numbers

列表类型(list)可以存储一个有序的字符串列表,内部实现是 双向链表 , 借助列表类型,Redis还可以作为队列使用, 且与散列类型键最多能容纳的字段数量相同, 一个列表类型键最多能容纳2^32-1个元素 常用命令 向左\向右插入元素 lpush key value [value....] 从左边向列表增加元素 rpush key value [value....] 从右边向列表增加元素 lpush numbers 1 lpush numbers 2 3 此时为 3 2 1 rpush numbers 0 rpush numbers -1 此时为 3 2 1 0 -1 从左\从右弹出元素(返回元素的值) lpop key 从左弹出 rpop key 从右弹出 此时列表numbers为 3 2 1 0 -1 lpop numbers "3" rpop numbers "-1" 一些其他的用法1 获得列表中元素的个数: llen key llen number 3 获得列表片段: lrange key start stop lrange number 0 2 1) "2" 2) "1" 3) "0" 此命令也支持负索引 lrange number -2 -1 1) "1" 2) "0" 表示右边第二个元素到右边第一个元素 若要展示number全部存储的字符串并且不知道数量

Let's say I create some number A , of the order 10^4 : A = 81472.368639; disp(A) 8.1472e+04 That wasn't what I wanted. Where are my decimals? There should be six decimals more. Checking the variable editor shows me this: Again, I lost my decimals. How do I keep these for further calculations? Scientific notation, or why you didn't lose any decimals You didn't lose any decimals, this is just MATLAB's way of displaying large numbers. MATLAB rounds the display of numbers, both in the command window and in the variable editor, to one digit before the dot and four after that, using scientific

Hello I want to create a list of strings where each string is a number. Fox example given the number 4 I would like to create a function that returns a list with elements '0','1','2','3','4'. In C/C++ this can be done by using the Ascii code of 0 and then increase them. I am new to python and I don’t know how to do it. Is there any way to do it? You can do this with str and range like this map(str, range(number + 1)) When number = 4 , print(map(str, range(4 + 1))) # ['0', '1', '2', '3', '4'] map being frowned upon by many python developers, here is the comprehension list equivalent: [str(x)

How can one choose a number with a specific probability p ? Say we must choose between {0, 1} and the probability p stands for choosing 1 . So when p=0.8 we choose 1 with 80% and 0 with 20%. Is there a simple solution in R for this? Take a look at sample function. > set.seed(1) > sample(c(0,1), size=10, replace=TRUE, prob=c(0.2,0.8)) [1] 1 1 1 0 1 0 0 1 1 1 From the helpfile you can read: sample takes a sample of the specified size from the elements of x using either with or without replacement. and the argument prob in sample acts as ... A vector of probability weights for obtaining the

I have just observed that the parseInt function doesn't take care about the decimals in case of integers (numbers containing the e character). Let's take an example: -3.67394039744206e-15 > parseInt(-3.67394039744206e-15) -3 > -3.67394039744206e-15.toFixed(19) -3.6739e-15 > -3.67394039744206e-15.toFixed(2) -0 > Math.round(-3.67394039744206e-15) 0 I expected that the parseInt will also return 0 . What's going on at lower level? Why does parseInt return 3 in this case (some snippets from the source code would be appreciated)? In this example I'm using node v0.12.1 , but I expect same to happen

is there a predefined format-function that rounds a number to the first 3 digits? (The start should be a numbers != 0) -0.02528498 to -0.0253 1.857403 to 1.86 2060943 to 2060000 0.00006513832 to 0.0000651 You can use the function signif : signif(-0.02528498, 3) # [1] -0.0253 signif(1.857403, 3) # [1] 1.86 signif(2060943, 3) # [1] 2060000 signif(0.00006513832, 3) # [1] 0.0000651 来源： https://stackoverflow.com/questions/29674349/round-number-to-the-first-3-digits-start-with-digit-0

Using Node.js, I'm evaluating the expression: 0 < Number.MIN_VALUE To my surprise, this returns true . Why is that? And: How can I get the smallest number available for which the comparison works as expected? Number.MIN_VALUE is 5e-324 , i.e. the smallest positive number that can be represented within float precision, i.e. that's as close as you can get to zero. It defines the best resolution floats give you. Now the overall smallest value is Number.NEGATIVE_INFINITY although that's not really numeric in the strict sense. Number.MIN_VALUE is equivalent to 5e-324 , which is greater than 0.