zero

问题 This may be a pretty basic question. I understand that there is a C convention to set the value of null pointers to zero. Is it possible that you can ever allocate space for a new variable in Windows, and the address of that allocated space happens to be zero? If not, what usually occupies that address region? 回答1: On MS-DOS the null pointer is a fairly valid pointer and due to the OS running in real mode it was actually possible to overwrite the 0x0 address with garbage and corrupt the

问题 How can I replace all 0's with the last non-zero value per ID in R? Example: Input: df <- data.frame(ID = c(1,1,1,1,1,1,1,2,2,2,2), Var1 = c(0,10, 30, 0, 0,50,80,0, 0, 57, 0)) Output: df <- data.frame(ID = c(1,1,1,1,1,1,1,2,2,2,2), Var1 = c(0,10, 30, 0, 0,50,80,0, 0, 57, 0), res = c(0,10,30,30,30,50,80,0,0,57,57)) Is there an easy way with lag function? 回答1: Here's a tidyverse approach: library(tidyverse) df %>% group_by(ID) %>% mutate(x = replace(Var1, cumsum(Var1 !=0) > 0 & Var1 == 0, NA))

问题 How can I replace all 0's with the last non-zero value per ID in R? Example: Input: df <- data.frame(ID = c(1,1,1,1,1,1,1,2,2,2,2), Var1 = c(0,10, 30, 0, 0,50,80,0, 0, 57, 0)) Output: df <- data.frame(ID = c(1,1,1,1,1,1,1,2,2,2,2), Var1 = c(0,10, 30, 0, 0,50,80,0, 0, 57, 0), res = c(0,10,30,30,30,50,80,0,0,57,57)) Is there an easy way with lag function? 回答1: Here's a tidyverse approach: library(tidyverse) df %>% group_by(ID) %>% mutate(x = replace(Var1, cumsum(Var1 !=0) > 0 & Var1 == 0, NA))

问题 This is the code ` region<-"ARLINGTON" data<-read.csv("city_markets.csv") for(i in 1:length(data[[1]])){ if(grep(region,as.character(data[[3]][i]),ignore.case=TRUE)==1){ for(j in 1:length(data)){ write(data[[j]][i],"analyzed.txt",append=TRUE) } } } ` now what i'm trying to do here is I'm accessing the csv's column(3rd one) and comparing it with the region specified! i keep getting the error Error in if (grep(region, as.character(data[[3]][i]), ignore.case = TRUE) == : argument is of length

问题 I use Hibernate to access to a legacy DB. For some tables the parent-child reference integrity is not enforced, and the long 0 value is used instead of NULL for some "parent" columns in child tables to denote "no parent". I still want to use these relations in @ManyToOne and @OneToMany fields, but get EntityNotFound error since the 0 value does not correspond to any record in master table. What are my options? 回答1: Use the NotFound annotation: @NotFound(action = NotFoundAction.IGNORE) See

问题 The float format (IEEE) has 32 bits. First bit for the sign, after that 8 bits for a biased exponent and after that another 23 bits for the mantissa. In this mantissa is the first 1 (is always 1) always hidden which leads me to my question: how does the number 0 look like in this format? because if the exponents 0 the number will always be 1. plus the mantissa is always minimum 1 right? If they are only zeros in the mantissa it will count as '1.0'... I really don't get this. 回答1: Wikipedia

问题 Let's say I have a table that could have various numbers in it, for example it could look like the following: ExampleA: MyTable -10, -3, 5, 10, ExampleB: MyTable -10, -5, 3, 10, So if I queried the table in ExampleA I'd want it to return "-3" (The value closet to 0) Likewise if I queried the table in ExampleB I'd want it to return "3" (The value closest to 0) I always want to find the value that is closest to zero regardless of the numbers in the table, how can I do this? Also, how could I

问题 I have an oracle column(artnr) contains a length of 1 which is of type number(9). I want to update the number as following... Example : If number is 0 then it should be 00000 If number is 1 then it should be 00001 If number is 12 the it should be 00012 Remember : here 00000,0000, and 00012 are of number datatypes The following are the methods I have tried but failed.. UPDATE pitb.toestel b SET b.artnr = LPAD (b.artnr, 5, 0) WHERE b.idinventaris = 403743; Failed because Lpad can only be

问题 I was wondering if there is a way, using PHP, to change this date format: 01.08.86 (January 8, 1986) to this format: 1.8.86. 回答1: <?php $date = "01.08.86"; $unix = strtotime($date); echo date('n.j.y', $unix); 回答2: How about a regex based solution: $str = '01.08.86'; $a = array('/^0(\d+)/','/\.0(\d+)/'); $b = array('\1','.\1'); $str = preg_replace($a,$b,$str); // $str is now '1.8.86' 来源： https://stackoverflow.com/questions/2480112/remove-leading-zeros-from-abbreviated-date-with-php

问题 The MSDN documentation mentions that double type includes negative zero. However, both -1.0 / double.PositiveInfinity and -double.Epsilon / 2 appear to return normal 0 (and compare equal to it). How can I get -0? 回答1: Here is a practical example of differentiating between the two without examining the bits. MSDN links here and here assisted me in constructing this example. static void Main(string[] args) { float a = 5 / float.NegativeInfinity; float b = 5 / float.PositiveInfinity; float c = 1