Answer Key

How to use this key

Work the colab first, then come here. The code below is one correct answer, not the only one. This exercise gives you less scaffolding than an end-of-day activity does, so there are more ways to reach each number than usual. If your code looks different but produces the same numbers, you were right.

The written answers matter more than the code. You can already tell whether your code ran. What you cannot check on your own is whether you read the result correctly, and that is what the green Answer boxes are for. Compare your markdown cells against them.

⬅️ Back to the colab

Setup

Code

import pandas as pd

base = 'https://eds-217-essential-python.github.io/data/'

temp = pd.read_csv(base + 'monthly_temperature_data.csv')
co2 = pd.read_csv(base + 'monthly_co2_concentration.csv')

Part 1: One table out of two

1. How many rows does each table have, and what columns? What is the earliest and latest Date in each? (.min() and .max() on the Date column will do it, since these dates are text in %Y-%m-%d form and text in that form sorts correctly.)

Code

print(temp.shape, list(temp.columns))
print(co2.shape, list(co2.columns))
print(temp['Date'].min(), temp['Date'].max())
print(co2['Date'].min(), co2['Date'].max())

(1736, 2) ['Date', 'MonthlyAnomaly']
(796, 2) ['Date', 'CO2Concentration']
1880-01-01 2024-08-01
1958-04-01 2024-07-01

✅ Answer

Temperature: 1736 rows, two columns (Date, MonthlyAnomaly), January 1880 to August 2024. CO₂: 796 rows, two columns (Date, CO2Concentration), April 1958 to July 2024.

Both tables carry one row per month, and both row counts are exactly what the date spans require: 1880-01 through 2024-08 is 144 years and 8 months, which is 1736 months, and 1958-04 through 2024-07 is 796 months. Neither record has an internal gap, so every difference you find between the two tables from here on is a difference at the ends and not a hole in the middle.

Note the second date pair as well as the first. The records start 78 years apart, but they also stop one month apart, and that second fact matters in question 4.

2. Merge the two tables on Date, with the default how=. How many rows come back?

Code

inner = pd.merge(temp, co2, on='Date')
inner.shape

(796, 3)

✅ Answer

796 rows, which is exactly the CO₂ table’s row count. The default is how='inner', so a row survives only when the same Date appears in both tables.

That the answer equals 796 and not something smaller tells you every CO₂ month has a matching temperature row. The temperature record fully contains the CO₂ record in time, so the inner merge is limited entirely by the shorter of the two series. The 66 years of overlap are the CO₂ record, and nothing else.

3. Merge them again with how='left', putting temp on the left. How many rows now, and how many nulls, in which column?

Code

left = pd.merge(temp, co2, on='Date', how='left')
print(left.shape)
print(left.isnull().sum())

(1736, 3)
Date                  0
MonthlyAnomaly        0
CO2Concentration    940
dtype: int64

✅ Answer

1736 rows, with 940 nulls, all of them in CO2Concentration. A left merge keeps every row of the left table whether or not a match exists, so the row count is the temperature table’s row count unchanged.

The nulls are all in one column, and that is the signature of a left merge: the left table’s columns are complete by construction, and only the columns brought in from the right can be empty. 1736 minus 940 is 796, the inner merge’s answer, which is a useful arithmetic check that you merged what you thought you merged.

4. In a markdown cell: the two merges differ by 940 rows. Say in one sentence what those 940 rows are, without looking anything up, since your answer to question 1 already told you.

Code

missing = left[left['CO2Concentration'].isnull()]
print(missing['Date'].min(), missing['Date'].max())

1880-01-01 2024-08-01

✅ Answer

They are the months when a temperature reading exists and a CO₂ reading does not, which is the 939 months from January 1880 through March 1958, before Keeling’s instrument was running, plus one more at the far end: August 2024.

That last row is the part most pairs miss. It is easy to picture the gap as a single block at the start of the record, but the temperature series runs one month further than the CO₂ series does, so the unmatched months are at both ends. Your question 1 output said so: 2024-08-01 against 2024-07-01. 939 plus 1 is 940, and if you predicted 939 you had the right idea and read only one of the two date pairs.

5. A decision, and there is no right answer that does not depend on the question. You are writing one of the following two papers. For each one, say which merge you would use and why:

“Atmospheric CO₂ and global temperature move together”
“The global surface temperature record, 1880 to the present”

✅ Answer

The first paper takes the inner merge, the second takes the left merge.

The first paper’s claim is about the two variables together. A month with no CO₂ value cannot contribute anything to a statement about how the two move in relation to each other, so those 940 rows are not data for this question. The inner merge’s 796 rows are the honest sample, and the paper’s title should really say 1958 to 2024, not 1880.

The second paper is about temperature and mentions CO₂ nowhere. Throwing away 939 months of perfectly good thermometer readings because a different instrument had not been built yet would be a serious error, so the left merge is right. Better still, this paper does not need a merge at all; temp on its own is the dataset, and the CO₂ column is only there if you want it for context.

The general point is that the how= argument is a decision about which question you are asking, not a technical detail of pandas. Choose it after you know what the analysis is for.

Part 2: Making the dates real

6 and 7. Build climate, parse the dates, add year and month.

Code

climate = pd.merge(temp, co2, on='Date').copy()
climate['date'] = pd.to_datetime(climate['Date'], format='%Y-%m-%d')
climate['date'].dtype

dtype('<M8[ns]')

Code

climate['year'] = climate['date'].dt.year
climate['month'] = climate['date'].dt.month
climate[['Date', 'date', 'year', 'month']].head()

	Date	date	year	month
0	1958-04-01	1958-04-01	1958	4
1	1958-05-01	1958-05-01	1958	5
2	1958-06-01	1958-06-01	1958	6
3	1958-07-01	1958-07-01	1958	7
4	1958-08-01	1958-08-01	1958	8

✅ Answer

The dtype prints as datetime64[ns], which is the proof the question asked for. If it still says object, the parse silently did not happen and every .dt call after it will fail.

Keep both columns for now. Date is text and date is a real timestamp, and they look identical when printed side by side, which is exactly why the dtype check is worth doing rather than eyeballing the head of the table.

8. Build a table of annual means: group by year and report the count of months, the mean MonthlyAnomaly, and the mean CO2Concentration. Show the first three rows and the last three.

Code

annual = climate.groupby('year').agg({'Date': 'count',
                                      'MonthlyAnomaly': 'mean',
                                      'CO2Concentration': 'mean'})
print(annual.head(3))
print(annual.tail(3))

      Date  MonthlyAnomaly  CO2Concentration
year                                        
1958     9        0.004444        315.184444
1959    12        0.030833        315.981667
1960    12       -0.025000        316.908333
      Date  MonthlyAnomaly  CO2Concentration
year                                        
2022    12        0.893333        418.528333
2023    12        1.169167        421.075833
2024     7        1.287143        425.522857

9. Two of the sixty-seven years in that table are not twelve months long. Which two, and why? Filter them out into a table called full_years and say how many years survive.

Code

full_years = annual[annual['Date'] == 12]
print(annual[annual['Date'] != 12])
print(full_years.shape)

      Date  MonthlyAnomaly  CO2Concentration
year                                        
1958     9        0.004444        315.184444
2024     7        1.287143        425.522857
(65, 3)

✅ Answer

1958, with 9 months, and 2024, with 7. Sixty-five years survive, 1959 through 2023.

Both are the ends of the CO₂ record rather than anything wrong with the data. Keeling’s first reading is April 1958, so 1958 contributes April through December, and the file was downloaded after July 2024, so 2024 contributes January through July.

The count column is doing the work here. A mean never tells you how many values it averaged, so a nine-month mean and a twelve-month mean print in the same format, to the same number of decimal places, and look equally trustworthy. Asking for count alongside every mean is the habit that catches this, and it costs one extra line.

10. In a markdown cell: what would the 2024 row have done to a claim about the warmest year on record, if you had not checked the count? Two sentences.

✅ Answer

2024 would have taken first place, at a mean anomaly of 1.287 °C against 2023’s 1.169 °C, and the claim would have been built on seven months of data compared against sixty-five twelve-month averages.

The comparison is not sound even though 2024 really was warm, because a partial year is not the same quantity as a full year. 2023 makes the point on its own: its January to July mean is 1.036 °C and its August to December mean is 1.356 °C, so which months you happen to have changes the answer by more than the gap between the top two years. full_years exists so that every number you compare is built from the same twelve slots.

Part 3: The shape that shows the sawtooth

11. Build a wide table with year down the rows, month across the columns, and CO2Concentration in the cells. Show the first three rows and the last three.

Code

co2_grid = climate.pivot_table(index='year', columns='month', values='CO2Concentration')
print(co2_grid.head(3).round(2))
print(co2_grid.tail(3).round(2))

month      1       2       3       4       5       6       7       8       9   \
year                                                                            
1958      NaN     NaN     NaN  317.45  317.51  317.27  315.87  314.93  313.21   
1959   315.58  316.49  316.65  317.72  318.29  318.15  316.54  314.80  313.84   
1960   316.43  316.98  317.58  319.03  320.03  319.58  318.18  315.90  314.17   

month      10      11      12  
year                           
1958   312.42  313.33  314.67  
1959   313.33  314.81  315.58  
1960   313.83  315.00  316.19  
month      1       2       3       4       5       6       7       8       9   \
year                                                                            
2022   418.13  419.24  418.76  420.19  420.97  420.94  418.85  417.15  415.91   
2023   419.47  420.31  420.99  423.31  424.00  423.68  421.83  419.68  418.50   
2024   422.80  424.55  425.38  426.57  426.90  426.91  425.55     NaN     NaN   

month      10      11      12  
year                           
2022   415.74  417.47  418.99  
2023   418.82  420.46  421.86  
2024      NaN     NaN     NaN

12. That one table holds two different patterns at once. Read it down a single column, then read it across a single row. In a markdown cell, describe both in one sentence each.

✅ Answer

Down a column is the trend: May goes from 318.29 ppm in 1959 to 424.00 ppm in 2023, a rise of 105.71 ppm over 64 years, and the same rise appears in whichever month you pick.

Across a row is the seasonal cycle: within 2023 the value climbs to a peak of 424.00 ppm in May and falls to a trough of 418.50 ppm in September before starting to climb again.

The pivot separates the two patterns because it puts one on each axis. In the raw monthly column they are added together and you see a jagged rising line, which is harder to measure. Reshaping data is often what makes a comparison possible, and the NaN values in the 1958 and 2024 rows are the same two part years from question 9 showing up in a new shape.

13. Use the wide table to measure the second pattern. For every year, the difference between its largest monthly value and its smallest is the size of the annual cycle. Compute that difference for 1959 and for 2023, and report both.

Code

print(co2_grid.loc[1959].max() - co2_grid.loc[1959].min())
print(co2_grid.loc[2023].max() - co2_grid.loc[2023].min())

4.960000000000036
5.5

✅ Answer

4.96 ppm in 1959 and 5.50 ppm in 2023, an increase of 0.54 ppm in the size of the annual cycle across 64 years. That is about 11 percent larger, on a cycle that was already the most regular feature of the record.

Report these to two decimal places. The 1959 figure prints as 4.960000000000036, which is floating-point arithmetic showing through and not a real precision claim: the underlying measurements have two decimals, so their difference has two decimals.

14. In a markdown cell: the annual cycle in CO₂ is caused by the northern hemisphere’s plants, which take carbon out of the air in the growing season and put it back in the autumn. Given that, say what your two numbers from question 13 might mean, and say honestly how confident you are in a comparison of two single years.

Code

for y in [1959, 1960, 1965, 1991, 2000, 2016, 2023]:
    print(y, round(co2_grid.loc[y].max() - co2_grid.loc[y].min(), 2))

✅ Answer

The reading is that the northern hemisphere’s growing season now moves more carbon each year than it did in 1959, and the confidence you should have in it, from two years, is very low.

The mechanism supports the reading. A larger swing means plants drew down more carbon between spring and autumn and released more of it again, which is what you would expect from a longer growing season, a warmer high-latitude spring, or simply more vegetation taking part.

The evidence does not support it. Run the same subtraction on a few more years and the year to year scatter is larger than the change you are trying to see: 1960 is 6.20 ppm, larger than 2023’s 5.50, and 2000 is 4.83, smaller than 1959’s 4.96. 1959 is the third smallest cycle in all 65 full years, so question 13 compared a year with an unusually small cycle against an ordinary one. The honest statement is that two years cannot separate a trend from noise, and that measuring this properly means averaging the cycle over a decade at each end, or fitting a line through all 65 values.

15. Build the same wide table for MonthlyAnomaly. Does the temperature record have a comparable seasonal cycle in it? Say why or why not in one sentence. (The answer is in what the word anomaly means.)

Code

temp_grid = climate.pivot_table(index='year', columns='month', values='MonthlyAnomaly')
print(temp_grid.tail(3).round(2))

month    1     2     3     4     5     6     7     8     9     10    11    12
year                                                                         
2022   0.91  0.89  1.05  0.84  0.84  0.92  0.94  0.95  0.89  0.96  0.73  0.80
2023   0.87  0.96  1.22  0.99  0.94  1.08  1.19  1.19  1.48  1.34  1.42  1.35
2024   1.24  1.43  1.40  1.32  1.16  1.25  1.21   NaN   NaN   NaN   NaN   NaN

Code

temp_grid.mean().round(3)

month
1     0.395
2     0.414
3     0.446
4     0.385
5     0.364
6     0.357
7     0.358
8     0.350
9     0.350
10    0.365
11    0.369
12    0.360
dtype: float64

✅ Answer

No, and it is missing by construction. An anomaly is already a difference: each month’s temperature minus that same month’s long-term average. January is compared against Januaries and July against Julys, so the seasonal cycle has been subtracted out before the number reaches your table.

The column means show it. Averaged over the whole table the twelve months span 0.350 °C in August to 0.446 °C in March, a spread of about 0.10 °C, and some of even that comes from the two incomplete years contributing to some columns and not others. Compare that with CO₂, where a single year swings 5 ppm out of a total range of about 110 ppm.

Individual rows do vary. 2023 runs from 0.87 °C in January to 1.48 °C in September, a spread of 0.61 °C. That is real variation in how far each month sat above its own baseline, not a repeating seasonal shape, and it does not line up with the same months from one year to the next.

Part 4: The claim

16. Using full_years, compare the 1960s with the 2010s. Compute the mean of each column for year between 1960 and 1969, and again for year between 2010 and 2019, and report the change in each.

Code

sixties = full_years[(full_years.index >= 1960) & (full_years.index <= 1969)]
tens = full_years[(full_years.index >= 2010) & (full_years.index <= 2019)]

print(sixties[['MonthlyAnomaly', 'CO2Concentration']].mean())
print(tens[['MonthlyAnomaly', 'CO2Concentration']].mean())

MonthlyAnomaly       -0.030250
CO2Concentration    320.287167
dtype: float64
MonthlyAnomaly        0.804583
CO2Concentration    400.410583
dtype: float64

✅ Answer

CO₂ rose from 320.29 ppm to 400.41 ppm, an increase of 80.12 ppm, and the temperature anomaly rose from -0.030 °C to 0.805 °C, an increase of 0.835 °C.

Both decades are ten complete years drawn from full_years, which is what makes the two means comparable. The 1960s anomaly is very slightly negative, meaning that decade sat a hundredth of a degree below the twentieth-century baseline, so the 2010s figure is a rise from roughly zero rather than a rise from something already elevated. In relative terms the CO₂ increase is 25.0 percent of the 1960s level.

17. In a markdown cell of four or five sentences: state the two changes you measured, with units. Then state clearly what this analysis does not show. You have put two curves in the same table and found that both went up; name one thing a scientist would need before calling that a causal claim.

There is no single right answer here. A strong response quotes both changes with their units, says plainly that the analysis is a comparison of two rising series, and names something concrete that would be needed to go further. Here is one that would earn full marks.

✅ A model answer

Between the 1960s and the 2010s, atmospheric CO₂ measured at Mauna Loa rose by 80.12 ppm, from a decadal mean of 320.29 ppm to 400.41 ppm, and the global temperature anomaly rose by 0.835 °C, from -0.030 °C to 0.805 °C. Both decades are built from ten complete twelve-month years, so the two means are comparable and neither is distorted by a partial year.

What this analysis shows is that two quantities both increased over the same 66 years. It does not show that one caused the other. Any pair of series that both rise steadily will agree in this test, including many with no physical connection at all, and this comparison has no way to tell them apart. The 796-row overlap is also a single sample of a single planet, so there is no control case in which CO₂ stayed flat and we could see what temperature did.

Before calling this causal, a scientist would need a physical mechanism that predicts the size of the effect in advance, and here that mechanism exists: the radiative absorption properties of CO₂ were measured in the laboratory in the nineteenth century, and they predict roughly this much warming for roughly this much CO₂. The prediction came first and the observation matched it, and that is what makes the argument strong. It would also help to rule out the alternatives, showing that solar output and volcanic aerosols over the same period do not account for the trend. My two columns are consistent with the accepted explanation, and consistency is not evidence of cause on its own.

Where the marks are

If you compare your notebook against this key, look for these four things before you look at anything else.

You know why the inner merge returned exactly 796 rows. Not “because inner keeps matches”, but because 796 is the CO₂ table’s own row count, which means the shorter record set the limit and nothing was lost from it.
Your question 4 answer counts both ends. 939 months at the start and one month, August 2024, at the finish. Question 1 printed the evidence for both.
Every mean you reported came with a count, or from full_years. The 2024 row is the cautionary case: a seven-month mean that would have topped the warmest-year ranking, with nothing in the output to show that it came from seven months.
Your question 14 answer is honest about two years. Naming the mechanism is the easy half. Saying that 1959 against 2023 cannot separate a trend from year to year scatter is the half that earns the marks, and the extra years above are how you check it.

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