last

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Using Spark 1.5.1, I've been trying to forward fill null values with the last known observation for one column of my DataFrame. It is possible to start with a null value and for this case I would to backward fill this null value with the first knwn observation. However, If that too complicates the code, this point can be skipped. In this post , a solution in Scala was provided for a very similar problem by zero323 . But, I don't know Scala and I don't succeed to ''translate'' it in Pyspark API code. It's possible to do it with Pyspark ?

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Let's say we have a 1d numpy array filled with some int values. And let's say that some of them are 0 . Is there any way, using numpy array's power, to fill all the 0 values with the last non-zero values found? for example: arr = np.array([1, 0, 0, 2, 0, 4, 6, 8, 0, 0, 0, 0, 2]) fill_zeros_with_last(arr) print arr [1 1 1 2 2 4 6 8 8 8 8 8 2] A way to do it would be with this function: def fill_zeros_with_last(arr): last_val = None # I don't really care about the initial value for i in range(arr.size): if arr[i]: last_val = arr[i] elif last

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: What's the most efficient way to calculate the last day of the prior quarter? Example: given the date 11/19/2008, I want to return 9/30/2008. Platform is SQL Server 回答1: If @Date has the date in question Select DateAdd(day, -1, dateadd(qq, DateDiff(qq, 0, @Date), 0)) EDIT: Thanks to @strEagle below, simpler still is: Select dateadd(qq, DateDiff(qq, 0, @Date), -1) 回答2: Actually simpler is: SELECT DATEADD(qq, DATEDIFF(qq, 0, GETDATE()), -1) 回答3: Get the current date SELECT CONVERT(DATE,GETDATE()) [Current Date] Get the 1st date of the quarter

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I have a question about the coin change problem where we not only have to print the number of ways to change $n with the given coin denominations for eg {1,5,10,25}, but also print the ways For example if the target = $50, and the coins are {1,5,10,25} , then the ways to actually get use the coins to get the target are 2 × $25 1 × $25 + 2 × $10 + 1 × $5 etc. What is the best time complexity we could get to solve this problem? I tried to modify the dynamic programming solution for the coin change problem where we only need the

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: Is there an easy way to index a numpy multidimensional array along the last dimension, using an array of indices? For example, take an array a of shape (10, 10, 20) . Let's assume I have an array of indices b , of shape (10, 10) so that the result would be c[i, j] = a[i, j, b[i, j]] . I've tried the following example: a = np . ones (( 10 , 10 , 20 )) b = np . tile ( np . arange ( 10 ) + 10 , ( 10 , 1 )) c = a [ b ] However, this doesn't work because it then tries to index like a[b[i, j], b[i, j]] , which is not the same as a[i, j,

关于trie ​ 其实字典树和以上两种算法有很大不同,但是hash由于其优秀的应用,导致有些字符串查找用hash也是可行的. ​ 字典树中支持添加,查找,区间查询(可持久化字典树),而且在异或操作上有更加好的操作; 前置知识 ​ 树的基本构造; 入坑 ​ 字典树是通过动态建点,而形成的树,基本数组有两维, \(tr[x][to]\) 中第一维存的是节点标号,而第二维存的是当字符为 \(to\) 时通向的节点; 基本操作 ​ 我当时入门是学的是 这道题 ; 给你一些初始字符串,询问,给你一个字符串,这个字符串在这个初始字符串中是否存在 ​ 当时使用hash写的,但是没过; ​ 现在我们可以用字典树先存一下初始字符串,然后在树上匹配,单次时间复杂度 \(O(n)\) ; code #include<bits/stdc++.h> using namespace std; int n,m,trie[300007][27],num[300007],sz; bool vis[300007]; void build(char a[]){ int now=0; for(int i=0;i<strlen(a);i++){ if(!trie[now][a[i]-'a']) trie[now][a[i]-'a']=++sz; now=trie[now][a[i]-'a']; } num[now]++;

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: How can I find the last number in any big string? For eg in the following string I want 47 as the output: 'tr bgcolor="aa77bb"td>font face="verdana"color="white" size="2">b>Total/b>/font>/td>\td>font face="verdana"color="white" size="2">b>47/b>/font>/td>/tr>' PS: We don't know the number. The number 47 is just an example. It can be any number from 0 to 900. 回答1: >>> import re >>> text = 'tr bgcolor="aa77bb"td>font face="verdana"color="white" size="2">b>Total/b>/font>/td>\td>font face="verdana"color="white" size="2">b>47/b>/font>/td

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: We have a simple ul <ul> <li> some text </li> <li> some some text </li> <li> text more </li> <li> text here </li> </ul> ul { text-align: center; width: 100px; } li { width: auto; display: inline; } So the ul can have several lines. How do I take last li of the each line inside ul ? 回答1: If by last li on the each line of the ul you mean the last li in each ul , then: $ ( 'ul li:last-child' ); However, if you mean that you have your li 's within the same ul written up on several lines in your source code, and you now want to get the

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am using Spark 2.0 with the Python API. I have a dataframe with a column of type DateType(). I would like to add a column to the dataframe containing the most recent Monday. I can do it like this: reg_schema = pyspark.sql.types.StructType([ pyspark.sql.types.StructField('AccountCreationDate', pyspark.sql.types.DateType(), True), pyspark.sql.types.StructField('UserId', pyspark.sql.types.LongType(), True) ]) reg = spark.read.schema(reg_schema).option('header', True).csv(path_to_file) reg = reg.withColumn('monday', pyspark.sql.functions.when

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Debian Linux (Debian Stable, UBUNTU LTS, etc.) and others use apt-get as good and reliable installer tool. I need to use only apt-get ... The problem is that I do the simple sudo apt-get install nodejs but as result I have nodejs --version (or node --version ) 0.10.41 (!) It is not the today's most updated stable Node.JS version , 5.2.0 I also try node -p process.versions.v8 that results in 3.14.5.9 so, also is very less than 5.2 . NOTES I am using UBUNTU 14 LTS.... And try a workaround but not works, sudo apt-get remove --purge nodejs sudo