minimum

given 5 numbers, what is the minimum number of comparisons needed to find the median?

馋奶兔 提交于 2019-12-06 04:34:11
how do you setup minimum number of comparisons in general? James McNellis To cite Donald Knuth (by way of Wikipedia, since I don't have my copy of TAOCP at the moment), the lower bound for the number of comparisons is six: http://en.wikipedia.org/wiki/Selection_algorithm (scroll down to the section entitled "Lower Bounds"). Your goal is, effectively, to find the k lowest values where k is half the size of the list, rounded up, (so, k = 3; n = 5) and then take the max of those. Plugging that into the formula there on the page, you get: (5 - 3) + 1 + 1 + 2 = 6 In this case, the actual minimum

How to find the minimum value in a numpy matrix?

会有一股神秘感。 提交于 2019-12-05 20:22:50
问题 Hey this is a quick and easy question... How would i find the minimum value of this matrix, excluding 0? As in, 8 arr = numpy.array([[ 0., 56., 20., 44.], [ 68., 0., 56., 8.], [ 32., 56., 0., 44.], [ 68., 20., 56., 0.]]) 回答1: As you're using numpy , you could use arr[arr>0].min() for the case you posted. but if your array could have negative values, then you should use arr[arr != 0].min() 来源: https://stackoverflow.com/questions/11764260/how-to-find-the-minimum-value-in-a-numpy-matrix

Better way to find a minimum value that fits a condition?

久未见 提交于 2019-12-05 20:11:17
I can't get past the feeling that I am missing something obvious. Is there a clearer or more idiomatic way to do what the following function does? closest.preceding <- function(candidates, limit) { # return the value in candidates that is closest to but less than the limit return(limit - min(limit-candidates[candidates < limit])) } Thanks for any insight. You could do: max(candidates[candidates<limit]) Which first filters out just those candidates that are (strictly) less than the limit, and then takes the max of those (which much be the closest). I'm sure there are other ways too. 来源: https:/

Swift 4 - Setting a minimum deployment target

主宰稳场 提交于 2019-12-05 01:37:13
The current deployment target is 11.0 which is fine. However, I would like to know how I would set a minimum of 8.0? You can set the deployment target in your project's target's general settings. You can read more about this in Apple documentation "Setting Deployment Target" . 来源: https://stackoverflow.com/questions/46465408/swift-4-setting-a-minimum-deployment-target

Set highcharts y-axis min value to 0, unless there is negative data

倾然丶 夕夏残阳落幕 提交于 2019-12-04 23:47:01
I'm having an issue with highcharts where I have a number of different charts being generated by JSON calls. For the majority of these charts I need the minimum y-axis value to be set at 0, however there are a couple of occasions where negative values need to be shown. How can I tell highcharts to have a minimum y-axis value of 0 only if there are no negative values in the data, is this even possible? Thanks The option that you're looking for is called softMin and was introduced in version 5.0.1 . The docs describe it as follows: A soft minimum for the axis. If the series data minimum is

Angular ng-repeat filter with greater than

狂风中的少年 提交于 2019-12-04 21:31:00
I'm using angular to create a search property form, and I've got a select box with amount of bedrooms, how can I create a greater then filter? (for example: when 2 bedrooms selected, display all properties with 2 bedrooms or more) The select box: <select class="form-control" name="bedrooms" ng-model="Bedrooms" ng-options="property.data.SubType as (property.data.Bedrooms + ' Bedrooms ('+ property.count +')') for property in properties | unique:'Bedrooms' | orderBy:'data.Bedrooms'"> <option value="">Minimum Bedrooms</option> </select> And the filter: <tr ng-repeat="property in filtered =

R getting the minimum value for each row in a matrix, and returning the row and column name

笑着哭i 提交于 2019-12-04 21:08:20
问题 I have a matrix like so: Only in reality it is hundreds or thousands of values. What I need to do is return the minimum value for each row, along with the row/col name. So for row 1 in the example, "BAC", the minimum is 0.92 for BAC/CSCO, so I need to return something like: BAC/CSCO 0.92 And then repeat this for each row in the matrix. Assistance is greatly appreciated. I think apply is the trick, but I can't quite get the right combination. 回答1: X <- matrix(runif(20), nrow=4) rownames(X) <-

Set minimum iPhone OS version for app?

烂漫一生 提交于 2019-12-04 20:58:58
问题 I'm about to publish an app on the app store, and I'm looking to set the minimum OS version as it appears in iTunes as "Requires iPhone OS 3.x or later". 2 questions: 1) Where do I set this in my Xcode project? 2) I'm aware of the UITableViewCell numberOfLines property that is present only in OS > 3.1. If I set my minimum as OS 3.0, will people who have 3.1 be able to see the number of lines properly as I coded? (Obviously people on 3.0 won't be able to) Thanks. 回答1: What you need to do is

I have need the N minimum (index) values in a numpy array

99封情书 提交于 2019-12-04 15:52:23
问题 Hi I have an array with X amount of values in it I would like to locate the indexs of the ten smallest values. In this link they calculated the maximum effectively, How to get indices of N maximum values in a numpy array? however I cant comment on links yet so I'm having to repost the question. I'm not sure which indices i need to change to achieve the minimum and not the maximum values. This is their code In [1]: import numpy as np In [2]: arr = np.array([1, 3, 2, 4, 5]) In [3]: arr.argsort(

Select the minimum value for each row join by another table

房东的猫 提交于 2019-12-04 10:13:51
问题 I have the following table: Table1 Table2 CardNo ID Record Date ID Name Dept 1 101 8.00 11/7/2013 101 Danny Green 2 101 13.00 11/7/2013 102 Tanya Red 3 101 15.00 11/7/2013 103 Susan Blue 4 102 11.00 11/7/2013 104 Gordon Blue 5 103 12.00 11/7/2013 6 104 12.00 11/7/2013 7 104 18.00 11/7/2013 8 101 1.00 12/7/2013 9 101 10.00 12/7/2013 10 102 0.00 12/7/2013 11 102 1.00 12/7/2013 12 104 3.00 12/7/2013 13 104 4.00 12/7/2013 i want the result to be like this: Name Dept Record Danny Green 8.00 Tanya