DataFrame
The underlying idea of a DataFrame is based on spreadsheets. We can see the data structure of a DataFrame as tabular and spreadsheet-like. A DataFrame logically corresponds to a "sheet" of an Excel document. A DataFrame has both a row and a column index.
Like a spreadsheet or Excel sheet, a DataFrame object contains an ordered collection of columns. Each column consists of a unique data typye, but different columns can have different types, e.g. the first column may consist of integers, while the second one consists of boolean values and so on.
There is a close connection between the DataFrames and the Series of Pandas. A DataFrame can be seen as a concatenation of Series, each Series having the same index, i.e. the index of the DataFrame.
We will demonstrate this in the following example.
We define the following three Series:
import pandas as pd years = range(2014, 2018) shop1 = pd.Series([2409.14, 2941.01, 3496.83, 3119.55], index=years) shop2 = pd.Series([1203.45, 3441.62, 3007.83, 3619.53], index=years) shop3 = pd.Series([3412.12, 3491.16, 3457.19, 1963.10], index=years)
What happens, if we concatenate these "shop" Series? Pandas provides a concat function for this purpose:
pd.concat([shop1, shop2, shop3])The previous Python code returned the following result:
2014 2409.14 2015 2941.01 2016 3496.83 2017 3119.55 2014 1203.45 2015 3441.62 2016 3007.83 2017 3619.53 2014 3412.12 2015 3491.16 2016 3457.19 2017 1963.10 dtype: float64
This result is not what we have intended or expected. The reason is that concat used 0 as the default for the axis parameter. Let's do it with "axis=1":
shops_df = pd.concat([shop1, shop2, shop3], axis=1) shops_dfThis gets us the following:
| 0 | 1 | 2 | |
|---|---|---|---|
| 2014 | 2409.14 | 1203.45 | 3412.12 |
| 2015 | 2941.01 | 3441.62 | 3491.16 |
| 2016 | 3496.83 | 3007.83 | 3457.19 |
| 2017 | 3119.55 | 3619.53 | 1963.10 |
Let's do some fine sanding by giving names to the columns:
cities = ["Zürich", "Winterthur", "Freiburg"] shops_df.columns = cities print(shops_df) # alternative way: give names to series: shop1.name = "Zürich" shop2.name = "Winterthur" shop3.name = "Freiburg" print("------") shops_df2 = pd.concat([shop1, shop2, shop3], axis=1) print(shops_df2)
Zürich Winterthur Freiburg
2014 2409.14 1203.45 3412.12
2015 2941.01 3441.62 3491.16
2016 3496.83 3007.83 3457.19
2017 3119.55 3619.53 1963.10
------
Zürich Winterthur Freiburg
2014 2409.14 1203.45 3412.12
2015 2941.01 3441.62 3491.16
2016 3496.83 3007.83 3457.19
2017 3119.55 3619.53 1963.10
This was nice, but what kind of data type is our result?
print(type(shops_df))
<class 'pandas.core.frame.DataFrame'>
This means, we can arrange or concat Series into DataFrames!
DataFrames from Dictionaries
A DataFrame has a row and column index; it's like a dict of Series with a common index.
cities = {"name": ["London", "Berlin", "Madrid", "Rome", "Paris", "Vienna", "Bucharest", "Hamburg", "Budapest", "Warsaw", "Barcelona", "Munich", "Milan"], "population": [8615246, 3562166, 3165235, 2874038, 2273305, 1805681, 1803425, 1760433, 1754000, 1740119, 1602386, 1493900, 1350680], "country": ["England", "Germany", "Spain", "Italy", "France", "Austria", "Romania", "Germany", "Hungary", "Poland", "Spain", "Germany", "Italy"]} city_frame = pd.DataFrame(cities) city_frameThe above Python code returned the following result:
| country | name | population | |
|---|---|---|---|
| 0 | England | London | 8615246 |
| 1 | Germany | Berlin | 3562166 |
| 2 | Spain | Madrid | 3165235 |
| 3 | Italy | Rome | 2874038 |
| 4 | France | Paris | 2273305 |
| 5 | Austria | Vienna | 1805681 |
| 6 | Romania | Bucharest | 1803425 |
| 7 | Germany | Hamburg | 1760433 |
| 8 | Hungary | Budapest | 1754000 |
| 9 | Poland | Warsaw | 1740119 |
| 10 | Spain | Barcelona | 1602386 |
| 11 | Germany | Munich | 1493900 |
| 12 | Italy | Milan | 1350680 |
Retrieving the Column Names
It's possible to get the names of the columns as a list:
city_frame.columns.valuesWe received the following result:
array(['country', 'name', 'population'], dtype=object)
Custom Index
We can see that an index (0,1,2, ...) has been automatically assigned to the DataFrame. We can also assign a custom index to the DataFrame object:
ordinals = ["first", "second", "third", "fourth", "fifth", "sixth", "seventh", "eigth", "ninth", "tenth", "eleventh", "twelvth", "thirteenth"] city_frame = pd.DataFrame(cities, index=ordinals) city_frameAfter having executed the Python code above we received the following:
| country | name | population | |
|---|---|---|---|
| first | England | London | 8615246 |
| second | Germany | Berlin | 3562166 |
| third | Spain | Madrid | 3165235 |
| fourth | Italy | Rome | 2874038 |
| fifth | France | Paris | 2273305 |
| sixth | Austria | Vienna | 1805681 |
| seventh | Romania | Bucharest | 1803425 |
| eigth | Germany | Hamburg | 1760433 |
| ninth | Hungary | Budapest | 1754000 |
| tenth | Poland | Warsaw | 1740119 |
| eleventh | Spain | Barcelona | 1602386 |
| twelvth | Germany | Munich | 1493900 |
| thirteenth | Italy | Milan | 1350680 |
Rearranging the Order of Columns
We can also define and rearrange the order of the columns at the time of creation of the DataFrame. This makes also sure that we will have a defined ordering of our columns, if we create the DataFrame from a dictionary. Dictionaries are not ordered, as you have seen in our chapter on Dictionaries in our Python tutorial, so we cannot know in advance what the ordering of our columns will be:
city_frame = pd.DataFrame(cities, columns=["name", "country", "population"]) city_frameThe previous code returned the following result:
| name | country | population | |
|---|---|---|---|
| 0 | London | England | 8615246 |
| 1 | Berlin | Germany | 3562166 |
| 2 | Madrid | Spain | 3165235 |
| 3 | Rome | Italy | 2874038 |
| 4 | Paris | France | 2273305 |
| 5 | Vienna | Austria | 1805681 |
| 6 | Bucharest | Romania | 1803425 |
| 7 | Hamburg | Germany | 1760433 |
| 8 | Budapest | Hungary | 1754000 |
| 9 | Warsaw | Poland | 1740119 |
| 10 | Barcelona | Spain | 1602386 |
| 11 | Munich | Germany | 1493900 |
| 12 | Milan | Italy | 1350680 |
We change both the column order and the ordering of the index with the function reindex with the following code:
city_frame.reindex(index=[0, 2, 4, 6, 8, 10, 12, 1, 3, 5, 7, 9, 11], columns=['country', 'name', 'population'])The above Python code returned the following result:
| country | name | population | |
|---|---|---|---|
| 0 | England | London | 8615246 |
| 2 | Spain | Madrid | 3165235 |
| 4 | France | Paris | 2273305 |
| 6 | Romania | Bucharest | 1803425 |
| 8 | Hungary | Budapest | 1754000 |
| 10 | Spain | Barcelona | 1602386 |
| 12 | Italy | Milan | 1350680 |
| 1 | Germany | Berlin | 3562166 |
| 3 | Italy | Rome | 2874038 |
| 5 | Austria | Vienna | 1805681 |
| 7 | Germany | Hamburg | 1760433 |
| 9 | Poland | Warsaw | 1740119 |
| 11 | Germany | Munich | 1493900 |
Now, we want to rename our columns. For this purpose, we will use the DataFrame method 'rename'. This method supports two calling conventions
- (index=index_mapper, columns=columns_mapper, ...)
- (mapper, axis={'index', 'columns'}, ...)
We will rename the columns of our DataFrame into Romanian names in the following example. We set the parameter inplace to True so that our DataFrame will be changed instead of returning a new DataFrame, if inplace is set to False, which is the default!
city_frame.rename(columns={"name":"Nume", "country":"țară", "population":"populație"}, inplace=True) city_frameThe above code returned the following:
| Nume | țară | populație | |
|---|---|---|---|
| 0 | London | England | 8615246 |
| 1 | Berlin | Germany | 3562166 |
| 2 | Madrid | Spain | 3165235 |
| 3 | Rome | Italy | 2874038 |
| 4 | Paris | France | 2273305 |
| 5 | Vienna | Austria | 1805681 |
| 6 | Bucharest | Romania | 1803425 |
| 7 | Hamburg | Germany | 1760433 |
| 8 | Budapest | Hungary | 1754000 |
| 9 | Warsaw | Poland | 1740119 |
| 10 | Barcelona | Spain | 1602386 |
| 11 | Munich | Germany | 1493900 |
| 12 | Milan | Italy | 1350680 |
Existing Column as the Index of a DataFrame
We want to create a more useful index in the following example. We will use the country name as the index, i.e. the list value associated to the key "country" of our cities dictionary:
city_frame = pd.DataFrame(cities, columns=["name", "population"], index=cities["country"]) city_frameThe previous code returned the following:
| name | population | |
|---|---|---|
| England | London | 8615246 |
| Germany | Berlin | 3562166 |
| Spain | Madrid | 3165235 |
| Italy | Rome | 2874038 |
| France | Paris | 2273305 |
| Austria | Vienna | 1805681 |
| Romania | Bucharest | 1803425 |
| Germany | Hamburg | 1760433 |
| Hungary | Budapest | 1754000 |
| Poland | Warsaw | 1740119 |
| Spain | Barcelona | 1602386 |
| Germany | Munich | 1493900 |
| Italy | Milan | 1350680 |
Alternatively, we can change an existing DataFrame. We can us the method set_index to turn a column into an index. "set_index" does not work in-place, it returns a new data frame with the chosen column as the index:
city_frame = pd.DataFrame(cities) city_frame2 = city_frame.set_index("country") print(city_frame2)
name population country England London 8615246 Germany Berlin 3562166 Spain Madrid 3165235 Italy Rome 2874038 France Paris 2273305 Austria Vienna 1805681 Romania Bucharest 1803425 Germany Hamburg 1760433 Hungary Budapest 1754000 Poland Warsaw 1740119 Spain Barcelona 1602386 Germany Munich 1493900 Italy Milan 1350680
We saw in the previous example that the set_index method returns a new DataFrame object and doesn't change the original DataFrame. If we set the optional parameter "inplace" to True, the DataFrame will be changed in place, i.e. no new object will be created:
city_frame = pd.DataFrame(cities) city_frame.set_index("country", inplace=True) print(city_frame)
name population country England London 8615246 Germany Berlin 3562166 Spain Madrid 3165235 Italy Rome 2874038 France Paris 2273305 Austria Vienna 1805681 Romania Bucharest 1803425 Germany Hamburg 1760433 Hungary Budapest 1754000 Poland Warsaw 1740119 Spain Barcelona 1602386 Germany Munich 1493900 Italy Milan 1350680
Label-Indexing on the Rows
So far we have indexed DataFrames via the columns. We will demonstrate now, how we can access rows from DataFrames via the locators 'loc' and 'iloc'. ('ix' is deprecated and will be removed in the future)
city_frame = pd.DataFrame(cities, columns=("name", "population"), index=cities["country"]) print(city_frame.loc["Germany"])
name population Germany Berlin 3562166 Germany Hamburg 1760433 Germany Munich 1493900
print(city_frame.loc[["Germany", "France"]])
name population Germany Berlin 3562166 Germany Hamburg 1760433 Germany Munich 1493900 France Paris 2273305
print(city_frame.loc[city_frame.population>2000000])
name population England London 8615246 Germany Berlin 3562166 Spain Madrid 3165235 Italy Rome 2874038 France Paris 2273305
Sum and Cumulative Sum
We can calculate the sum of all the columns of a DataFrame or the sum of certain columns:
print(city_frame.sum())
name LondonBerlinMadridRomeParisViennaBucharestHamb... population 33800614 dtype: object
city_frame["population"].sum()After having executed the Python code above we received the following:
33800614
We can use "cumsum" to calculate the cumulative sum:
x = city_frame["population"].cumsum() print(x)
England 8615246 Germany 12177412 Spain 15342647 Italy 18216685 France 20489990 Austria 22295671 Romania 24099096 Germany 25859529 Hungary 27613529 Poland 29353648 Spain 30956034 Germany 32449934 Italy 33800614 Name: population, dtype: int64
Assigning New Values to Columns
x is a Pandas Series. We can reassign the previously calculated cumulative sums to the population column:
city_frame["population"] = x print(city_frame)
name population England London 8615246 Germany Berlin 12177412 Spain Madrid 15342647 Italy Rome 18216685 France Paris 20489990 Austria Vienna 22295671 Romania Bucharest 24099096 Germany Hamburg 25859529 Hungary Budapest 27613529 Poland Warsaw 29353648 Spain Barcelona 30956034 Germany Munich 32449934 Italy Milan 33800614
Instead of replacing the values of the population column with the cumulative sum, we want to add the cumulative population sum as a new culumn with the name "cum_population".
city_frame = pd.DataFrame(cities, columns=["country", "population", "cum_population"], index=cities["name"]) city_frameThe previous Python code returned the following output:
| country | population | cum_population | |
|---|---|---|---|
| London | England | 8615246 | NaN |
| Berlin | Germany | 3562166 | NaN |
| Madrid | Spain | 3165235 | NaN |
| Rome | Italy | 2874038 | NaN |
| Paris | France | 2273305 | NaN |
| Vienna | Austria | 1805681 | NaN |
| Bucharest | Romania | 1803425 | NaN |
| Hamburg | Germany | 1760433 | NaN |
| Budapest | Hungary | 1754000 | NaN |
| Warsaw | Poland | 1740119 | NaN |
| Barcelona | Spain | 1602386 | NaN |
| Munich | Germany | 1493900 | NaN |
| Milan | Italy | 1350680 | NaN |
We can see that the column "cum_population" is set to Nan, as we haven't provided any data for it.
We will assign now the cumulative sums to this column:
city_frame["cum_population"] = city_frame["population"].cumsum() city_frameAfter having executed the Python code above we received the following result:
| country | population | cum_population | |
|---|---|---|---|
| London | England | 8615246 | 8615246 |
| Berlin | Germany | 3562166 | 12177412 |
| Madrid | Spain | 3165235 | 15342647 |
| Rome | Italy | 2874038 | 18216685 |
| Paris | France | 2273305 | 20489990 |
| Vienna | Austria | 1805681 | 22295671 |
| Bucharest | Romania | 1803425 | 24099096 |
| Hamburg | Germany | 1760433 | 25859529 |
| Budapest | Hungary | 1754000 | 27613529 |
| Warsaw | Poland | 1740119 | 29353648 |
| Barcelona | Spain | 1602386 | 30956034 |
| Munich | Germany | 1493900 | 32449934 |
| Milan | Italy | 1350680 | 33800614 |
We can also include a column name which is not contained in the dictionary, when we create the DataFrame from the dictionary. In this case, all the values of this column will be set to NaN:
city_frame = pd.DataFrame(cities, columns=["country", "area", "population"], index=cities["name"]) print(city_frame)
country area population London England NaN 8615246 Berlin Germany NaN 3562166 Madrid Spain NaN 3165235 Rome Italy NaN 2874038 Paris France NaN 2273305 Vienna Austria NaN 1805681 Bucharest Romania NaN 1803425 Hamburg Germany NaN 1760433 Budapest Hungary NaN 1754000 Warsaw Poland NaN 1740119 Barcelona Spain NaN 1602386 Munich Germany NaN 1493900 Milan Italy NaN 1350680
Accessing the Columns of a DataFrame
There are two ways to access a column of a DataFrame. The result is in both cases a Series:
# in a dictionary-like way: print(city_frame["population"])
London 8615246 Berlin 3562166 Madrid 3165235 Rome 2874038 Paris 2273305 Vienna 1805681 Bucharest 1803425 Hamburg 1760433 Budapest 1754000 Warsaw 1740119 Barcelona 1602386 Munich 1493900 Milan 1350680 Name: population, dtype: int64
# as an attribute print(city_frame.population)
London 8615246 Berlin 3562166 Madrid 3165235 Rome 2874038 Paris 2273305 Vienna 1805681 Bucharest 1803425 Hamburg 1760433 Budapest 1754000 Warsaw 1740119 Barcelona 1602386 Munich 1493900 Milan 1350680 Name: population, dtype: int64
print(type(city_frame.population))
<class 'pandas.core.series.Series'>
city_frame.populationThe above Python code returned the following result:
London 8615246 Berlin 3562166 Madrid 3165235 Rome 2874038 Paris 2273305 Vienna 1805681 Bucharest 1803425 Hamburg 1760433 Budapest 1754000 Warsaw 1740119 Barcelona 1602386 Munich 1493900 Milan 1350680 Name: population, dtype: int64
From the previous example, we can see that we have not copied the population column. "p" is a view on the data of city_frame.
Assigning New Values to a Column
The column area is still not defined. We can set all elements of the column to the same value:
city_frame["area"] = 1572 print(city_frame)
country area population London England 1572 8615246 Berlin Germany 1572 3562166 Madrid Spain 1572 3165235 Rome Italy 1572 2874038 Paris France 1572 2273305 Vienna Austria 1572 1805681 Bucharest Romania 1572 1803425 Hamburg Germany 1572 1760433 Budapest Hungary 1572 1754000 Warsaw Poland 1572 1740119 Barcelona Spain 1572 1602386 Munich Germany 1572 1493900 Milan Italy 1572 1350680
In this case, it will be definitely better to assign the exact area to the cities. The list with the area values needs to have the same length as the number of rows in our DataFrame.
# area in square km: area = [1572, 891.85, 605.77, 1285, 105.4, 414.6, 228, 755, 525.2, 517, 101.9, 310.4, 181.8] # area could have been designed as a list, a Series, an array or a scalar city_frame["area"] = area print(city_frame)
country area population London England 1572.00 8615246 Berlin Germany 891.85 3562166 Madrid Spain 605.77 3165235 Rome Italy 1285.00 2874038 Paris France 105.40 2273305 Vienna Austria 414.60 1805681 Bucharest Romania 228.00 1803425 Hamburg Germany 755.00 1760433 Budapest Hungary 525.20 1754000 Warsaw Poland 517.00 1740119 Barcelona Spain 101.90 1602386 Munich Germany 310.40 1493900 Milan Italy 181.80 1350680
Sorting DataFrames
Let's sort our DataFrame according to the city area:
city_frame = city_frame.sort_values(by="area", ascending=False) print(city_frame)
country area population London England 1572.00 8615246 Rome Italy 1285.00 2874038 Berlin Germany 891.85 3562166 Hamburg Germany 755.00 1760433 Madrid Spain 605.77 3165235 Budapest Hungary 525.20 1754000 Warsaw Poland 517.00 1740119 Vienna Austria 414.60 1805681 Munich Germany 310.40 1493900 Bucharest Romania 228.00 1803425 Milan Italy 181.80 1350680 Paris France 105.40 2273305 Barcelona Spain 101.90 1602386
Let's assume, we have only the areas of London, Hamburg and Milan. The areas are in a series with the correct indices. We can assign this series as well:
city_frame = pd.DataFrame(cities, columns=["country", "area", "population"], index=cities["name"]) some_areas = pd.Series([1572, 755, 181.8], index=['London', 'Hamburg', 'Milan']) city_frame['area'] = some_areas print(city_frame)
country area population London England 1572.0 8615246 Berlin Germany NaN 3562166 Madrid Spain NaN 3165235 Rome Italy NaN 2874038 Paris France NaN 2273305 Vienna Austria NaN 1805681 Bucharest Romania NaN 1803425 Hamburg Germany 755.0 1760433 Budapest Hungary NaN 1754000 Warsaw Poland NaN 1740119 Barcelona Spain NaN 1602386 Munich Germany NaN 1493900 Milan Italy 181.8 1350680
Inserting new columns into existing DataFrames
In the previous example we have added the column area at creation time. Quite often it will be necessary to add or insert columns into existing DataFrames. For this purpose the DataFrame class provides a method "insert", which allows us to insert a column into a DataFrame at a specified location:
insert(self, loc, column, value, allow_duplicates=False)`
The parameters are specified as:
| Parameter | Meaning |
|---|---|
| loc | int |
| This value should be within the range 0 <= loc <= len(columns) |
|
| column | the column name |
| value | can be a list, a Series an array or a scalar |
| allow_duplicates | If allow_duplicates is False,an Exception will be raised, if column is already contained in the DataFrame. |
city_frame = pd.DataFrame(cities, columns=["country", "population"], index=cities["name"]) idx = 1 city_frame.insert(loc=idx, column='area', value=area) city_frameThe above code returned the following output:
| country | area | population | |
|---|---|---|---|
| London | England | 1572.00 | 8615246 |
| Berlin | Germany | 891.85 | 3562166 |
| Madrid | Spain | 605.77 | 3165235 |
| Rome | Italy | 1285.00 | 2874038 |
| Paris | France | 105.40 | 2273305 |
| Vienna | Austria | 414.60 | 1805681 |
| Bucharest | Romania | 228.00 | 1803425 |
| Hamburg | Germany | 755.00 | 1760433 |
| Budapest | Hungary | 525.20 | 1754000 |
| Warsaw | Poland | 517.00 | 1740119 |
| Barcelona | Spain | 101.90 | 1602386 |
| Munich | Germany | 310.40 | 1493900 |
| Milan | Italy | 181.80 | 1350680 |
DataFrame from Nested Dictionaries
A nested dictionary of dicts can be passed to a DataFrame as well. The indices of the outer dictionary are taken as the the columns and the inner keys. i.e. the keys of the nested dictionaries, are used as the row indices:
growth = {"Switzerland": {"2010": 3.0, "2011": 1.8, "2012": 1.1, "2013": 1.9}, "Germany": {"2010": 4.1, "2011": 3.6, "2012": 0.4, "2013": 0.1}, "France": {"2010":2.0, "2011":2.1, "2012": 0.3, "2013": 0.3}, "Greece": {"2010":-5.4, "2011":-8.9, "2012":-6.6, "2013": -3.3}, "Italy": {"2010":1.7, "2011": 0.6, "2012":-2.3, "2013":-1.9} }
growth_frame = pd.DataFrame(growth) growth_frameWe received the following result:
| France | Germany | Greece | Italy | Switzerland | |
|---|---|---|---|---|---|
| 2010 | 2.0 | 4.1 | -5.4 | 1.7 | 3.0 |
| 2011 | 2.1 | 3.6 | -8.9 | 0.6 | 1.8 |
| 2012 | 0.3 | 0.4 | -6.6 | -2.3 | 1.1 |
| 2013 | 0.3 | 0.1 | -3.3 | -1.9 | 1.9 |
You like to have the years in the columns and the countries in the rows? No problem, you can transpose the data:
growth_frame.TWe received the following output:
| 2010 | 2011 | 2012 | 2013 | |
|---|---|---|---|---|
| France | 2.0 | 2.1 | 0.3 | 0.3 |
| Germany | 4.1 | 3.6 | 0.4 | 0.1 |
| Greece | -5.4 | -8.9 | -6.6 | -3.3 |
| Italy | 1.7 | 0.6 | -2.3 | -1.9 |
| Switzerland | 3.0 | 1.8 | 1.1 | 1.9 |
growth_frame = growth_frame.T growth_frame2 = growth_frame.reindex(["Switzerland", "Italy", "Germany", "Greece"]) print(growth_frame2)
2010 2011 2012 2013 Switzerland 3.0 1.8 1.1 1.9 Italy 1.7 0.6 -2.3 -1.9 Germany 4.1 3.6 0.4 0.1 Greece -5.4 -8.9 -6.6 -3.3
Filling a DataFrame with random values:
import numpy as np names = ['Frank', 'Eve', 'Stella', 'Guido', 'Lara'] index = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"] df = pd.DataFrame((np.random.randn(12, 5)*1000).round(2), columns=names, index=index) dfAfter having executed the Python code above we received the following output:
| Frank | Eve | Stella | Guido | Lara | |
|---|---|---|---|---|---|
| January | -368.36 | 331.54 | -724.05 | -461.15 | -243.98 |
| February | 688.92 | -46.79 | 590.58 | 433.88 | 353.69 |
| March | -45.12 | -30.79 | 185.01 | -1860.11 | 483.82 |
| April | 1640.49 | 824.92 | -254.50 | 18.88 | -422.22 |
| May | -456.31 | 306.34 | -479.41 | -1339.50 | 1343.17 |
| June | 199.46 | -265.89 | -1972.07 | -1170.65 | -953.38 |
| July | 628.83 | -1412.50 | -276.20 | 1242.49 | -814.17 |
| August | 1000.72 | 19.01 | 664.81 | 261.63 | 1622.36 |
| September | 836.30 | -946.11 | 433.53 | 499.66 | 2267.20 |
| October | 388.00 | -891.88 | 267.87 | -642.04 | 1049.18 |
| November | -23.38 | 346.94 | 815.81 | 597.91 | 2209.41 |
| December | 594.83 | -2009.37 | 626.31 | -511.81 | 280.32 |