根据两个列表之间的不同名称删除组件 [英] Removing components based on the different names between two lists
本文介绍了根据两个列表之间的不同名称删除组件的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
ID。我想使用列表中每个组件分配的名称删除多余的ID。我的实际数据集有一大堆ID，因此我想实质上创建一个函数，该函数允许我使用另一个列表中的名称作为索引来删除任何具有不同名称的ID。
我怎么能着手做这件事呢？在本例中，我将从列表中删除IDD，这不是因为D中的数据框，而是因为july2中的名称与july中的名称不同。
我已尝试使用setdiff，但结果只是返回了我放在第一个参数中的列表。
> setdiff(july, july2)
<list_of<
tbl_df<
date : date
x : double
y : double
ID : character
jDate: double
Month: double
new : date
>
>[12]>
$A
# A tibble: 16 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-04 60161. 800440. A 14794 7 2010-07-01
2 2010-07-08 61139. 825947. A 14798 7 2010-07-01
3 2010-07-04 60161. 800440. A 14794 7 2010-07-01
4 2010-07-08 61139. 825947. A 14798 7 2010-07-01
5 2010-07-04 60161. 800440. A 14794 7 2010-07-01
6 2010-07-08 61139. 825947. A 14798 7 2010-07-01
7 2010-07-04 60161. 800440. A 14794 7 2010-07-01
8 2010-07-08 61139. 825947. A 14798 7 2010-07-01
9 2010-07-04 60161. 800440. A 14794 7 2010-07-01
10 2010-07-08 61139. 825947. A 14798 7 2010-07-01
11 2010-07-04 60161. 800440. A 14794 7 2010-07-01
12 2010-07-08 61139. 825947. A 14798 7 2010-07-01
13 2010-07-04 60161. 800440. A 14794 7 2010-07-01
14 2010-07-08 61139. 825947. A 14798 7 2010-07-01
15 2010-07-04 60161. 800440. A 14794 7 2010-07-01
16 2010-07-08 61139. 825947. A 14798 7 2010-07-01
$A
# A tibble: 24 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-12 66502. 804956. A 14802 7 2010-07-11
2 2010-07-16 79728. 858097. A 14806 7 2010-07-11
3 2010-07-20 77342. 830852. A 14810 7 2010-07-11
4 2010-07-12 66502. 804956. A 14802 7 2010-07-11
5 2010-07-16 79728. 858097. A 14806 7 2010-07-11
6 2010-07-20 77342. 830852. A 14810 7 2010-07-11
7 2010-07-12 66502. 804956. A 14802 7 2010-07-11
8 2010-07-16 79728. 858097. A 14806 7 2010-07-11
9 2010-07-20 77342. 830852. A 14810 7 2010-07-11
10 2010-07-12 66502. 804956. A 14802 7 2010-07-11
# ... with 14 more rows
$A
# A tibble: 16 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-24 75483. 828763. A 14814 7 2010-07-21
2 2010-07-28 69508. 806470. A 14818 7 2010-07-21
3 2010-07-24 75483. 828763. A 14814 7 2010-07-21
4 2010-07-28 69508. 806470. A 14818 7 2010-07-21
5 2010-07-24 75483. 828763. A 14814 7 2010-07-21
6 2010-07-28 69508. 806470. A 14818 7 2010-07-21
7 2010-07-24 75483. 828763. A 14814 7 2010-07-21
8 2010-07-28 69508. 806470. A 14818 7 2010-07-21
9 2010-07-24 75483. 828763. A 14814 7 2010-07-21
10 2010-07-28 69508. 806470. A 14818 7 2010-07-21
11 2010-07-24 75483. 828763. A 14814 7 2010-07-21
12 2010-07-28 69508. 806470. A 14818 7 2010-07-21
13 2010-07-24 75483. 828763. A 14814 7 2010-07-21
14 2010-07-28 69508. 806470. A 14818 7 2010-07-21
15 2010-07-24 75483. 828763. A 14814 7 2010-07-21
16 2010-07-28 69508. 806470. A 14818 7 2010-07-21
$B
# A tibble: 24 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-01 72826. 888060. B 14791 7 2010-07-01
2 2010-07-05 67469. 807307. B 14795 7 2010-07-01
3 2010-07-09 77834. 868002. B 14799 7 2010-07-01
4 2010-07-01 72826. 888060. B 14791 7 2010-07-01
5 2010-07-05 67469. 807307. B 14795 7 2010-07-01
6 2010-07-09 77834. 868002. B 14799 7 2010-07-01
7 2010-07-01 72826. 888060. B 14791 7 2010-07-01
8 2010-07-05 67469. 807307. B 14795 7 2010-07-01
9 2010-07-09 77834. 868002. B 14799 7 2010-07-01
10 2010-07-01 72826. 888060. B 14791 7 2010-07-01
# ... with 14 more rows
$B
# A tibble: 16 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-13 74643. 845222. B 14803 7 2010-07-11
2 2010-07-17 78530. 807316. B 14807 7 2010-07-11
3 2010-07-13 74643. 845222. B 14803 7 2010-07-11
4 2010-07-17 78530. 807316. B 14807 7 2010-07-11
5 2010-07-13 74643. 845222. B 14803 7 2010-07-11
6 2010-07-17 78530. 807316. B 14807 7 2010-07-11
7 2010-07-13 74643. 845222. B 14803 7 2010-07-11
8 2010-07-17 78530. 807316. B 14807 7 2010-07-11
9 2010-07-13 74643. 845222. B 14803 7 2010-07-11
10 2010-07-17 78530. 807316. B 14807 7 2010-07-11
11 2010-07-13 74643. 845222. B 14803 7 2010-07-11
12 2010-07-17 78530. 807316. B 14807 7 2010-07-11
13 2010-07-13 74643. 845222. B 14803 7 2010-07-11
14 2010-07-17 78530. 807316. B 14807 7 2010-07-11
15 2010-07-13 74643. 845222. B 14803 7 2010-07-11
16 2010-07-17 78530. 807316. B 14807 7 2010-07-11
$B
# A tibble: 24 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-21 61332. 840310. B 14811 7 2010-07-21
2 2010-07-25 69102. 809024. B 14815 7 2010-07-21
3 2010-07-29 66088. 817887. B 14819 7 2010-07-21
4 2010-07-21 61332. 840310. B 14811 7 2010-07-21
5 2010-07-25 69102. 809024. B 14815 7 2010-07-21
6 2010-07-29 66088. 817887. B 14819 7 2010-07-21
7 2010-07-21 61332. 840310. B 14811 7 2010-07-21
8 2010-07-25 69102. 809024. B 14815 7 2010-07-21
9 2010-07-29 66088. 817887. B 14819 7 2010-07-21
10 2010-07-21 61332. 840310. B 14811 7 2010-07-21
# ... with 14 more rows
$C
# A tibble: 24 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-02 71110. 898586. C 14792 7 2010-07-01
2 2010-07-06 78769. 821287. C 14796 7 2010-07-01
3 2010-07-10 62446. 874366. C 14800 7 2010-07-01
4 2010-07-02 71110. 898586. C 14792 7 2010-07-01
5 2010-07-06 78769. 821287. C 14796 7 2010-07-01
6 2010-07-10 62446. 874366. C 14800 7 2010-07-01
7 2010-07-02 71110. 898586. C 14792 7 2010-07-01
8 2010-07-06 78769. 821287. C 14796 7 2010-07-01
9 2010-07-10 62446. 874366. C 14800 7 2010-07-01
10 2010-07-02 71110. 898586. C 14792 7 2010-07-01
# ... with 14 more rows
$C
# A tibble: 16 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-14 77316. 882468. C 14804 7 2010-07-11
2 2010-07-18 65028. 815016. C 14808 7 2010-07-11
3 2010-07-14 77316. 882468. C 14804 7 2010-07-11
4 2010-07-18 65028. 815016. C 14808 7 2010-07-11
5 2010-07-14 77316. 882468. C 14804 7 2010-07-11
6 2010-07-18 65028. 815016. C 14808 7 2010-07-11
7 2010-07-14 77316. 882468. C 14804 7 2010-07-11
8 2010-07-18 65028. 815016. C 14808 7 2010-07-11
9 2010-07-14 77316. 882468. C 14804 7 2010-07-11
10 2010-07-18 65028. 815016. C 14808 7 2010-07-11
11 2010-07-14 77316. 882468. C 14804 7 2010-07-11
12 2010-07-18 65028. 815016. C 14808 7 2010-07-11
13 2010-07-14 77316. 882468. C 14804 7 2010-07-11
14 2010-07-18 65028. 815016. C 14808 7 2010-07-11
15 2010-07-14 77316. 882468. C 14804 7 2010-07-11
16 2010-07-18 65028. 815016. C 14808 7 2010-07-11
$C
# A tibble: 24 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-22 65117. 866750. C 14812 7 2010-07-21
2 2010-07-26 78462. 823259. C 14816 7 2010-07-21
3 2010-07-30 69577. 848118. C 14820 7 2010-07-21
4 2010-07-22 65117. 866750. C 14812 7 2010-07-21
5 2010-07-26 78462. 823259. C 14816 7 2010-07-21
6 2010-07-30 69577. 848118. C 14820 7 2010-07-21
7 2010-07-22 65117. 866750. C 14812 7 2010-07-21
8 2010-07-26 78462. 823259. C 14816 7 2010-07-21
9 2010-07-30 69577. 848118. C 14820 7 2010-07-21
10 2010-07-22 65117. 866750. C 14812 7 2010-07-21
# ... with 14 more rows
$D
# A tibble: 16 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-03 77586. 819905. D 14793 7 2010-07-01
2 2010-07-07 76249. 848582. D 14797 7 2010-07-01
3 2010-07-03 77586. 819905. D 14793 7 2010-07-01
4 2010-07-07 76249. 848582. D 14797 7 2010-07-01
5 2010-07-03 77586. 819905. D 14793 7 2010-07-01
6 2010-07-07 76249. 848582. D 14797 7 2010-07-01
7 2010-07-03 77586. 819905. D 14793 7 2010-07-01
8 2010-07-07 76249. 848582. D 14797 7 2010-07-01
9 2010-07-03 77586. 819905. D 14793 7 2010-07-01
10 2010-07-07 76249. 848582. D 14797 7 2010-07-01
11 2010-07-03 77586. 819905. D 14793 7 2010-07-01
12 2010-07-07 76249. 848582. D 14797 7 2010-07-01
13 2010-07-03 77586. 819905. D 14793 7 2010-07-01
14 2010-07-07 76249. 848582. D 14797 7 2010-07-01
15 2010-07-03 77586. 819905. D 14793 7 2010-07-01
16 2010-07-07 76249. 848582. D 14797 7 2010-07-01
$D
# A tibble: 24 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-11 61531. 883305. D 14801 7 2010-07-11
2 2010-07-15 69514. 867063. D 14805 7 2010-07-11
3 2010-07-19 69178. 890183. D 14809 7 2010-07-11
4 2010-07-11 61531. 883305. D 14801 7 2010-07-11
5 2010-07-15 69514. 867063. D 14805 7 2010-07-11
6 2010-07-19 69178. 890183. D 14809 7 2010-07-11
7 2010-07-11 61531. 883305. D 14801 7 2010-07-11
8 2010-07-15 69514. 867063. D 14805 7 2010-07-11
9 2010-07-19 69178. 890183. D 14809 7 2010-07-11
10 2010-07-11 61531. 883305. D 14801 7 2010-07-11
# ... with 14 more rows
$D
# A tibble: 24 x 7
date x y ID jDate Month new
<date> <dbl> <dbl> <chr> <dbl> <dbl> <date>
1 2010-07-23 74554. 898077. D 14813 7 2010-07-21
2 2010-07-27 77455. 834715. D 14817 7 2010-07-21
3 2010-07-31 77461. 873993. D 14821 7 2010-07-21
4 2010-07-23 74554. 898077. D 14813 7 2010-07-21
5 2010-07-27 77455. 834715. D 14817 7 2010-07-21
6 2010-07-31 77461. 873993. D 14821 7 2010-07-21
7 2010-07-23 74554. 898077. D 14813 7 2010-07-21
8 2010-07-27 77455. 834715. D 14817 7 2010-07-21
9 2010-07-31 77461. 873993. D 14821 7 2010-07-21
10 2010-07-23 74554. 898077. D 14813 7 2010-07-21
# ... with 14 more rows
ID <- rep(c("A","B","C", "D"), 1000)
ID2 <- rep(c("A", "B", "C"), 1000)
date <- rep_len(seq(dmy("01-01-2010"), dmy("31-12-2013"), by = "days"), 500)
x <- runif(length(date), min = 60000, max = 80000)
y <- runif(length(date), min = 800000, max = 900000)
df <- data.frame(date = date,
x = x,
y =y,
ID)
df2 <- data.frame(date = date,
x = x,
y =y,
ID2)
df2$jDate <- julian(as.Date(df2$date), origin = as.Date("1970-01-01"))
df2$Month <- month(df2$date)
july <- df %>%
# Creates a new column assigning the first day in the 10-day interval in which
# the date falls under (e.g., 01-03-2021 would be in the first 10-day interval
# so the `floor_date` assigned to it would be 01-01-2021)
mutate(new = floor_date(date, "10 days")) %>%
# For any months that has 31 days, the 31st day would normally be assigned its
# own interval. The code below takes the 31st day and joins it with the
# previous interval.
group_by(ID) %>%
mutate(new = if_else(day(new) == 31, new - days(10), new)) %>%
group_by(new, .add = TRUE) %>%
# Filter the data by the season based on the `season_categ` column
filter(Month == "7") %>%
group_split()
july2 <- df2 %>%
# Creates a new column assigning the first day in the 10-day interval in which
# the date falls under (e.g., 01-03-2021 would be in the first 10-day interval
# so the `floor_date` assigned to it would be 01-01-2021)
mutate(new = floor_date(date, "10 days")) %>%
# For any months that has 31 days, the 31st day would normally be assigned its
# own interval. The code below takes the 31st day and joins it with the
# previous interval.
group_by(ID2) %>%
mutate(new = if_else(day(new) == 31, new - days(10), new)) %>%
group_by(new, .add = TRUE) %>%
# Filter the data by the season based on the `season_categ` column
filter(Month == "7") %>%
group_split()
names(july) <- sapply(july, function(x) paste(x$ID[1]))
names(july2) <- sapply(july2, function(x) paste(x$ID2[1]))
推荐答案
听起来您正在尝试删除7月列表中没有7月2日列表中姓名的条目。如果是这样，将下面的代码添加到您的堆栈中将会起到作用：
# Get the unique names of the `july` list
jul_names <- unique(names(july))
# Find out which names are shared between the two lists
same_names <- jul_names[jul_names%in%unique(names(july2))]
# Subset the july list to only keep those entries with specific names
july <- july[names(july)%in%same_names]
如果这不是您所希望的，那么我们需要有关该问题的更多详细信息。正如一位评论者指出的那样，你的Reprex中有几个错误，所以我对你想要做的事情做出了最好的猜测：
july <- df %>%
mutate(new = floor_date(date, "10 days")) %>%
mutate(new = if_else(day(new) == 31, new - days(10), new)) %>%
group_by(ID, new) %>%
filter(month(date) == 7) %>%
group_split()
这篇关于根据两个列表之间的不同名称删除组件的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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