Как найти отклонение от средней температуры

Calculating average temperatures gives you a more accurate picture of the temperature in a specific location than a single measurement ever could. Temperatures fluctuate throughout the day, over the course of a week, month to month and year to year, as well as varying substantially depending on where exactly you are. Making sense of this and coming up with one figure for your purposes requires calculating the mean temperature, which is a specific type of average. You do this by adding up all of your individual measurements and dividing by the number of measurements.

Decide What You Want the Average Of

Plan your calculations based on what exactly you want to work out. Do you want the average temperature for the week in one location, for the day over multiple locations or something else? The calculations are essentially the same in most cases, but this determines what data you need to collect for the calculation.

Take Your Measurements or Obtain Your Data

Take your measurements or find the data you need from an online source. (For example, the National Centers for Environmental Information has data for the U.S. See Resources for a link.) If you’re looking for data to determine the average temperature for the week, you can collect data each day (preferably at the same time in the day in the same location), but if you’re looking for data covering a longer period or over a larger area, it will probably be easier to use existing data.

Convert the Data Into the Same Unit

Celsius, Kelvin and Fahrenheit all express temperature, but all of your data has to be in the same unit if you want to calculate the average. To convert from Celsius to Kelvin, simply add 273 to the temperature in Celsius:

text{temperature in Kelvin}=text{temperature in Celsius}+273

Use the following formulas to convert from Celsius to Fahrenheit or vice-versa:

text{temperature in Fahrenheit}=1.8(text{temperature in Celsius})+32

text{temperature in Celsius}=frac{(text{temperature in fahrenheit}-32)}{1.8}

Add Up Your Individual Measurements

Start calculating your average by taking the sum of your individual measurements, all in the same unit of temperature. Imagine you have the following data for measurements taken over a week, all in degrees Fahrenheit: 70, 68, 79, 78, 73, 69 and 72. Take the sum as follows:

text{sum} = 70 + 68 + 79 + 78 + 73 + 69 + 72=509

Divide by the Number of Measurements

Divide the total from the previous step by the number of measurement to find the average temperature. In the example, there were seven measurements taken, so you divide by 7 to find the average:

text{average temperature}=frac{text{sum of measured temperatures}}{text{number of measurements}}

The result from the previous step gives:

text{average temperature}=frac{509}{7}=72.7text{ degrees Fahrenheit}

Expand this approach as needed for other situations. For example, you can take an average of measurements taken at different times in the day or at different locations to find the average temperature for each day. Then you can find the mean of these results to come up with an average temperature for the whole week.

Я работаю над набором данных, который будет универсальным для нескольких климатических станций для анализа температуры и осадков. Я наткнулся на кирпичную стену, проектируя «климатические нормы», успешно рассчитал среднесуточную температуру.TAVG, среднемесячная темп. AVG_TAVG, и подвел итоги PRCP а также SNOW для ежемесячных итогов.

Там, где я зашла в тупик, подсчитываю отклонение от нормы, в настоящее время данные с 1981 по 2010 год считаются климатическими нормами.

Вот как сейчас выглядит мой набор данных:

mso_light
    
    year month day  date       PRCP SNOW  SNWD TMAX TMIN  TAVG
1   1948    1   1   1948-01-01    0    0   102   44 -122 -39.0
2   1948    1   2   1948-01-02    3    0    51   44    6  25.0
3   1948    1   3   1948-01-03    0    0    25   44  -39   2.5
4   1948    1   4   1948-01-04   38   64    76   33  -56 -11.5
5   1948    1   5   1948-01-05    0    0    76   -6  -83 -44.5
6   1948    1   6   1948-01-06  107    0    51   22  -61 -19.5
7   1948    1   7   1948-01-07  147    0    25   28  -17   5.5
8   1948    1   8   1948-01-08    8   13    25   39  -83 -22.0
9   1948    1   9   1948-01-09    0    0    25   -6 -117 -61.5
10  1948    1   10  1948-01-10    8   10    25  -11 -156 -83.5

Поэтому я изначально чувствовал, что мне нужен date для сортировки я удалю его, если он не понадобится в будущем.

Далее я хотел бы добавить столбец для DepNormT, который рассчитывается путем взятия 1 января — 31 декабря с 1981 по 2010 гг. и усреднения TAVGчтобы найти нормальную среднюю температуру. затемDepNormT будет разница между собой и TAVG для всего набора данных.

Я пробовал несколько способов добиться этого, вот две версии:

mso_DeptT <- mso_light %>%
  group_by(month, day) %>%
  mean(mso_light$TAVG[1981:2010], na.rm = T) %>%
  ungroup()

Это дает мне следующую ошибку:

  no applicable method for 'ungroup' applied to an object of class "c('double', 'numeric')"
In addition: Warning message:
In mean.default(., mso_light$TAVG[1981:2010], na.rm = T) :
  argument is not numeric or logical: returning NA

Это еще одна версия:

##mso_DeptT <- filter(mso_light, year >= "1981", year <= "2010") %>%
##  group_by(day, month) %>%
##  mutate(daily_DeptT = mean(TAVG, na.rm = T)) %>%
##  ungroup()

mso_sum <- mso_light %>%
  group_by(month, year) %>% 
  summarize(AVG_TAVG=mean(TAVG, na.rm = TRUE),
          T_PRCP=sum(PRCP, na.rm=TRUE),
          T_SNOW=sum(SNOW, na.rm=TRUE)) %>% 
  ungroup()

## To find monthly normal precipitation and snowfall - using dataset mso_sum

cli_Avg <- filter(mso_sum, year >= "1981", year <= "2010") %>%
  group_by(month) %>%
  summarize(Mon_Precip = mean(T_PRCP, na.rm = T),
            Mon_Snow = mean(T_SNOW, na.rm = T))

Это дало мне среднее значение за 30 лет, которое было равно среднему значению за каждый день. TAVG. Например:

    year month day  date       PRCP SNOW  SNWD TMAX TMIN  TAVG  DepNormT
1   1948    1   1   1948-01-01    0    0   102   44 -122 -39.0      -39.0
2   1948    1   2   1948-01-02    3    0    51   44    6  25.0       25.0
3   1948    1   3   1948-01-03    0    0    25   44  -39   2.5        2.5
4   1948    1   4   1948-01-04   38   64    76   33  -56 -11.5        ect
5   1948    1   5   1948-01-05    0    0    76   -6  -83 -44.5          .
6   1948    1   6   1948-01-06  107    0    51   22  -61 -19.5          .
7   1948    1   7   1948-01-07  147    0    25   28  -17   5.5          .
8   1948    1   8   1948-01-08    8   13    25   39  -83 -22.0
9   1948    1   9   1948-01-09    0    0    25   -6 -117 -61.5
10  1948    1   10  1948-01-10    8   10    25  -11 -156 -83.5

Спасибо за предложения.