divide

问题 I'm currently computing the binomial coefficient of two natural numbers by write a tail recursion in Scala. But my code has something wrong with the dividing numbers, integer division by k like I did as that will give you a non-zero remainder and hence introduce rounding errors. So could anyone help me figure it out, how to fix it ? def binom(n: Int, k: Int): Int = { require(0 <= k && k <= n) def binomtail(n: Int, k: Int, ac: Int): Int = { if (n == k || k == 0) ac else binomtail(n - 1, k - 1,

问题 I was reading relational algebra from one of the textbook. I came across DIVIDE operation. From Wikipedia: The division is a binary operation that is written as R ÷ S. The result consists of the restrictions of tuples in R to the attribute names unique to R, i.e., in the header of R but not in the header of S, for which it holds that all their combinations with tuples in S are present in R. Thus if R is: +----+----+ | A | B | +----+----+ | a1 | b1 | | a2 | b1 | | a3 | b1 | | a4 | b1 | | a1 |

问题 var result = parseFloat($('.lot2').text()) / parseFloat($('span.Price').text()); $('.result').text(result); }); How can I convert selected values from comma separated values to dot separated values? There is this function str.replace, but I don't know how to put it in the function. 回答1: How can I convert selected values from comma separated values to dot separated values? From that, I take it that the values you get from $('.lot2').text() and $('span.Price').text() will use , as the decimal

问题 I have a dataframe like this: import pandas as pd dic = {'A':[100,200,250,300], 'B':['ci','ci','po','pa'], 'C':['s','t','p','w']} df = pd.DataFrame(dic) My goal is to separate the row in 2 dataframes: df1 = contains all the rows that do not repeat values along column B (unque rows). df2 = containts only the rows who repeat themeselves. The result should look like this: df1 = A B C df2 = A B C 0 250 po p 0 100 ci s 1 300 pa w 1 250 ci t Note: the dataframes could be in general very big and

问题 Following up this question and dataframes, I am trying to convert this Into this (I know it looks the same, but refer to the next code line to see the difference): In pandas, I used the line code teste_2 = (value/value.groupby(level=0).sum()) and in pyspark I tried several solutions; the first one was: df_2 = (df/df.groupby(["age"]).sum()) However, I am getting the following error: TypeError: unsupported operand type(s) for /: 'DataFrame' and 'DataFrame' The second one was: df_2 = (df.filter

问题 Let's say I have $t0 , and I'd like to divide its integer contents by two, and store it in $t1 . My gut says: srl $t1, $t0, 2 ... but wouldn't that be a problem if... say... the right-most bit was 1? Or does it all come out in the wash because the right-most bit (if positive) makes $t0 an odd number, which becomes even when divided? Teach me, O wise ones... 回答1: Use instruction sra: Shift right arithmetic !! sra $t1, $t0, 1 Divides the content of $t0 by the first power of 2. Description:

问题 Let's say I have $t0 , and I'd like to divide its integer contents by two, and store it in $t1 . My gut says: srl $t1, $t0, 2 ... but wouldn't that be a problem if... say... the right-most bit was 1? Or does it all come out in the wash because the right-most bit (if positive) makes $t0 an odd number, which becomes even when divided? Teach me, O wise ones... 回答1: Use instruction sra: Shift right arithmetic !! sra $t1, $t0, 1 Divides the content of $t0 by the first power of 2. Description:

来源： https://stackoverflow.com/questions/44040355/how-to-select-round-up-number-in-sql-server

来源： https://stackoverflow.com/questions/5560772/apply-and-calculating-proportion-of-first-row-for-all-dataframe-rows

来源： https://stackoverflow.com/questions/5560772/apply-and-calculating-proportion-of-first-row-for-all-dataframe-rows