L), class = "data.frame")
我使用此命令过滤了03/1975至01/1976之间的数据，但无法正常工作。
library(dplyr)
datawork %>% filter(Year>=1975 & Month>=5 & Year <=1976 & Month <=1 )
我在做什么错了？
答案 0 :(得分：3)
给出Year和Month只是整数列，与Month相关的逻辑谓词将返回一个空数据帧。一种解决方案是将其转换为日期格式，并在dplyr::between行中使用filter。 （请参见下面的可复制示例）。
如@TCZhang所述，如果不进行日期转换，则可以将Year = 1976 Month = 3视为Year = 1976 + 3/12 Years。因此，通过将分数相乘可以得到（1976 * 12）+ 3，您可以直接在dplyr::filter
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
library(lubridate)
#>
#> Attaching package: 'lubridate'
#> The following object is masked from 'package:base':
#>
#> date
datawork <- structure(list(Year = c(1975L, 1975L, 1975L, 1975L, 1975L, 1975L,
1975L, 1975L, 1975L, 1975L, 1975L, 1975L, 1976L, 1976L), Month = c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 1L, 2L), A = c(0.419727177,
0.411522634, 0.415627598, 0.425350915, 0.431778929, 0.455892409,
0.464252553, 0.473933649, 0.48947626, 0.481231954, 0.495049505,
0.49419323, 0.4927322, 0.493778392), log.S.grb = c(-0.86815035734372,
-0.887891256732457, -0.877965616764487, -0.854840768392932, -0.839841560146749,
-0.785498442482632, -0.767326579572197, -0.746687948097975, -0.714419316655485,
-0.731405892265036, -0.703097511313113, -0.704828684428783, -0.707789457396815,
-0.705668461635217), D = c(10.2641, 9.7704, 9.3694, 9.2403, 9.4459,
9.4826, 10.4272, 10.3805, 10.4835, 11.4103, 10.988, 10.7708,
9.2987, 8.6161), E = c(5.68, 5.4, 5.53, 5.5, 5.2, 5.86, 6.25,
6.36, 6.58, 5.51, 5.54, 5.2, 4.73, 5), J = c(1.15663289, 1.05923536,
0.938740721, 0.890710069, 1.012043355, 0.843618397, 0.850583558,
0.957856493, 0.888553262, 1.391339534, 1.309574432, 1.322714922,
1.247746749, 0.894350421), st3_st.grb = c(NA, NA, NA, 1.33095889507878,
4.80496965857083, 9.24671742818556, 8.75141888207357, 9.31536120487736,
7.10791258271466, 3.59206873071611, 4.35904367848617, 0.959063222670167,
2.36164348682203, -0.257095032210358), idiff_3m.grb = c(1.08206512614859,
1.03439401057093, 0.909556730440464, 0.886569121738501, 1.00676358283349,
0.856864819463321, 0.983497311932147, 0.946390153374566, 0.917819573656709,
1.38813518035337, 1.28343598984861, 1.31491300418167, 1.08566149047631,
0.860507535403032), EXCESS = c("NA", "NA", "NA", "NA", "NA",
"NA", "NA", "NA", "NA", "NA", "NA", "NA", "NA", "NA")), row.names = c(NA,
14L), class = "data.frame")
datawork %>%
mutate(date = make_date(year = Year, month = Month)) %>%
filter(between(date, as.Date("1975-03-01"), as.Date("1976-01-01")))
#> Year Month A log.S.grb D E J st3_st.grb
#> 1 1975 3 0.4156276 -0.8779656 9.3694 5.53 0.9387407 NA
#> 2 1975 4 0.4253509 -0.8548408 9.2403 5.50 0.8907101 1.3309589
#> 3 1975 5 0.4317789 -0.8398416 9.4459 5.20 1.0120434 4.8049697
#> 4 1975 6 0.4558924 -0.7854984 9.4826 5.86 0.8436184 9.2467174
#> 5 1975 7 0.4642526 -0.7673266 10.4272 6.25 0.8505836 8.7514189
#> 6 1975 8 0.4739336 -0.7466879 10.3805 6.36 0.9578565 9.3153612
#> 7 1975 9 0.4894763 -0.7144193 10.4835 6.58 0.8885533 7.1079126
#> 8 1975 10 0.4812320 -0.7314059 11.4103 5.51 1.3913395 3.5920687
#> 9 1975 11 0.4950495 -0.7030975 10.9880 5.54 1.3095744 4.3590437
#> 10 1975 12 0.4941932 -0.7048287 10.7708 5.20 1.3227149 0.9590632
#> 11 1976 1 0.4927322 -0.7077895 9.2987 4.73 1.2477467 2.3616435
#> idiff_3m.grb EXCESS date
#> 1 0.9095567 NA 1975-03-01
#> 2 0.8865691 NA 1975-04-01
#> 3 1.0067636 NA 1975-05-01
#> 4 0.8568648 NA 1975-06-01
#> 5 0.9834973 NA 1975-07-01
#> 6 0.9463902 NA 1975-08-01
#> 7 0.9178196 NA 1975-09-01
#> 8 1.3881352 NA 1975-10-01
#> 9 1.2834360 NA 1975-11-01
#> 10 1.3149130 NA 1975-12-01
#> 11 1.0856615 NA 1976-01-01
由reprex package（v0.2.0）于2018-08-31创建。