division

问题 Normally the dtype is hidden when it's equivalent to the native type: >>> import numpy as np >>> np.arange(5) array([0, 1, 2, 3, 4]) >>> np.arange(5).dtype dtype('int32') >>> np.arange(5) + 3 array([3, 4, 5, 6, 7]) But somehow that doesn't apply to floor division or modulo: >>> np.arange(5) // 3 array([0, 0, 0, 1, 1], dtype=int32) >>> np.arange(5) % 3 array([0, 1, 2, 0, 1], dtype=int32) Why is there a difference? Python 3.5.4, NumPy 1.13.1, Windows 64bit 回答1: You actually have multiple

问题 Normally the dtype is hidden when it's equivalent to the native type: >>> import numpy as np >>> np.arange(5) array([0, 1, 2, 3, 4]) >>> np.arange(5).dtype dtype('int32') >>> np.arange(5) + 3 array([3, 4, 5, 6, 7]) But somehow that doesn't apply to floor division or modulo: >>> np.arange(5) // 3 array([0, 0, 0, 1, 1], dtype=int32) >>> np.arange(5) % 3 array([0, 1, 2, 0, 1], dtype=int32) Why is there a difference? Python 3.5.4, NumPy 1.13.1, Windows 64bit 回答1: You actually have multiple

问题 I have two dataframes : df1 and df2 df1 : TIMESTAMP eq1 eq2 eq3 2016-05-10 13:20:00 40 30 10 2016-05-10 13:40:00 40 10 20 df2 : TIMESTAMP eq1 eq2 eq3 2016-05-10 13:20:00 10 20 30 2016-05-10 13:40:00 10 20 20 I would like to divide df1 by df2 : each column of df1 by all column of df2 to get this result df3 : TIMESTAMP eq1 eq2 eq3 2016-05-10 13:20:00 40/(10+10) 30/(20+20) 10/(30+20) 2016-05-10 13:40:00 40/(10+10) 10/(20+20) 20/(30+20) Any idea please? 回答1: You can use div, but before set_index

问题 Suppose you have this array: In [29]: a = array([[10, 20, 30, 40, 50], [14, 28, 42, 56, 70], [18, 36, 54, 72, 90]]) Out[30]: a array([[ 0, 0, 0, 0, 0], [14, 28, 42, 56, 70], [18, 36, 54, 72, 90]]) Now divide the third row by the first one (using from future import division) In [32]: a[0]/a[2] Out[32]: array([ 0.55555556, 0.55555556, 0.55555556, 0.55555556, 0.55555556]) Now do the same with each row in a loop: In [33]: for i in range(3): print a[i]/a[2] [ 0.55555556 0.55555556 0.55555556 0

问题 I have problem with find a solution to the problem. Divisible/2 predicate examines whether a number N is divisible by one of numbers in the list divisible([H|_],N) :- N mod H =:= 0. divisible([H|T],N) :- N mod H =\= 0, divisible(T,N). I need to build a predicate find that will find Number < N that are not divisible by the list of numbers example input/output: ?- find(5, [3,5],Num). output is : Num = 4; Num = 2; Num = 1. False Here N is 5 and list of number is [3,5] Current Code: findNum(1, LN

问题 I have problem with find a solution to the problem. Divisible/2 predicate examines whether a number N is divisible by one of numbers in the list divisible([H|_],N) :- N mod H =:= 0. divisible([H|T],N) :- N mod H =\= 0, divisible(T,N). I need to build a predicate find that will find Number < N that are not divisible by the list of numbers example input/output: ?- find(5, [3,5],Num). output is : Num = 4; Num = 2; Num = 1. False Here N is 5 and list of number is [3,5] Current Code: findNum(1, LN

问题 So I’ve been doing something with primes in python, and I’m currently using this def isDivisible(number,divisor): if number % divisor == 0: return True return False to check if a number is divisible by the divisor. So I was wondering if there was a faster way to do this? 回答1: What about: return (number % divisor == 0) 回答2: A speed test shows that checking not() is faster than a != 0 solution: %%timeit not(8 % 3) # 19 ns ± 0.925 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each) %

问题 There is integer variable, voltage in millivolts. signed int voltage_mv = 134; //134mV I have 2-segment display and I want to display hundredths of volts. How can I convert milivolts to hundredths volts in one operation? Without IF statement, without function? 134 => 13 135 => 14 回答1: How about simple rounding: int millivoltToDisplay (int millivolts) { return (millivolts+5)/10; } (written as a function for clarity) 回答2: For the same of completeness, if the denominator is odd, then instead of

问题 I'm trying to test if a float i.e( 54.10 ) is divisible by 0.10 . 54.10 % .10 returns .10 and not 0 , why is that and how can I get it to do what I want it to do? 回答1: You can use the decimal module to avoid the floating point precision problem: >>> import decimal >>> decimal.Decimal('54.10') % decimal.Decimal('0.10') Decimal('0.00') 回答2: The tried and true method here is to multiply your divisor and dividend by a power of 10. Effectively, 54.10 becomes 541 and 0.10 becomes 1. Then you can

问题 I am an experienced php developer just starting to learn Java. I am following some Lynda courses at the moment and I'm still really early stages. I'm writing sample programs that ask for user input and do simple calculation and stuff. Yesterday I came across this situation: double result = 1 / 2; With my caveman brain I would think result == 0.5 , but no, not in Java. Apparantly 1 / 2 == 0.0 . Yes, I know that if I change one of the operands to a double the result would also be a double. This