我的数据看起来类似于以下内容：
const input = [
[1, 2, 3, 4],
[12, 13, 14, 5],
[11, 16, 15, 6],
[10, 9, 8, 7]
];
function snail(array) {
var res = [];
if (!array.length) return res;
var next = array.shift();
if (next) res = res.concat(next);
for (var i = 0; i < array.length; i++) {
res.push(array[i].pop());
}
next = array.pop()
if (next) res = res.concat(next.reverse());
for (var i = array.length - 1; i >= 0; i--) {
res.push(array[i].shift());
}
return res.concat(snail(array));
}
console.log(snail(input));我试图在$`2013 Jul`
# A tibble: 2 x 12
AAPL AMD ADI ABBV A APD AA CF NVDA HOG WMT AMZN
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 -0.00252 0.00385 0.000314 0.00148 0.000231 0.000644 -0.00107 -0.00137 0.000886 0.000806 0.000689 0.000615
2 1 5 2 5 2 3 1 1 4 4 3 3
$`2013 Aug`
# A tibble: 2 x 12
AAPL AMD ADI ABBV A APD AA CF NVDA HOG WMT AMZN
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 0.0000471 0.00297 0.00111 0.00187 0.0000542 0.00186 -0.000795 -0.00109 0.00140 0.000700 0.000965 0.00101
2 1 5 3 5 2 4 1 1 4 2 3 3
列表中的case_when，dplyr和mutate函数中使用map。在这里，我将所有包含5或1的列表替换为0.5，将所有其他列表都替换为0。
我正在尝试tibble的变体，但不知道如何将mutate_all与case_when一起使用。我也愿意就如何解决这个问题提出其他建议。
mutate_all数据
x[1] %>%
mutate_all(case_when(... = 5 ~ 0.5))
编辑：
是否可以确保数字总和为1？例如，当前我们有：
list(`2013 Jul` = structure(list(AAPL = c(-0.00252413896048252,
1), AMD = c(0.00385385230384388, 5), ADI = c(0.000313658814841043,
2), ABBV = c(0.00148473194650269, 5), A = c(0.000231372267065186,
2), APD = c(0.000644411609229898, 3), AA = c(-0.00106999405402468,
1), CF = c(-0.00136811540143579, 1), NVDA = c(0.000886436095375894,
4), HOG = c(0.000806051331850114, 4), WMT = c(0.000689490484865284,
3), AMZN = c(0.000614708184565435, 3)), row.names = c(NA, -2L
), class = c("tbl_df", "tbl", "data.frame")), `2013 Aug` = structure(list(
AAPL = c(0.0000471064768722691, 1), AMD = c(0.00297250845145986,
5), ADI = c(0.00110927645875706, 3), ABBV = c(0.00186505842086247,
5), A = c(0.0000542259939665846, 2), APD = c(0.00186178155179209,
4), AA = c(-0.000794925865044543, 1), CF = c(-0.00109320436559941,
1), NVDA = c(0.00139874293303574, 4), HOG = c(0.000699507074667968,
2), WMT = c(0.000964557826996342, 3), AMZN = c(0.00100980845937234,
3)), row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"
)), `2013 Sep` = structure(list(AAPL = c(0.000874550640770086,
3), AMD = c(0.00212896308150426, 5), ADI = c(0.000297401899798995,
1), ABBV = c(0.00126327568847214, 4), A = c(0.00097767693668047,
3), APD = c(0.00143416399096666, 5), AA = c(-0.000734440361937234,
1), CF = c(-0.000254998800234454, 1), NVDA = c(0.00127259109916125,
4), HOG = c(0.00105093597431519, 3), WMT = c(0.00038339075327491,
2), AMZN = c(0.000479002073488143, 2)), row.names = c(NA, -2L
), class = c("tbl_df", "tbl", "data.frame")), `2013 Oct` = structure(list(
AAPL = c(0.000682565466572836, 2), AMD = c(0.00313699867162714,
5), ADI = c(0.000209923665516306, 1), ABBV = c(0.000865756791407934,
2), A = c(0.00161631482825611, 4), APD = c(0.00168294315897343,
5), AA = c(-0.000319519044240903, 1), CF = c(0.00096163857613333,
3), NVDA = c(0.00158604314072248, 4), HOG = c(0.00151424115101764,
3), WMT = c(0.00000265229900199134, 1), AMZN = c(0.00124777917896926,
3)), row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"
)), `2013 Nov` = structure(list(AAPL = c(0.00138847413611967,
4), AMD = c(0.00131189086851618, 3), ADI = c(0.000998905149605624,
2), ABBV = c(0.00053428429850944, 1), A = c(0.0016278252466143,
4), APD = c(0.0018596391559994, 5), AA = c(0.000727945791304539,
1), CF = c(0.00128641077503917, 3), NVDA = c(0.000839077531026809,
2), HOG = c(0.00128443125529569, 3), WMT = c(-0.00000406995915300601,
1), AMZN = c(0.00279573900270221, 5)), row.names = c(NA, -2L), class = c("tbl_df",
"tbl", "data.frame")), `2013 Dec` = structure(list(AAPL = c(0.00176889092052374,
5), AMD = c(-0.000742603775364103, 1), ADI = c(0.00044132637464973,
1), ABBV = c(0.00113925715965696, 3), A = c(0.00135042334177499,
4), APD = c(0.00122867453428789, 3), AA = c(0.00102055404174894,
2), CF = c(0.00128611035428346, 3), NVDA = c(0.000674203471055744,
2), HOG = c(0.00164877495332821, 4), WMT = c(0.000671450466059644,
1), AMZN = c(0.00299158521138261, 5)), row.names = c(NA, -2L), class = c("tbl_df",
"tbl", "data.frame")))
这是正确的。但是，我现在看到有四个$`2017 Aug`
# A tibble: 2 x 12
AAPL AMD ADI ABBV A APD AA CF NVDA HOG WMT AMZN
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0.5 0 0 0 0 0.5 0.5 0.5 0.5 0
。是否可以将输出获取为：
0-5答案 0 :(得分：1)
因为它是list，所以我们可以遍历list
map(lst1, ~ .x %>%
mutate_all(~ case_when(. %in% c(5, 1) ~ .5,
TRUE ~ 0)) )
在这种情况下，我们也可以在没有case_when的情况下做到这一点
map(lst1, ~ .x %>%
mutate_all( ~ c(0, .5)[(. %in% c(5, 1)) + 1]))
如果我们需要将列的总和设为1，对于没有sum等于1（因为只有两行）的情况，一种选择是从1中减去第一个数字
map(lst1, ~ .x %>%
mutate_all(~ case_when(. %in% c(5, 1) ~ .5,
TRUE ~ 0) ) %>%
mutate_all(~ if(sum(.) != 1) replace(., 1, 1 - sum(.)) else .) )
或者如果我们需要规范化行
map(lst1, ~ .x %>%
mutate_all(~ case_when(. %in% c(5, 1) ~ .5,
TRUE ~ 0) ) %>%
pmap_dfr(., ~
{v1 <- c(...)
v2 <- if(sum(v1) > 1) replace(v1, v1 != 0, 1/sum(v1!=0)) else v1
as.list(v2)}))
#$`2013 Jul`
# A tibble: 2 x 12
# AAPL AMD ADI ABBV A APD AA CF NVDA HOG WMT AMZN
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 0 0 0 0 0 0 0 0 0 0 0 0
#2 0.2 0.2 0 0.2 0 0 0.2 0.2 0 0 0 0
#$`2013 Aug`
# A tibble: 2 x 12
# AAPL AMD ADI ABBV A APD AA CF NVDA HOG WMT AMZN
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 0 0 0 0 0 0 0 0 0 0 0 0
#2 0.2 0.2 0 0.2 0 0 0.2 0.2 0 0 0 0
#$`2013 Sep`
# A tibble: 2 x 12
# AAPL AMD ADI ABBV A APD AA CF NVDA HOG WMT AMZN
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 0 0 0 0 0 0 0 0 0 0 0 0
#2 0 0.2 0.2 0 0 0.2 0.2 0.2 0 0 0 0
#$`2013 Oct`
# A tibble: 2 x 12
# AAPL AMD ADI ABBV A APD AA CF NVDA HOG WMT AMZN
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 0 0 0 0 0 0 0 0 0 0 0 0
#2 0 0.2 0.2 0 0 0.2 0.2 0 0 0 0.2 0
$`2013 Nov`
# A tibble: 2 x 12
AAPL AMD ADI ABBV A APD AA CF NVDA HOG WMT AMZN
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0.2 0 0.2 0.2 0 0 0 0.2 0.2
$`2013 Dec`
# A tibble: 2 x 12
AAPL AMD ADI ABBV A APD AA CF NVDA HOG WMT AMZN
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 0 0 0 0 0 0 0 0 0 0 0 0
2 0.2 0.2 0.2 0 0 0 0 0 0 0 0.2 0.2