#### Exercise 0: Environment and libraries ##### The exercise is validated is all questions of the exercise are validated. ##### Activate the virtual environment. If you used `conda` run `conda activate your_env` ##### Run `python --version`. ###### Does it print `Python 3.x`? x >= 8 ##### Does `import jupyter`, `import numpy` and `import pandas` run without any error? --- --- #### Exercise 1: Your first DataFrame ##### The exercise is validated is all questions of the exercise are validated. ##### The solution of question 1 is accepted if the DataFrame created is the same as the "model" DataFrame. Check that the index is not 1,2,3,4,5. ##### The solution of question 2 is accepted if the columns' types are as below and if the types of the first value of the columns are as below: ```console ``` ```console ``` --- --- #### Exercise 2: Electric power consumption ##### The exercise is validated is all questions of the exercise are validated ##### The solution of question 1 is accepted if `drop` is used with `axis=1`.`inplace=True` may be useful to avoid to affect the result to a variable. A solution that could be accepted too (even if it's not a solution I recommend) is `del`. ##### The solution of question 2 is accepted if the DataFrame returns the output below. If the type of the index is not `dtype='datetime64[ns]'` the solution is not accepted. I recommend to use `set_index` with `inplace=True` to do so. ```python Input: df.head().index Output: DatetimeIndex(['2006-12-16', '2006-12-16','2006-12-16', '2006-12-16','2006-12-16'], dtype='datetime64[ns]', name='Date', freq=None) ``` ##### The solution of question 3 is accepted if all the types are `float64` as below. The preferred solution is `pd.to_numeric` with `coerce=True`. ```python Input: df.dtypes Output: Global_active_power float64 Global_reactive_power float64 Voltage float64 Global_intensity float64 Sub_metering_1 float64 dtype: object ``` ##### The solution of question 4 is accepted if you use `df.describe()`. ##### The solution of question 5 is accepted if `dropna` is used and if the number of missing values is equal to 0. It is important to notice that 25979 rows contain missing values (for a total of 129895). `df.isna().sum()` allows to check the number of missing values and `df.dropna()` with `inplace=True` allows to remove the rows with missing values. ##### The solution of question 6 is accepted if one of the two approaches below were used: ```python #solution 1 df.loc[:,'A'] = (df['A'] + 1) * 0.06 #solution 2 df.loc[:,'A'] = df.loc[:,'A'].apply(lambda x: (x+1)*0.06) ``` You may wonder `df.loc[:,'A']` is required and if `df['A'] = ...` works too. **The answer is no**. This is important in Pandas. Depending on the version of Pandas, it may return a warning. The reason is that you are affecting a value to a **copy** of the DataFrame and not in the DataFrame. More details: https://stackoverflow.com/questions/20625582/how-to-deal-with-settingwithcopywarning-in-pandas ##### The solution of question 7 is accepted as long as the output of `print(filtered_df.head().to_markdown())` is as below and if the number of rows is equal to **449667**. | Date | Global_active_power | Global_reactive_power | |:--------------------|----------------------:|------------------------:| | 2008-12-27 00:00:00 | 0.996 | 0.066 | | 2008-12-27 00:00:00 | 1.076 | 0.162 | | 2008-12-27 00:00:00 | 1.064 | 0.172 | | 2008-12-27 00:00:00 | 1.07 | 0.174 | | 2008-12-27 00:00:00 | 0.804 | 0.184 | ##### The solution of question 8 is accepted if the output is: ```console Global_active_power 0.254 Global_reactive_power 0.000 Voltage 238.350 Global_intensity 1.200 Sub_metering_1 0.000 Name: 2007-02-16 00:00:00, dtype: float64 ``` ##### The solution of question 9 if the output is `Timestamp('2009-02-22 00:00:00')`. ##### The solution of question 10 if the output of `print(sorted_df.tail().to_markdown())` is: | Date | Global_active_power | Global_reactive_power | Voltage | |:--------------------|----------------------:|------------------------:|----------:| | 2008-08-28 00:00:00 | 0.076 | 0 | 234.88 | | 2008-08-28 00:00:00 | 0.076 | 0 | 235.18 | | 2008-08-28 00:00:00 | 0.076 | 0 | 235.4 | | 2008-08-28 00:00:00 | 0.076 | 0 | 235.64 | | 2008-12-08 00:00:00 | 0.076 | 0 | 236.5 | ##### The solution of question 11 is accepted if the output is as below. The solution is based on `groupby` which creates groups based on the index `Date` and aggregates the groups using the `mean`. ```console Date 2006-12-16 3.053475 2006-12-17 2.354486 2006-12-18 1.530435 2006-12-19 1.157079 2006-12-20 1.545658 ... 2010-12-07 0.770538 2010-12-08 0.367846 2010-12-09 1.119508 2010-12-10 1.097008 2010-12-11 1.275571 Name: Global_active_power, Length: 1433, dtype: float64 ``` --- --- #### Exercise 3: E-commerce purchases ##### The exercise is validated is all questions of the exercise are validated. ##### To validate this exercise all answers should return the expected numerical value given in the correction AND uses Pandas. For example using NumPy to compute the mean doesn't respect the philosophy of the exercise which is to use Pandas. ##### The solution of question 1 is accepted if it contains **10000 entries** and **14 columns**. There many solutions based on: shape, info, describe. ##### The solution of question 2 is accepted if the answer is **50.34730200000025**. Even if `np.mean` gives the solution, `df['Purchase Price'].mean()` is preferred ##### The solution of question 3 is accepted if the min is `0`and the max is `99.989999999999995` ##### The solution of question 4 is accepted if the answer is **1098** ##### The solution of question 5 is accepted if the answer is **30** ##### The solution of question 6 is accepted if the are `4932` people that made the purchase during the `AM` and `5068` people that made the purchase during `PM`. There many ways to the solution but the goal of this question was to make you use `value_counts` ##### The solution of question 7 is accepted if the answer is as below. There many ways to the solution but the goal of this question was to use `value_counts` Interior and spatial designer 31 Lawyer 30 Social researcher 28 Purchasing manager 27 Designer, jewellery 27 ##### The solution of question 8 is accepted if the purchase price is **75.1** ##### The solution of question 9 is accepted if the email address is **bondellen@williams-garza.com** ##### The solution of question 10 is accepted if the answer is **39**. The preferred solution is based on this: `df[(df['A'] == X) & (df['B'] > Y)]` ##### The solution of question 11 is accepted if the answer is **1033**. The preferred solution is based on the usage of `apply` on a `lambda` function that slices the string that contains the expiration date. ##### The solution of question 12 is accepted if the answer is as below. The preferred solution is based on the usage of `apply` on a `lambda` function that slices the string that contains the email. The `lambda` function uses `split` to split the string on `@`. Finally, `value_counts` is used to count the occurrences. - hotmail.com 1638 - yahoo.com 1616 - gmail.com 1605 - smith.com 42 - williams.com 37 --- --- #### Exercise 4: Handling missing values ##### The exercise is validated is all questions of the exercise are validated (except the bonus question) ##### The solution of question 1 is accepted if the two steps are implemented in that order. First, convert the numerical columns to `float` and then fill the missing values. The first step may involve `pd.to_numeric(df.loc[:,col], errors='coerce')`. The second step is validated if you eliminated all missing values. However there are many possibilities to fill the missing values. Here is one of them: example: ```python df.fillna({0:df.sepal_length.mean(), 2:df.sepal_width.median(), 3:0, 4:0}) ``` ##### The solution of question 2 is accepted if the solution is `df.loc[:,col].fillna(df[col].median())`. ##### The solution of bonus question is accepted if you find out this answer: Once we filled the missing values as suggested in the first question, `df.describe()` returns this interesting summary. We notice that the mean is way higher than the median. It means that there are maybe some outliers in the data. The quantile 75 and the max confirm that: 75% of the flowers have a sepal length smaller than 6.4 cm, but the max is 6900 cm. If you check on the internet you realise this small flower can't be that big. The outliers have a major impact on the mean which equals to 56.9. Filling this value for the missing value is not correct since it doesn't correspond to the real size of this flower. That is why in that case the best strategy to fill the missing values is the median. The truth is that I modified the data set ! But real data sets ALWAYS contains outliers. Always think about the meaning of the data transformation ! If you fill the missing values by zero, it means that you consider that the length or width of some flowers may be 0. It doesn't make sense. | | sepal_length | sepal_width | petal_length | petal_width | | :---- | -----------: | ----------: | -----------: | ----------: | | count | 146 | 141 | 120 | 147 | | mean | 56.9075 | 52.6255 | 15.5292 | 12.0265 | | std | 572.222 | 417.127 | 127.46 | 131.873 | | min | -4.4 | -3.6 | -4.8 | -2.5 | | 25% | 5.1 | 2.8 | 2.725 | 0.3 | | 50% | 5.75 | 3 | 4.5 | 1.3 | | 75% | 6.4 | 3.3 | 5.1 | 1.8 | | max | 6900 | 3809 | 1400 | 1600 | ##### The solution of bonus question is accepted if the presence of negative values and huge values have been detected. A good data scientist always check abnormal values in the dataset. **YOU SHOULD ALWAYS TRY TO UNDERSTAND YOUR DATA**. Print the row with index 122 ;-) This week, we will have the opportunity to focus on the data pre-processing to understand how the outliers can be handled.