Whether financial, political, or social -- data's true power lies in its ability to answer questions definitively. In this project, we try to answer a fundamental question: "What's the weather like as we approach the equator?"

Now, we know what you may be thinking: "Duh. It gets hotter..."

But, if pressed, how would you prove it?

Please refer WeatherPy.ipynb and VacationPy.ipynb for the detailed implementation.

• Latitude values are measured relative to the equator and range from -90° at the South Pole to 90° at the North Pole. Longitude values are measured relative to the prime meridian. They range from -180° when traveling west to 180° when traveling east.Please checkout geographic coordinate system for further details.

• Generate a set of representation latitude and longitude values

  # Range of latitudes and longitudes
  lat_range = (-90, 90)
  lng_range = (-180, 180)
  
  #Create a seed
  np.random.seed(1000)
  
  # Create a set of random lat and lng combinations
  lats = np.random.uniform(lat_range[0], lat_range[1], size=1600)
  lngs = np.random.uniform(lng_range[0], lng_range[1], size=1600)

• Find the closest city for each of the representational latitude and longitude values using python citipy library

  # Incorporate citipy to determine city based on latitude and longitude
  from citipy import citipy
  cities = []
  lat_lngs = zip(lats, lngs)
  
  # Identify nearest city for each lat, lng combination
  for lat_lng in lat_lngs:
      city = citipy.nearest_city(lat_lng[0], lat_lng[1]).city_name
      # If the city is unique, then add it to a our cities list
      if city not in cities:
          cities.append(city)

  Note:- Some latitude, longitude combination will not have nearest city (eg:- in the ocean). Hence, a larger set of lat,long was kept initially to get more than 500 cities

• Next, we perform weather check on each city in the list, using a series of successive API calls to OpenWeatherMap API and extract ['City','Lat', 'Lng', 'Max Temp', 'Humidity', 'Cloudiness', 'Wind Speed', 'Country', 'Date']. This extracted data is kept in a DataFrame.

   #Create a placeholder DF for the extracted data from API calls
   weather_DF = pd.DataFrame(columns=['City','Lat', 'Lng', 'Max Temp', 'Humidity', 'Cloudiness', 'Wind Speed', 'Country', 'Date']) 
  
   #Data to get extracted
   summary = ['name', 'coord.lat', 'coord.lon', 'main.temp_max', 'main.humidity', 'clouds.all', 'wind.speed', 'sys.country', 'dt']             
  
   #Parms to pass to the API call
   params = {'units': 'imperial',
             'appid' : weather_api_key}
  
   #Iteratively call openweathermap api using python wrapper
   print("Beginning Data Retrieval\n\
   -----------------------------")
   count=0 #Successful queries
   for index, city in enumerate(cities):
       try:
           result = owm.get_current(city,**params)
           weather_DF.loc[count] = result(*summary)
           print(f"Processed Record {index} | {city}")
           count =1
       except:
           print(f"Record {index}: City {city} not found. Skipping...") 
       time.sleep(1) #1 sec delay between API calls
   print("-----------------------------\n\
   Data Retrieval Complete\n\
   -----------------------------")         

• Create a series of scatter plots to showcase the following relationships:

  • Temperature (F) vs. Latitude

    

  • Humidity (%) vs. Latitude

    

  • Cloudiness (%) vs. Latitude

    

  • Wind Speed (mph) vs. Latitude

    

• Write a function that creates the linear regression plots

    def linregress_plots(DF, xl, yl, xlabel='Latitude', ylabel='', title='', figname='plot.png'):
  
    m, c, r, p, _ = linregress(DF[xl], DF[yl])
    print(f"The r-squared is: {r**2}")
    
  
    #Create a new figure
    _=plt.figure()
  
    #Scatter plot
    ax = DF.plot(x=xl, 
              y=yl,
              kind='scatter',
              s=30,
              title=title,
              ylim = (min(DF[yl])-5, max(DF[yl] 15))
              )            
  
    _=ax.set_xlabel(xlabel)
    _=ax.set_ylabel(ylabel)
  
    #Regression Line
    y=m*DF[xl]   c
    _=ax.plot(DF[xl], y, 'r-')
    
    
    pos=((0.15, 0.2) if m<=-0.4 else ((0.15, 0.75) if m>0.4 else (0.5, 0.80))) #Annotate position
    
    #A way to dynamically finds the number of decimal positions if there is avery small value Eg:- 0.000000067
    #We don't want to denote it as 0.00
    val = m*100
    digits = 2
    while int(val)==0:
        val*=10
        digits =1
    
    s = "{:." f"{digits}" "f}"
    format_string = "y = " s "x   {:.2f}"
    linear_eqn = format_string.format(m, c)
    _=ax.annotate(linear_eqn,
            xy=pos, xycoords='figure fraction', fontsize=15, color='r')
  
    plt.savefig(f"../Images/{figname}")
    _=plt.show()
    
    return(r, p)
  
    #This function returns the r value, and p value
    #r value: Pearson Correlation Coefficient
    #p value: is a measure of the significance of the gradient. If p value is < 0.01 (Significance level),
    #it means that, we cannot independent variable affects dependant variable

• Run linear regression on each relationship, only this time separating them into Northern Hemisphere (greater than or equal to 0 degrees latitude) and Southern Hemisphere (less than 0 degrees latitude):

  • Northern Hemisphere - Temperature (F) vs. Latitude

    

  • Southern Hemisphere - Temperature (F) vs. Latitude

    

      Temperature depends on the distance from equator. 
      * Please observe the p value of the linear regression estimator << 0. This means that slope is NOT zero
      * In both hemispheres, a high correlation between latitude and temperature
      * We can observe a pattern in scatter plot also
      As we move towards equator, temperature increases in both sides of the hemisphere
      From the data, it looks like, temperatures at cities equidistant from equator in both the sides might not be same.
        * For instance, 
            . At latitude  30, temperature is approximated as -0.57*30 90.47=73.37F
            . At latitude -30, temperature is approximated as 0.65*-30 78.31 = 58.81F. 
        * This is because, most of the northern hemisphere is land and most of the southern hemisphere is ocean and ocean is likely to be colder
    

  • Northern Hemisphere - Humidity (%) vs. Latitude

    

  • Southern Hemisphere - Humidity (%) vs. Latitude

    

    - Humidity(%) doesn't correlate with the distance from equator. 
      * Please observe that p value of the linear regression estimator >> 0 (>significance level(typically 0.05)). This means that WE CANNOT say that slope is NOT zero.
      * In both hemispheres, a near to ZERO correlation between latitude and humidity.
      * No pattern in scatter plot.
    - Humidity is centered around different values in both hemispheres.
        * In northern hemisphere, most of the cities are having humidity around 67%.
        * In southern hemisphere, most of the cities are having humidity around 73%.
    

  • Northern Hemisphere - Cloudiness (%) vs. Latitude

    

  • Southern Hemisphere - Cloudiness (%) vs. Latitude

    

    - Cloudiness(%) doesn't correlate with the distance from equator. 
      * Please observe that p value of the linear regression estimator > significance level (typically 0.05). This means that WE CANNOT say that slope is NOT zero.
      * In both hemispheres, a weak correlation between latitude and cloudiness.
      * No pattern in scatter plot.
    - Cloudiness is centered around different values in both hemispheres.
        * Northern hemisphere has average cloudiness around 53%.
        * Southern hemisphere has average cloudiness around 46%.
    

  • Northern Hemisphere - Wind Speed (mph) vs. Latitude

    

  • Southern Hemisphere - Wind Speed (mph) vs. Latitude

    

    - Windspeed doesn't correlate with the distance from equator. 
      * Please observe that p value of the linear regression estimator > significance level (typically 0.05).
          This means that WE CANNOT say that slope is NOT zero.
      * In both hemispheres, a weak correlation between latitude and Windspeed.
      * No pattern in scatter plot.
    - Windspeed is centered around different but close values in both hemispheres.
        * Northern hemisphere has average windspeed around 8.6 mph.
        * Southern hemisphere has average windspeed around 7.9 mph.
    

• Create a heat map that displays the humidity for every city from the part I of the homework.

  

• Narrow down the DataFrame to find your ideal weather condition. For example:

  • A max temperature lower than 80 degrees but higher than 72.

  • Wind speed less than 10 mph.

  • Zero cloudiness.

  Drop any rows that don't contain all three conditions. You want to be sure the weather is ideal.

    DF_IDEAL = DF.drop(DF[~((DF['Max Temp']<80.0) & (DF['Max Temp']>70.0) & (DF['Wind Speed']<10.0) & (DF['Cloudiness']==0))].index)
  
    DF_IDEAL.info()
    
    <class 'pandas.core.frame.DataFrame'>
    Int64Index: 9 entries, 37 to 536
    Data columns (total 8 columns):
     #   Column      Non-Null Count  Dtype  
    ---  ------      --------------  -----  
     0   City        9 non-null      object 
     1   Country     9 non-null      object 
     2   Lat         9 non-null      float64
     3   Lng         9 non-null      float64
     4   Max Temp    9 non-null      float64
     5   Humidity    9 non-null      float64
     6   Cloudiness  9 non-null      float64
     7   Wind Speed  9 non-null      float64
    dtypes: float64(6), object(2)
    memory usage: 648.0  bytes

• Using Google Places API to find the first hotel for each city located within 5000 meters of your coordinates (The result is sorted based on popularity)

  hotel_df = DF_IDEAL.iloc[:,:4].copy()
  hotel_df['Hotel Name'] = ""
  
  base_url = 'https://maps.googleapis.com/maps/api/place/textsearch/json'
  
  for index, row in hotel_df.iterrows():
      params = {
              "location": f"{row['Lat']},{row['Lng']}",
              "query": 'hotel',
              "radius": 5000,
              "key": g_key
              }
      try:
          result = requests.get(base_url, params).json()
          hotel_df.loc[index, "Hotel Name"] = result['results'][0]['name']
  
      except:
          print(f"Couldn't retrive hotel for {row['City']} at index {index}..Skipping")     

• Plot the hotels on top of the humidity heatmap with each pin containing the Hotel Name, City, and Country.