I want a random string of characters only (uppercase or lowercase), no numbers, in Go. What is the fastest and simplest way to do this?
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- dqtiober37522 2015-08-05 12:18关注
Paul's solution provides a simple, general solution.
The question asks for the "the fastest and simplest way". Let's address the fastest part too. We'll arrive at our final, fastest code in an iterative manner. Benchmarking each iteration can be found at the end of the answer.
All the solutions and the benchmarking code can be found on the Go Playground. The code on the Playground is a test file, not an executable. You have to save it into a file named
XX_test.go
and run it withgo test -bench . -benchmem
Foreword:
The fastest solution is not a go-to solution if you just need a random string. For that, Paul's solution is perfect. This is if performance does matter. Although the first 2 steps (Bytes and Remainder) might be an acceptable compromise: they do improve performance by like 50% (see exact numbers in the II. Benchmark section), and they don't increase complexity significantly.
Having said that, even if you don't need the fastest solution, reading through this answer might be adventurous and educational.
I. Improvements
1. Genesis (Runes)
As a reminder, the original, general solution we're improving is this:
func init() { rand.Seed(time.Now().UnixNano()) } var letterRunes = []rune("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ") func RandStringRunes(n int) string { b := make([]rune, n) for i := range b { b[i] = letterRunes[rand.Intn(len(letterRunes))] } return string(b) }
2. Bytes
If the characters to choose from and assemble the random string contains only the uppercase and lowercase letters of the English alphabet, we can work with bytes only because the English alphabet letters map to bytes 1-to-1 in the UTF-8 encoding (which is how Go stores strings).
So instead of:
var letters = []rune("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
we can use:
var letters = []bytes("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
Or even better:
const letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
Now this is already a big improvement: we could achieve it to be a
const
(there arestring
constants but there are no slice constants). As an extra gain, the expressionlen(letters)
will also be aconst
! (The expressionlen(s)
is constant ifs
is a string constant.)And at what cost? Nothing at all.
string
s can be indexed which indexes its bytes, perfect, exactly what we want.Our next destination looks like this:
const letterBytes = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ" func RandStringBytes(n int) string { b := make([]byte, n) for i := range b { b[i] = letterBytes[rand.Intn(len(letterBytes))] } return string(b) }
3. Remainder
Previous solutions get a random number to designate a random letter by calling
rand.Intn()
which delegates toRand.Intn()
which delegates toRand.Int31n()
.This is much slower compared to
rand.Int63()
which produces a random number with 63 random bits.So we could simply call
rand.Int63()
and use the remainder after dividing bylen(letterBytes)
:func RandStringBytesRmndr(n int) string { b := make([]byte, n) for i := range b { b[i] = letterBytes[rand.Int63() % int64(len(letterBytes))] } return string(b) }
This works and is significantly faster, the disadvantage is that the probability of all the letters will not be exactly the same (assuming
rand.Int63()
produces all 63-bit numbers with equal probability). Although the distortion is extremely small as the number of letters52
is much-much smaller than1<<63 - 1
, so in practice this is perfectly fine.To make this understand easier: let's say you want a random number in the range of
0..5
. Using 3 random bits, this would produce the numbers0..1
with double probability than from the range2..5
. Using 5 random bits, numbers in range0..1
would occur with6/32
probability and numbers in range2..5
with5/32
probability which is now closer to the desired. Increasing the number of bits makes this less significant, when reaching 63 bits, it is negligible.4. Masking
Building on the previous solution, we can maintain the equal distribution of letters by using only as many of the lowest bits of the random number as many is required to represent the number of letters. So for example if we have 52 letters, it requires 6 bits to represent it:
52 = 110100b
. So we will only use the lowest 6 bits of the number returned byrand.Int63()
. And to maintain equal distribution of letters, we only "accept" the number if it falls in the range0..len(letterBytes)-1
. If the lowest bits are greater, we discard it and query a new random number.Note that the chance of the lowest bits to be greater than or equal to
len(letterBytes)
is less than0.5
in general (0.25
on average), which means that even if this would be the case, repeating this "rare" case decreases the chance of not finding a good number. Aftern
repetition, the chance that we sill don't have a good index is much less thanpow(0.5, n)
, and this is just an upper estimation. In case of 52 letters the chance that the 6 lowest bits are not good is only(64-52)/64 = 0.19
; which means for example that chances to not have a good number after 10 repetition is1e-8
.So here is the solution:
const letterBytes = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ" const ( letterIdxBits = 6 // 6 bits to represent a letter index letterIdxMask = 1<<letterIdxBits - 1 // All 1-bits, as many as letterIdxBits ) func RandStringBytesMask(n int) string { b := make([]byte, n) for i := 0; i < n; { if idx := int(rand.Int63() & letterIdxMask); idx < len(letterBytes) { b[i] = letterBytes[idx] i++ } } return string(b) }
5. Masking Improved
The previous solution only uses the lowest 6 bits of the 63 random bits returned by
rand.Int63()
. This is a waste as getting the random bits is the slowest part of our algorithm.If we have 52 letters, that means 6 bits code a letter index. So 63 random bits can designate
63/6 = 10
different letter indices. Let's use all those 10:const letterBytes = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ" const ( letterIdxBits = 6 // 6 bits to represent a letter index letterIdxMask = 1<<letterIdxBits - 1 // All 1-bits, as many as letterIdxBits letterIdxMax = 63 / letterIdxBits // # of letter indices fitting in 63 bits ) func RandStringBytesMaskImpr(n int) string { b := make([]byte, n) // A rand.Int63() generates 63 random bits, enough for letterIdxMax letters! for i, cache, remain := n-1, rand.Int63(), letterIdxMax; i >= 0; { if remain == 0 { cache, remain = rand.Int63(), letterIdxMax } if idx := int(cache & letterIdxMask); idx < len(letterBytes) { b[i] = letterBytes[idx] i-- } cache >>= letterIdxBits remain-- } return string(b) }
6. Source
The Masking Improved is pretty good, not much we can improve on it. We could, but not worth the complexity.
Now let's find something else to improve. The source of random numbers.
There is a
crypto/rand
package which provides aRead(b []byte)
function, so we could use that to get as many bytes with a single call as many we need. This wouldn't help in terms of performance ascrypto/rand
implements a cryptographically secure pseudorandom number generator so it's much slower.So let's stick to the
math/rand
package. Therand.Rand
uses arand.Source
as the source of random bits.rand.Source
is an interface which specifies aInt63() int64
method: exactly and the only thing we needed and used in our latest solution.So we don't really need a
rand.Rand
(either explicit or the global, shared one of therand
package), arand.Source
is perfectly enough for us:var src = rand.NewSource(time.Now().UnixNano()) func RandStringBytesMaskImprSrc(n int) string { b := make([]byte, n) // A src.Int63() generates 63 random bits, enough for letterIdxMax characters! for i, cache, remain := n-1, src.Int63(), letterIdxMax; i >= 0; { if remain == 0 { cache, remain = src.Int63(), letterIdxMax } if idx := int(cache & letterIdxMask); idx < len(letterBytes) { b[i] = letterBytes[idx] i-- } cache >>= letterIdxBits remain-- } return string(b) }
Also note that this last solution doesn't require you to initialize (seed) the global
Rand
of themath/rand
package as that is not used (and ourrand.Source
is properly initialized / seeded).One more thing to note here: package doc of
math/rand
states:The default Source is safe for concurrent use by multiple goroutines.
So the default source is slower than a
Source
that may be obtained byrand.NewSource()
, because the default source has to provide safety under concurrent access / use, whilerand.NewSource()
does not offer this (and thus theSource
returned by it is more likely to be faster).7. Utilizing
strings.Builder
All previous solutions return a
string
whose content is first built in a slice ([]rune
in Genesis, and[]byte
in subsequent solutions), and then converted tostring
. This final conversion has to make a copy of the slice's content, becausestring
values are immutable, and if the conversion would not make a copy, it could not be guaranteed that the string's content is not modified via its original slice. For details, see How to convert utf8 string to []byte? and golang: []byte(string) vs []byte(*string).Go 1.10 introduced
strings.Builder
.strings.Builder
a new type we can use to build contents of astring
similar tobytes.Buffer
. It does it internally using a[]byte
, and when we're done, we can obtain the finalstring
value using itsBuilder.String()
method. But what's cool in it is that it does this without performing the copy we just talked about above. It dares to do so because the byte slice used to build the string's content is not exposed, so it is guaranteed that no one can modify it unintentionally or maliciously to alter the produced "immutable" string.So our next idea is to not build the random string in a slice, but with the help of a
strings.Builder
, so once we're done, we can obtain and return the result without having to make a copy of it. This may help in terms of speed, and it will definitely help in terms of memory usage and allocations.func RandStringBytesMaskImprSrcSB(n int) string { sb := strings.Builder{} sb.Grow(n) // A src.Int63() generates 63 random bits, enough for letterIdxMax characters! for i, cache, remain := n-1, src.Int63(), letterIdxMax; i >= 0; { if remain == 0 { cache, remain = src.Int63(), letterIdxMax } if idx := int(cache & letterIdxMask); idx < len(letterBytes) { sb.WriteByte(letterBytes[idx]) i-- } cache >>= letterIdxBits remain-- } return sb.String() }
Do note that after creating a new
strings.Buidler
, we called itsBuilder.Grow()
method, making sure it allocates a big-enough internal slice (to avoid reallocations as we add the random letters).8. "Mimicing"
strings.Builder
with packageunsafe
strings.Builder
builds the string in an internal[]byte
, the same as we did ourselves. So basically doing it via astrings.Builder
has some overhead, the only thing we switched tostrings.Builder
for is to avoid the final copying of the slice.strings.Builder
avoids the final copy by using packageunsafe
:// String returns the accumulated string. func (b *Builder) String() string { return *(*string)(unsafe.Pointer(&b.buf)) }
The thing is, we can also do this ourselves, too. So the idea here is to switch back to building the random string in a
[]byte
, but when we're done, don't convert it tostring
to return, but do an unsafe conversion: obtain astring
which points to our byte slice as the string data.This is how it can be done:
func RandStringBytesMaskImprSrcUnsafe(n int) string { b := make([]byte, n) // A src.Int63() generates 63 random bits, enough for letterIdxMax characters! for i, cache, remain := n-1, src.Int63(), letterIdxMax; i >= 0; { if remain == 0 { cache, remain = src.Int63(), letterIdxMax } if idx := int(cache & letterIdxMask); idx < len(letterBytes) { b[i] = letterBytes[idx] i-- } cache >>= letterIdxBits remain-- } return *(*string)(unsafe.Pointer(&b)) }
(9. Using
rand.Read()
)Go 1.7 added a
rand.Read()
function and aRand.Read()
method. We should be tempted to use these to read as many bytes as we need in one step, in order to achieve better performance.There is one small "problem" with this: how many bytes do we need? We could say: as many as the number of output letters. We would think this is an upper estimation, as a letter index uses less than 8 bits (1 byte). But at this point we are already doing worse (as getting the random bits is the "hard part"), and we're getting more than needed.
Also note that to maintain equal distribution of all letter indices, there might be some "garbage" random data that we won't be able to use, so we would end up skipping some data, and thus end up short when we go through all the byte slice. We would need to further get more random bytes, "recursively". And now we're even losing the "single call to
rand
package" advantage...We could "somewhat" optimize the usage of the random data we acquire from
math.Rand()
. We may estimate how many bytes (bits) we'll need. 1 letter requiresletterIdxBits
bits, and we needn
letters, so we needn * letterIdxBits / 8.0
bytes rounding up. We can calculate the probability of a random index not being usable (see above), so we could request more that will "more likely" be enough (if it turns out it's not, we repeat the process). We can process the byte slice as a "bit stream" for example, for which we have a nice 3rd party lib:github.com/icza/bitio
(disclosure: I'm the author).But Benchmark code still shows we're not winning. Why is it so?
The answer to the last question is because
rand.Read()
uses a loop and keeps callingSource.Int63()
until it fills the passed slice. Exactly what theRandStringBytesMaskImprSrc()
solution does, without the intermediate buffer, and without the added complexity. That's whyRandStringBytesMaskImprSrc()
remains on the throne. Yes,RandStringBytesMaskImprSrc()
uses an unsynchronizedrand.Source
unlikerand.Read()
. But the reasoning still applies; and which is proven if we useRand.Read()
instead ofrand.Read()
(the former is also unsynchronzed).II. Benchmark
All right, it's time for benchmarking the different solutions.
Moment of truth:
BenchmarkRunes-4 2000000 723 ns/op 96 B/op 2 allocs/op BenchmarkBytes-4 3000000 550 ns/op 32 B/op 2 allocs/op BenchmarkBytesRmndr-4 3000000 438 ns/op 32 B/op 2 allocs/op BenchmarkBytesMask-4 3000000 534 ns/op 32 B/op 2 allocs/op BenchmarkBytesMaskImpr-4 10000000 176 ns/op 32 B/op 2 allocs/op BenchmarkBytesMaskImprSrc-4 10000000 139 ns/op 32 B/op 2 allocs/op BenchmarkBytesMaskImprSrcSB-4 10000000 134 ns/op 16 B/op 1 allocs/op BenchmarkBytesMaskImprSrcUnsafe-4 10000000 115 ns/op 16 B/op 1 allocs/op
Just by switching from runes to bytes, we immediately have 24% performance gain, and memory requirement drops to one third.
Getting rid of
rand.Intn()
and usingrand.Int63()
instead gives another 20% boost.Masking (and repeating in case of big indices) slows down a little (due to repetition calls): -22%...
But when we make use of all (or most) of the 63 random bits (10 indices from one
rand.Int63()
call): that speeds up big time: 3 times.If we settle with a (non-default, new)
rand.Source
instead ofrand.Rand
, we again gain 21%.If we utilize
strings.Builder
, we gain a tiny 3.5% in speed, but we also achieved 50% reduction in memory usage and allocations! That's nice!Finally if we dare to use package
unsafe
instead ofstrings.Builder
, we again gain a nice 14%.Comparing the final to the initial solution:
RandStringBytesMaskImprSrcUnsafe()
is 6.3 times faster thanRandStringRunes()
, uses one sixth memory and half as few allocations. Mission accomplished.本回答被题主选为最佳回答 , 对您是否有帮助呢?解决 无用评论 打赏 举报
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