例如，我有矢量
x=c(-1,-1,-1,-1,1,1,1,-1,-1,1,1,-1,-1,1,1,-1,1,-1,1).
以下列表：
Y =
list(1:7, 8:11, 12:15, 16:19)
如何根据列表x对向量Y进行排序？我的意思是排序前7个元素，接下来的4个元素，接下来的4个元素和最后4个元素。
应该是所需的输出
c(-1,-1,-1,-1,1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1)。
请注意，列表Y并不总是相同。
我尝试使用x[unlist(sapply(Y, sort))]，但它不起作用。
你有什么选择吗？
答案 0 :(得分：3)
unlist(lapply(lapply(Y, function(i) x[i]), sort))
# [1] -1 -1 -1 -1 1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1
这首先根据x中的索引将Y的元素提取到不同的列表项中，
lapply(Y, function(i) x[i])
然后用lapply(..., sort)独立地对每一个进行排序，然后将它们重新组合成一个带有unlist的向量。
使用此输入：
x = c(-1,-1,-1,-1,1,1,1,-1,-1,1,1,-1,-1,1,1,-1,1,-1,1)
Y = list(1:7, 8:11, 12:15, 16:19)
答案 1 :(得分：3)
不确定这是否要求你
/** Returns the version of the <a href="https://www.unicode.org/versions/enumeratedversions.html">
* Unicode standard</a> supported by the {@link java.lang.Character} class. The return value
* represents the major and minor version numbers (in low-order octets 1 and 0, respectively).
* The "update" version number is not represented, because this method cannot distinguish between
* update versions, such as 6.2.0 and 6.2.1.
* <p>
* This version number be can converted to a representation suitable for human interfaces
* with code such as <code>(version >> 8) + "." + (version & 0xFF)</code> or
* <code>System.out.printf("Unicode version: %d.%d%n", version >> 8, version & 0xFF)</code>.
* <p>
* This method is compatible with Java versions >= 1.1.
*
* @return The Unicode version number, with the major version number in low-order octet 1 and the
* minor version number in low-order octet 0.
*/
public static int getUnicodeVersion() {
/* Version determination here is based on testing support for the first
new code point added by each version of Unicode. (See
<https://www.unicode.org/Public/UCD/latest/ucd/DerivedAge.txt>.)
With regard to omission of the update version number, the document
"Unicode Standard Annex #44, Unicode Character Database" revision 21
(from Unicode 11.0.0 draft 1) states:
"Formally, the Age property is a catalog property whose enumerated
values correspond to a list of tuples consisting of a major version
integer and a minor version integer. The major version is a positive
integer constrained to the range 1..255. The minor version is a non-
negative integer constrained to the range 0..255. These range limit-
ations are specified so that implementations can be guaranteed that
all valid, assigned Age values can be represented in a sequence of two
unsigned bytes. A third value corresponding to the Unicode update
version is not required, because new characters are never assigned in
update versions of the standard."
See <https://www.unicode.org/reports/tr44/tr44-21.html#Character_Age>
for further details.
*/
// Preliminary Unicode 11 data obtained from <https://www.unicode.org/Public/11.0.0/ucd/DerivedAge-11.0.0d13.txt>.
if (Character.getType('\u0560') != Character.UNASSIGNED)
return 0xB00; // 11.0, June 2018.
if (Character.getType('\u0860') != Character.UNASSIGNED)
return 0xA00; // 10.0, June 2017.
if (Character.getType('\u08b6') != Character.UNASSIGNED)
return 0x900; // 9.0, June 2016.
if (Character.getType('\u08b3') != Character.UNASSIGNED)
return 0x800; // 8.0, June 2015.
if (Character.getType('\u037f') != Character.UNASSIGNED)
return 0x700; // 7.0, June 2014.
if (Character.getType('\u061c') != Character.UNASSIGNED)
return 0x603; // 6.3, September 2013.
if (Character.getType('\u20ba') != Character.UNASSIGNED)
return 0x602; // 6.2, September 2012.
if (Character.getType('\u058f') != Character.UNASSIGNED)
return 0x601; // 6.1, January 2012.
if (Character.getType('\u0526') != Character.UNASSIGNED)
return 0x600; // 6.0, October 2010.
if (Character.getType('\u0524') != Character.UNASSIGNED)
return 0x502; // 5.2, October 2009.
if (Character.getType('\u0370') != Character.UNASSIGNED)
return 0x501; // 5.1, March 2008.
if (Character.getType('\u0242') != Character.UNASSIGNED)
return 0x500; // 5.0, July 2006.
if (Character.getType('\u0237') != Character.UNASSIGNED)
return 0x401; // 4.1, March 2005.
if (Character.getType('\u0221') != Character.UNASSIGNED)
return 0x400; // 4.0, April 2003.
if (Character.getType('\u0220') != Character.UNASSIGNED)
return 0x302; // 3.2, March 2002.
if (Character.getType('\u03f4') != Character.UNASSIGNED)
return 0x301; // 3.1, March 2001.
if (Character.getType('\u01f6') != Character.UNASSIGNED)
return 0x300; // 3.0, September 1999.
if (Character.getType('\u20ac') != Character.UNASSIGNED)
return 0x201; // 2.1, May 1998.
if (Character.getType('\u0591') != Character.UNASSIGNED)
return 0x200; // 2.0, July 1996.
if (Character.getType('\u0000') != Character.UNASSIGNED)
return 0x101; // 1.1, June 1993.
return 0x100; // 1.0
}
输入
> unlist(lapply(y, function(z) sort(x[z])))
[1] -1 -1 -1 -1 1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1
答案 2 :(得分：2)
你也可以同时使用x和Y来避免循环和矢量化顺序（因为order允许在关系情况下按两个向量排序）
x[order(rep(seq_along(Y), lengths(Y)), x)]
# [1] -1 -1 -1 -1 1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1
插图的一些基准
set.seed(123)
N <- 1e5
x <- sample(N)
Y <- split(1:N, rep(1 : (N/5), each = 5))
microbenchmark::microbenchmark("Gregor" = unlist(lapply(lapply(Y, function(i) x[i]), sort)),
"Frank1" = ave(x, stack(setNames(Y, seq_along(Y)))$ind, FUN = sort),
"Frank2" = x[order(stack(setNames(Y, seq_along(Y)))$ind, x)],
"Jilber" = unlist(lapply(Y, function(z) sort(x[z]))),
"David" = x[order(rep(seq_along(Y), lengths(Y)), x)])
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# Gregor 904.277546 937.652137 958.911977 949.311012 961.324917 1164.024555 100 c
# Frank1 884.306262 922.496558 956.408754 941.394433 962.098976 1140.254656 100 c
# Frank2 27.384839 28.587845 30.806481 29.542219 31.684239 46.947814 100 b
# Jilber 923.135901 949.318532 967.981792 962.090176 976.574574 1137.115863 100 c
# David 2.184901 2.326817 2.622338 2.492732 2.524091 8.586322 100 a
矢量化解决方案~X500比循环更快