min

问题 I am in development of an android app.. actually i need to set a minimum and maximum value for an editext entry my minimum value is 18 and maximum is 65.I did the exact code of this package com.test; import android.text.InputFilter; import android.text.Spanned; public class InputFilterMinMax implements InputFilter { private int min, max; public InputFilterMinMax(int min, int max) { this.min = min; this.max = max; } public InputFilterMinMax(String min, String max) { this.min = Integer.parseInt

问题 For example [1,2,3,4,1,2] has min element 1, but it occurs for the last time at index 4. 回答1: a = [1,2,3,4,1,2] a.reverse() print len(a) - a.index(min(a)) - 1 Update after comment: The side effect can be removed by reversing again (but of course that is quite inefficient). a.reverse() 回答2: >>> values = [1,2,3,4,1,2] >>> -min((x, -i) for i, x in enumerate(values))[1] 4 No modification to the original list, works for arbitrary iterables, and only requires one pass. This creates an iterable of

问题 My data file looks like this: 3.6-band 6238 Over 0.5678 Over 0.6874 Over 0.7680 Over 0.7834 What I want to do is to pick out the smallest float and the word directly above it and print those two values. I have no idea what I'm doing. I've tried df=open('filepath') for line in df: df1=line.split() df2=min(df1) Which is my attempt at at least trying to isolate the smallest float. Problem is it's just giving me the last value. I think that's a problem with python not knowing to start over with

问题 Below is what I have so far: I don't know how to exclude 0 as a min number though. The assignment asks for 0 to be the exit number so I need to have the lowest number other than 0 appear in the min string. Any ideas? int min, max; Scanner s = new Scanner(System.in); System.out.print("Enter a Value: "); int val = s.nextInt(); min = max = val; while (val != 0) { System.out.print("Enter a Value: "); val = s.nextInt(); if (val < min) { min = val; } if (val > max) { max = val; } }; System.out

问题 Below is what I have so far: I don't know how to exclude 0 as a min number though. The assignment asks for 0 to be the exit number so I need to have the lowest number other than 0 appear in the min string. Any ideas? int min, max; Scanner s = new Scanner(System.in); System.out.print("Enter a Value: "); int val = s.nextInt(); min = max = val; while (val != 0) { System.out.print("Enter a Value: "); val = s.nextInt(); if (val < min) { min = val; } if (val > max) { max = val; } }; System.out

问题 In my Javascript program, I have a list of Person objects. For example [ "Michael": { "age": 45, "position": "manager", ... }, "Dwight": { "age": 36, "position": "assistant manager", ... }, .... ] I want to find the youngest Person . I've accomplished this by creating two arrays: one of all the Person s and one of all their ages, and getting the index of the lowest age and applying it to the first array. Like: var arrayOfPersons = [persons[0], persons[1], ....]; var arrayOfAges = [persons[0]

问题 I have applied index on embedded document field which is of type date. Due to unpredictable behavior of MongoDB query with $lt & $gt operators I am trying to use min() max() functions of Cursor which force both lower & upper bounds on the index. But when I used it, it gave me error: planner returned error: unable to find relevant index for max/min query" Query looks like: db.user.find().min({'record.date':ISODate("2014-12-01")}).max({'record.date':ISODate("2014-12-01")}).explain() I have

问题 I have applied index on embedded document field which is of type date. Due to unpredictable behavior of MongoDB query with $lt & $gt operators I am trying to use min() max() functions of Cursor which force both lower & upper bounds on the index. But when I used it, it gave me error: planner returned error: unable to find relevant index for max/min query" Query looks like: db.user.find().min({'record.date':ISODate("2014-12-01")}).max({'record.date':ISODate("2014-12-01")}).explain() I have

问题 I run through the following sequence of statements: >>> a = range(10) >>> min(a, key=lambda x: x < 5.3) 6 >>> max(a, key=lambda x: x < 5.3) 0 The min and max give the exact opposite of what I was expecting. The python documentation on min and max is pretty sketchy. Can anyone explain to me how the "key" works? 回答1: Explanation of the key argument Key works like this: a_list = ['apple', 'banana', 'canary', 'doll', 'elephant'] min(a_list, key=len) returns 'doll' , and max(a_list, key=len)

问题 Given this pandas dataframe with two columns, 'Values' and 'Intervals'. How do I get a third column 'MinMax' indicating whether the value is a maximum or a minimum within that interval? The challenge for me is that the interval length and the distance between intervals are not fixed, therefore I post the question. import pandas as pd import numpy as np data = pd.DataFrame([ [1879.289,np.nan],[1879.281,np.nan],[1879.292,1],[1879.295,1],[1879.481,1],[1879.294,1],[1879.268,1], [1879.293,1],[1879