counter

问题 I am chaining an undetermined amount of map reduce jobs together for a parallel BFS shortest path algorithm and when the path cannot be determined, my jobs loop infinitely without producing any records. I figured the best way to check this is to get the Map Output Bytes counter that is maintained by hadoop. How can I get access to this counter? 回答1: To get the map output bytes counter produced by the job, use long outputBytes = job.getCounters().findCounter("org.apache.hadoop.mapred.Task

问题 I have modeled 4 bit Ring Counter using D Flip Flop. The D flip flop is in separate file, included in my workspace. The D flip flop works correctly (gives correct output waveform). This is the code of ring counter: library ieee; use ieee.std_logic_1164.all; use ieee.std_logic_unsigned.all; entity ring4counter is port ( clk: std_logic; output: out std_logic_vector(3 downto 0)); end ring4counter; architecture ring4counter_arch of ring4counter is component dff port ( clk: std_logic; d: in std

问题 I'm trying to create a program which reads in a text file and find the number of individual words. I have worked out most of it but I am stuck on trying to get the counter to pick out words not letters as it is currently doing. import collections with open ("file.txt" ,"r") as myfile: data=myfile.read() [i.split(" ") for i in data] x=collections.Counter(data) print (x) My aim was to slip the list by whitespace which would result in each word being a object in the list. This however did not

问题 Problem Statement Given the below data set which has two columns Column1 & Column 2 , add another two more column called Counter and Counting time . The conditions to Initialize the counter and Counter time is as follows: The counter should be incremented only when value in Column1 > 1 and Column2 = 0 The Counter must start to increment after 2 values from the condition satisfied row The Counting time must contain the values of number of time the sequence has occurred (Sequence of data points

问题 I have a simple HTML Form that allows the user to enter some text which is then sent as an SMS/TXT Message via an SMS Gateway. The user enters the message text into a textarea: <textarea rows="10" cols="40" id="smsbody" validate = "required:true" name="MessageBody"></textarea> They can enter as little or as much text as they like, however as each SMS is limited to 160 characters I would like to display a character counter that shows both the number of characters entered and then also

问题 I have a CSV file in the following format key1,key2,key3,counter1,counter2,counter3,counter4 1,2,1,0,0,0,1 1,2,2,0,1,0,4 The CQL3 table has all value columns of type 'counter'. When I try to use the COPY command to load the CSV I get the usual error which asks for an UPDATE instead of an INSERT. The question is : how can I tell CQL to use an UPDATE ? Is there any other way to do this ? 回答1: using sstables solved this issue. Although a little slower than what i expected , it does the job 回答2:

问题 I need to make a routine that counts bits in a word that does not involve loops (only bit operations), and does not use large constants. int x = 0xFFFFFFFF; x += (~((x >> 1) & 0x55555555)+1); x = (((x >> 2) & 0x33333333) + (x & 0x33333333)); x = (((x >> 4) + x) & 0x0F0F0F0F); x += (x >> 8); x += (x >> 16); return(x & 0x0000003F); This I found on bit twiddling hacks, but the largest constant I can use is 0xFF... Not sure how to do this otherwise. Thanks folks. 回答1: You can for example use a

问题 I am using the Counter class in Python to count the number of words in a key of a large python dictionary. I am creating a new dictionary with the keys being incremented numbers and the values being the return of the counter function. This is what my new dictionary looks like: TP = {1:Counter({u'x':1, u'b':1, u'H':3}),2:Counter({u'j':1, u'm':4, u'e':2})...} What I want to do, if possible, is remove the Counter and the parentheses ( and ) so that my dictionary looks like: TP = {1:{u'x':1, u'b'

问题 I'm not so familiar with computing (software) theory, and I thought about this question - does the PC (program counter) always have to change (I guess, upon each new clock tick)? I searched a bit online, and found Commodore 64 Programmers Reference Manual ( heh :) ) that confirms it: " ...Commodore 64 (or, for that matter, any computer), the program counter is always changing " (as well as Chapter 6: Hard, soft or firm?); but I just wanted to have it commented here. I was thinking, if an

问题 This is my code so far for counting vowels. I need to to scan through a sentence, count and compare the vowels and then display the top occurring vowels. from collections import Counter vowelCounter = Counter() sentence=input("sentence") for word in sentence: vowelCounter[word] += 1 vowel, vowelCount= Counter(vowel for vowel in sentence.lower() if vowel in "aeiou").most_common(1)[0] Does anyone have a better way to do this ? 回答1: IMO, long lines are best avoided for the sake of clarity: #!