Neo4j密碼查詢中的按位操作替代

我需要在密碼查詢中按位「和」。似乎cypher不支持按位操作。任何建議的替代品？這就是我想要檢測的... 例如，268是（2^8 + 2^3 + 2^2），正如你所看到的，2^3 = 8是我原始數字的一部分。所以，如果我使用按位與（100001100）&（1000）= 1000，那麼我可以檢測8是否是268的一部分。如何在沒有按位支持的情況下執行此操作？有什麼建議麼？我需要在密碼中這樣做。Neo4j密碼查詢中的按位操作替代

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2016-02-05 Lina

謝謝戴夫。我試過你的解決方案，他們都工作。對我來說，找到另一種方法是一個很好的暗示。這是我解決它的方式。我用字符串比較。

with '100001100' as number , '100000000' as sub_number 
with number,sub_number,range(length (number)-1,length (number)-length(sub_number),-1) as tail,length (number)-length(sub_number) as difference 
unwind tail as i 
with i,sub_number,number, i - length (number) + length (sub_number) as sub_number_position 
with sub_number_position, (split(number,''))[i-1] as bit_mask , (split(sub_number,''))[sub_number_position] as sub_bit 
with collect(toInt(bit_mask) * toInt(sub_bit)) as result 

return result

顯然，數字和子編號可以有不同的值。

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2016-02-08 14:17:09 Lina

幾乎可以肯定失敗選擇在第一時間按位操作的目的，但如果你絕對需要和兩個二進制數的暗號，你可以做這樣的事情與收藏。

with split('100001100', '') as bin_term_1 
, split('000001000', '') as bin_term_2 
, toString(1) as one 
with bin_term_1, bin_term_2, one, range(0,size(bin_term_1)-1,1) as index 
unwind index as i 
with i, bin_term_1, bin_term_2, one, 
case 
    when (bin_term_1[i] = one) and (bin_term_2[i] = one) then 
    1 
    else 
    0 
    end as r 
return collect(r) as AND

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2016-02-05 15:29:13

如果你絕對必須做暗號的操作可能是一個更好的解決辦法是實現像@埃文的SO解決方案Alternative to bitwise operation使用暗號。

您可以通過使用暗號，看起來像這樣將您的數據開始...

// convert binary to a product of prime numbers 
// start with the number to conver an a collection of primes 
with '100001100' as number 
, [2,3,5,7,13,17,19,23,29,31,37] as primes 

// create an index based on the size of the binary number to convert 
// take a slice of the prime array that is the size of the number to convert 
with number 
, range(length(number)-1,0,-1) as index 
, primes[0..length(number)] as primes, decimals[0..length(number)] as decimals 

// iterate over the index and match the prime number to the bits in the number to convert 
unwind index as i 
with (split(number,''))[i] as binary_place_holder, primes[-i-1] as prime_place_holder, decimals[-i-1] as decimal_place_holder 

// collect the primes that are set by multiplying by the set bits 
with collect(toInt(binary_place_holder) * prime_place_holder) as prime_placeholders 

// filter out the zero bits 
with filter(p in prime_placeholders where p > 0) as prime_placeholders 

// return a product of primes of the set bits 
return prime_placeholders, reduce(pp = 1, p in prime_placeholders | pp * p) as prime_product

樣品上面查詢的輸出。該查詢可以適用於使用主要產品更新屬性。

這裏是當你想使用它，你可以使用素數的模量的位置轉換如何分解

然後屏幕帽你想測試的位。

// test if the fourth bit is set in the decimal 268 
// 268 is the equivalent of a prime product of 1015 
// a modulus 7 == 0 will indicate the bit is set 
with 1015 as prime_product 
, [2,3,5,7,13,17,19,23,29,31,37] as primes 
, 4 as bit_to_test 
with bit_to_test 
, prime_product 
, primes[bit_to_test-1] as prime 
, prime_product % primes[bit_to_test-1] as mod_remains 
with 
    case when mod_remains = 0 then 
    'bit ' + toString(bit_to_test) + ' set' 
    else 
    'bit ' + toString(bit_to_test) + ' NOT set' 
    end as bit_set 
return bit_set

來源

2016-02-06 15:56:57

使用密碼執行此類型測試的另一種方法是將您的十進制值轉換爲表示已設置位的小數的集合。

// convert the binary number to a collection of decimal parts 
// create an index the size of the number to convert 
// create a collection of decimals that correspond to the bit locations 
with '100001100' as number 
, [1,2,4,8,16,32,64,128,256,512,1024,2048,4096] as decimals 
with number 
, range(length(number)-1,0,-1) as index 
, decimals[0..length(number)] as decimals 

// map the bits to decimal equivalents 
unwind index as i 
with number, i, (split(number,''))[i] as binary_placeholder, decimals[-i-1] as decimal_placeholder 

// multiply the decimal value by the bits that are set 
with collect(decimal_placeholder * toInt(binary_placeholder)) as decimal_placeholders 

// filter out the zero values from the collection 
with filter(d in decimal_placeholders where d > 0) as decimal_placeholders 
return decimal_placeholders

下面是返回結果的示例。

然後，當你要測試的號碼是否是小數，你可以測試實際小數集合中存在。

with [4, 8, 256] as decimal_placeholders 
, 8 as decimal_to_test 
return 
    case 
    when decimal_to_test in decimal_placeholders then 
     toString(decimal_to_test) + ' value bit is set' 
    else 
     toString(decimal_to_test) + ' value bit is NOT set' 
    end as bit_set_test

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2016-02-06 16:11:12

或者，如果有一個APOC可用，他們可以使用apoc.bitwise.op這是一個圍繞java按位操作的包裝。

RETURN apoc.bitwise.op(268, "&",8) AS `268_AND_8`

其產生以下結果

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2018-01-11 20:30:50

Neo4j密碼查詢中的按位操作替代

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