I've got this string: "37:48.1234567"
(there's a colon and there's a decimal in there)
我有这个字符串:“37:48.1234567”(有一个冒号,那里有一个小数)
I need to remove everything from the decimal and past it.
我需要从小数中删除所有内容并将其删除。
The number before the decimal could be any length (ie. 1:23:39.12357
).
小数点前的数字可以是任意长度(即1:23:39.12357)。
3 个解决方案
#1
13
str.split(".")[0]
Hope that helps!
希望有所帮助!
#2
4
I made a little benchmark with the solutions up to now. I modified the solutions a bit.
到目前为止,我对解决方案做了一些基准测试。我稍微修改了一下解决方案。
One unclear problem: Whats the result for 1:23:39.123.57
? 1:23:39
or 1:23:39.123
?
一个不明确的问题:结果是1:23:39.123.57? 1:23:39或1:23:39.123?
I choosed 1:23:39
, only test regex has the other solution (see Tims answer)
我选择了1:23:39,只有测试正则表达式有另一种解决方案(见Tims回答)
The split
variant has 2x2 variants: [0]
or .first
and split with a 2nd parameter , 2
. The 2nd parameter of split limit the number of the splitted elements. We need only the first, so we may split in first and rest.
splitvariant具有2x2变体:[0]或.first并使用第二个参数2进行分割。分割的第二个参数限制分割元素的数量。我们只需要第一个,所以我们可以先分开休息。
The fastes solution: split('.', 2)
. If you use first
or [0]
makes no difference. I would prefer first.
紧固解决方案:拆分('。',2)。如果你先使用或[0]没有区别。我宁愿先。
The result ( ruby 1.9.2p290
):
结果(ruby 1.9.2p290):
Rehearsal -------------------------------------------------
regex 0.141000 0.000000 0.141000 ( 0.140625)
regex2 0.125000 0.000000 0.125000 ( 0.125000)
split[0] 0.031000 0.000000 0.031000 ( 0.031250)
split[0],2 0.047000 0.000000 0.047000 ( 0.046875)
split.first 0.031000 0.000000 0.031000 ( 0.031250)
split.first,2 0.031000 0.000000 0.031000 ( 0.031250)
---------------------------------------- total: 0.406000sec
user system total real
regex 0.125000 0.000000 0.125000 ( 0.125000)
regex2 0.109000 0.015000 0.124000 ( 0.125000)
split[0] 0.031000 0.000000 0.031000 ( 0.031250)
split[0],2 0.032000 0.000000 0.032000 ( 0.031250)
split.first 0.047000 0.000000 0.047000 ( 0.046875)
split.first,2 0.031000 0.000000 0.031000 ( 0.031250)
The result ( ruby 1.8.7p352
):
结果(ruby 1.8.7p352):
Rehearsal -------------------------------------------------
regex 0.060000 0.000000 0.060000 ( 0.060661)
regex2 0.050000 0.000000 0.050000 ( 0.049141)
split[0] 0.080000 0.000000 0.080000 ( 0.083703)
split[0],2 0.070000 0.000000 0.070000 ( 0.072780)
split.first 0.080000 0.000000 0.080000 ( 0.080419)
split.first,2 0.070000 0.000000 0.070000 ( 0.068370)
---------------------------------------- total: 0.410000sec
user system total real
regex 0.060000 0.000000 0.060000 ( 0.055427)
regex2 0.050000 0.000000 0.050000 ( 0.048538)
split[0] 0.080000 0.000000 0.080000 ( 0.084225)
split[0],2 0.070000 0.000000 0.070000 ( 0.071673)
split.first 0.070000 0.000000 0.070000 ( 0.079865)
split.first,2 0.060000 0.000000 0.060000 ( 0.068685)
The Benchmark:
基准:
require 'benchmark'
TESTDATA = %w{
37:48.1234567
1:23:39.12357
1:23:39.123.57
}
N = 10_000 #Number of Test loops
Benchmark.bmbm(10) {|b|
b.report('regex') { N.times { TESTDATA.each{|str| str.gsub(/\.[^.]*\Z/, '') }} }
b.report('regex2') { N.times { TESTDATA.each{|str| str.gsub(/\..*\Z/, '') }} }
b.report('split[0]') { N.times { TESTDATA.each{| str | str.split(".")[0] }} }
b.report('split[0],2') { N.times { TESTDATA.each{| str | str.split(".",2)[0] }} }
b.report('split.first') { N.times { TESTDATA.each{| str | str.split(".").first }} }
b.report('split.first,2') { N.times { TESTDATA.each{| str | str.split(".",2).first }} }
#~ b.report('') { N.times { TESTDATA.each{| str | }} }
#~ b.report('') { N.times { TESTDATA.each{|t| }} }
#~ b.report('') { N.times { TESTDATA.each{|t| }} }
} #Benchmark
#3
3
result = subject.gsub(/\.[^.]*\Z/, '')
does exactly this. It makes sure that if there is more than one decimal, only the last one will be affected.
正是如此。它确保如果有多个小数,则只会影响最后一个小数。
#1
13
str.split(".")[0]
Hope that helps!
希望有所帮助!
#2
4
I made a little benchmark with the solutions up to now. I modified the solutions a bit.
到目前为止,我对解决方案做了一些基准测试。我稍微修改了一下解决方案。
One unclear problem: Whats the result for 1:23:39.123.57
? 1:23:39
or 1:23:39.123
?
一个不明确的问题:结果是1:23:39.123.57? 1:23:39或1:23:39.123?
I choosed 1:23:39
, only test regex has the other solution (see Tims answer)
我选择了1:23:39,只有测试正则表达式有另一种解决方案(见Tims回答)
The split
variant has 2x2 variants: [0]
or .first
and split with a 2nd parameter , 2
. The 2nd parameter of split limit the number of the splitted elements. We need only the first, so we may split in first and rest.
splitvariant具有2x2变体:[0]或.first并使用第二个参数2进行分割。分割的第二个参数限制分割元素的数量。我们只需要第一个,所以我们可以先分开休息。
The fastes solution: split('.', 2)
. If you use first
or [0]
makes no difference. I would prefer first.
紧固解决方案:拆分('。',2)。如果你先使用或[0]没有区别。我宁愿先。
The result ( ruby 1.9.2p290
):
结果(ruby 1.9.2p290):
Rehearsal -------------------------------------------------
regex 0.141000 0.000000 0.141000 ( 0.140625)
regex2 0.125000 0.000000 0.125000 ( 0.125000)
split[0] 0.031000 0.000000 0.031000 ( 0.031250)
split[0],2 0.047000 0.000000 0.047000 ( 0.046875)
split.first 0.031000 0.000000 0.031000 ( 0.031250)
split.first,2 0.031000 0.000000 0.031000 ( 0.031250)
---------------------------------------- total: 0.406000sec
user system total real
regex 0.125000 0.000000 0.125000 ( 0.125000)
regex2 0.109000 0.015000 0.124000 ( 0.125000)
split[0] 0.031000 0.000000 0.031000 ( 0.031250)
split[0],2 0.032000 0.000000 0.032000 ( 0.031250)
split.first 0.047000 0.000000 0.047000 ( 0.046875)
split.first,2 0.031000 0.000000 0.031000 ( 0.031250)
The result ( ruby 1.8.7p352
):
结果(ruby 1.8.7p352):
Rehearsal -------------------------------------------------
regex 0.060000 0.000000 0.060000 ( 0.060661)
regex2 0.050000 0.000000 0.050000 ( 0.049141)
split[0] 0.080000 0.000000 0.080000 ( 0.083703)
split[0],2 0.070000 0.000000 0.070000 ( 0.072780)
split.first 0.080000 0.000000 0.080000 ( 0.080419)
split.first,2 0.070000 0.000000 0.070000 ( 0.068370)
---------------------------------------- total: 0.410000sec
user system total real
regex 0.060000 0.000000 0.060000 ( 0.055427)
regex2 0.050000 0.000000 0.050000 ( 0.048538)
split[0] 0.080000 0.000000 0.080000 ( 0.084225)
split[0],2 0.070000 0.000000 0.070000 ( 0.071673)
split.first 0.070000 0.000000 0.070000 ( 0.079865)
split.first,2 0.060000 0.000000 0.060000 ( 0.068685)
The Benchmark:
基准:
require 'benchmark'
TESTDATA = %w{
37:48.1234567
1:23:39.12357
1:23:39.123.57
}
N = 10_000 #Number of Test loops
Benchmark.bmbm(10) {|b|
b.report('regex') { N.times { TESTDATA.each{|str| str.gsub(/\.[^.]*\Z/, '') }} }
b.report('regex2') { N.times { TESTDATA.each{|str| str.gsub(/\..*\Z/, '') }} }
b.report('split[0]') { N.times { TESTDATA.each{| str | str.split(".")[0] }} }
b.report('split[0],2') { N.times { TESTDATA.each{| str | str.split(".",2)[0] }} }
b.report('split.first') { N.times { TESTDATA.each{| str | str.split(".").first }} }
b.report('split.first,2') { N.times { TESTDATA.each{| str | str.split(".",2).first }} }
#~ b.report('') { N.times { TESTDATA.each{| str | }} }
#~ b.report('') { N.times { TESTDATA.each{|t| }} }
#~ b.report('') { N.times { TESTDATA.each{|t| }} }
} #Benchmark
#3
3
result = subject.gsub(/\.[^.]*\Z/, '')
does exactly this. It makes sure that if there is more than one decimal, only the last one will be affected.
正是如此。它确保如果有多个小数,则只会影响最后一个小数。