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How do I get Min and max values to only print when "year" is entered?


How do I get Min and max values to only print when "year" is entered?

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How do I get Min and max values to only print when "year" is entered?
Tag : java , By : kennystone
Date : November 29 2020, 04:01 AM

around this issue Creating program where user can enter a month and see average temp for that month. User can enter "year" and see all months and avg temps for each month, as well as the yearly average and highest and lowest monthly average. So far I am having difficulty with the max/low average. It is printing when the month is entered and it shouldn't. I have tried to move the code around to different points but it doesn't seem to matter as it still prints when year or a month is entered. , The Min/Max has no if clause around it, so it always runs.
if( userMonthName.equalsIgnoreCase("year") )
{
    // User entered "year" - print out all values
    for( i=0; i < NUM_MONTHS; i++ ) 
    {
         System.out.println( monthName[i] + " " + avgMonthTemp[i] );
    }

     Arrays.sort( avgMonthTemp );
     System.out.println( "Minimum = " + avgMonthTemp[0] );
     System.out.println( "Maximum = " + avgMonthTemp[avgMonthTemp.length-1] ); 
}
else
{
    // Did the user enter a month name?  If so print just that
    for( i=0; i < NUM_MONTHS; i++ ) 
    {
        if ( userMonthName.equalsIgnoreCase( monthName[i] ) ) 
        {
             System.out.println( monthName[i] + " " + avgMonthTemp[i] );
        }
    }
}

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Read a sequence of words from input. Use "Quit" to terminate, print in order they were entered and don't print


Tag : cpp , By : Hitesh Prajapati
Date : March 29 2020, 07:55 AM
this one helps. Fixed a little, I didn't wanted to spend time to understand your for(j=i,m=0;j>=0,m<=i;m++) cycle so fully rewrited it.
#include<iostream>
#include<conio.h>
#include<string>

using namespace std;

int main()
{
    string word[100];
    string arr[100];
    int wordCount = 0;

    while(1)
    {
        cout<<"enter word: \n";
        cin>>word[wordCount];
        if(word[wordCount] == "quit")
        {  
            break;
        }
        ++wordCount;
    } 

    int arrCount = 0;
    for(int i = 0; i < wordCount; ++i)
    {                  
        bool found = false;   
        for (int j = 0; j < arrCount; ++j)
        {
            if (arr[j] == word[i])
            {
                found = true;
                break;
            }
        }
        if (!found)
        {
            arr[arrCount] = word;
            ++arrCount;
        }
    }                                                 

    for(int i = 0; i < arrCount; ++i)
    {
        cout<<arr[n]<<"\n";
    }

    getch();
}

In the context of year values how does "aaaa" differ from "yyyy"?


Tag : java , By : Ron
Date : March 29 2020, 07:55 AM
Does that help If you look at java.text.DateFormatSymbols you'll find the following field:
/**
 * Localized date-time pattern characters. For example, a locale may
 * wish to use 'u' rather than 'y' to represent years in its date format
 * pattern strings.
 * This string must be exactly 18 characters long, with the index of
 * the characters described by <code>DateFormat.ERA_FIELD</code>,
 * <code>DateFormat.YEAR_FIELD</code>, etc.  Thus, if the string were
 * "Xz...", then localized patterns would use 'X' for era and 'z' for year.
 * @serial
 */
String  localPatternChars = null;

django-stripe error: "Your card's expiration year is invalid" when testing, no matter year entered


Tag : python , By : sgmichelsen
Date : March 29 2020, 07:55 AM
I wish this help you I'm using a django-strip with a view to create a charge, and if that is successful save some local items to a db. ,
exp_year = request.POST.get('exp_year ', False)

Converting "year" and "week of year" columns to "date" in Pandas


Tag : python , By : AnToni00
Date : March 29 2020, 07:55 AM
I wish this help you You need combine %w for day of week - explanation with %W for week:
http://strftime.org/ for %W:
df = pd.DataFrame({'year':[2015, 2018],
                   'weekofyear':[10,12]})

dates = df.year*100+df.weekofyear
@adde
df['date'] = pd.to_datetime(dates.astype(str) + '0', format='%Y%W%w')
print (df)

   year  weekofyear  formatted_date       date
0  2015  10          201510         2015-03-15
1  2018  12          201812         2018-03-25
#added 0 only for demontration, you can remove it
df['formatted_date'] = df.year * 1000 + df.weekofyear * 10 + 0
df['date'] = pd.to_datetime(df['formatted_date'], format='%Y%W%w')
print (df)

   year  weekofyear  formatted_date       date
0  2015  10          2015100        2015-03-15
1  2018  12          2018120        2018-03-25

compare data of two columns in same table and if values match print "right" if not print"wrong" usin


Tag : r , By : Hans-Inge
Date : March 29 2020, 07:55 AM
I wish this help you , Base R solution (also please in future provide sample data):
 df$answer_check <- ifelse(df$correct_option == df$my_chosen_answers,
                              "right",
                              "wrong")

df$answer_count <- ave(df$answer_check, df$answer_check, FUN = length)
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