Need explanation on how pandas.drop is working here - pandas

I have a data frame, lets say xyz. I have written code to find out the % of null values each column possess in the dataframe. my code below:
round(100*(xyz.isnull().sum()/len(xyz.index)), 2)
let say i got following results:
abc 26.63
def 36.58
ghi 78.46
I want to drop column ghi because it has more than 70% of null values.
I achieved it using the following code:
xyz = xyz.drop(xyz.loc[:,round(100*(xyz.isnull().sum()/len(xyz.index)), 2)>70].columns, 1)
but , i did not understand how does this code works, can anyone please explain it?

the code is doing the following:
xyz.drop( [...], 1)
removes the specified elements for a given axis, either by row or by column. In this particular case, df.drop( ..., 1) means you're dropping by axis 1, i.e, column
xyz.loc[:, ... ].columns
will return a list with the column names resulting from your slicing condition
round(100*(xyz.isnull().sum()/len(xyz.index)), 2)>70
this instruction is counting the number of nulls, adding them up and normalizing by the number of rows, effectively computing the percentage of nan in each column. Then, the amount is rounded to have only 2 decimal positions and finally you return True is the number of nan is more than 70%. Hence, you get a mapping between columns and a True/False array.
Putting everything together: you're first producing a Boolean array that marks which columns have more than 70% nan, then, using .loc you use Boolean indexing to look only at the columns you want to drop ( nan % > 70%), then using .columns you recover the name of such columns, which then are used by the .drop instruction.
Hopefully this clear things up!

If you code is hard to understand , you can just check dropna with thresh, since pandas already cover this case.
df=df.dropna(axis=1,thresh=round(len(df)*0.3))

Related

Finding the mean of a column; but excluding a singular value

Imagine I have a dataset that is like so:
ID birthyear weight
0 619040 1962 0.1231231
1 600161 1963 0.981742
2 25602033 1963 1.3123124
3 624870 1987 10,000
and I want to get the mean of the column weight, but the obvious 10,000 is hindering the actual mean. In this situation I cannot change the value but must work around it, this is what I've got so far, but obviously it's including that last value.
avg_num_items = df_cleaned['trans_quantity'].mean()
translist = df_cleaned['trans_quantity'].tolist()
my dataframe is df_cleaned and the column I'm actually working with is 'trans_quantity' so how do I go about the mean while working around that value?
Since you added SQL in your tags, In SQL you'd want to exclude it in the WHERE clause:
SELECT AVG(trans_quantity)
FROM your_data_base
WHERE trans_quantity <> 10,000
In Pandas:
avg_num_items = df_cleaned[df_cleaned["trans_quantity"] != 10000]["trans_quantity"].mean()
You can also replace your value with a NAN and skip it in the mean:
avg_num_items = df_cleaned["trans_quantity"].replace(10000, np.nan).mean(skipna=True)
With pandas, ensure you have numeric data (10,000 is a string), filter the values above threshold and use the mean:
(pd.to_numeric(df['weight'], errors='coerce')
.loc[lambda x: x<10000]
.mean()
)
output: 0.8057258333333334

Pandas Replace_ column values

Hello,
I am analyzing the next dataset with this information .
The column ['program_number'] is an object but I want to change it to a integer colum.
I have tried to replace some values but it doesn´t work.
as you can see, some values like 6 is duplicate. like '6 ' and 6.
How can I resolve it? Many thanks
UPDATE
Didn't see 1X and 3X at first.
If you need those numbers and just want to remove the X then:
df["Program"] = df["Program"].str.strip(" X").astype(int)
If there is data in the column which aren't numbers or which shouldn't be converted, you can use pd.to_numeric with errors='corece'. If there are cells which can't be converted, you'll get NaN. Be aware that this will result in floating numbers.
df["Program"] = pd.to_numeric(df["Program"], errors="coerce")
old
You want to use str.strip() here, rather than replace.
Try this:
df1['program_number'] = df1['program_number'].str.strip().astype(int)

Pandas run function only on subset of whole Dataframe

Lets say i have Dataframe, which has 200 values, prices for products. I want to run some operation on this dataframe, like calculate average price for last 10 prices.
The way i understand it, right now pandas will go through every single row and calculate average for each row. Ie first 9 rows will be Nan, then from 10-200, it would calculate average for each row.
My issue is that i need to do a lot of these calculations and performance is an issue. For that reason, i would want to run the average only on say on last 10 values (dont need more) from all values, while i want to keep those values in the dataframe. Ie i dont want to get rid of those values or create new Dataframe.
I just essentially want to do calculation on less data, so it is faster.
Is something like that possible? Hopefully the question is clear.
Building off Chicodelarose's answer, you can achieve this in a more "pandas-like" syntax.
Defining your df as follows, we get 200 prices up to within [0, 1000).
df = pd.DataFrame((np.random.rand(200) * 1000.).round(decimals=2), columns=["price"])
The bit you're looking for, though, would the following:
def add10(n: float) -> float:
"""An exceptionally simple function to demonstrate you can set
values, too.
"""
return n + 10
df["price"].iloc[-12:] = df["price"].iloc[-12:].apply(add10)
Of course, you can also use these selections to return something else without setting values, too.
>>> df["price"].iloc[-12:].mean().round(decimals=2)
309.63 # this will, of course, be different as we're using random numbers
The primary justification for this approach lies in the use of pandas tooling. Say you want to operate over a subset of your data with multiple columns, you simply need to adjust your .apply(...) to contain an axis parameter, as follows: .apply(fn, axis=1).
This becomes much more readable the longer you spend in pandas. 🙂
Given a dataframe like the following:
Price
0 197.45
1 59.30
2 131.63
3 127.22
4 35.22
.. ...
195 73.05
196 47.73
197 107.58
198 162.31
199 195.02
[200 rows x 1 columns]
Call the following to obtain the mean over the last n rows of the dataframe:
def mean_over_n_last_rows(df, n, colname):
return df.iloc[-n:][colname].mean().round(decimals=2)
print(mean_over_n_last_rows(df, 2, "Price"))
Output:
178.67

Pandas identifying if any element is in a row

I have a data frame that is a single row of numerical values and I want to know if any of those values is greater than 2 and if so create a new column with the word 'Diff'
Col_,F_1,F_2
1,5,0
My dataframe is diff_df. Here is one thing I tried
c = diff_df >2
if c.any():
diff_df['difference']='Difference'
If I were to print c. it would be
Col_,F_1,F_2
False,True,False
I have tried c.all() and many iterations of other things. Clearly my inexperience is holding me back and google is not helping in this regards. Everything I try is either "The truth value of a Series (or Data Frame) is ambiguous use a.any(), a.all()...." Any help would be appreciated.
Since it is only one row, take the .max().max() of the dataframe. With one .max() you are going to get the .max() of each column. The second .max() takes the max of all the columns.
if diff_df.max().max() > 2: diff_df['difference']='Difference'
output:
Col_ F_1 F_2 difference
0 1 5 0 Difference
Use .loc accessor and .gt() to query and at the same time create new column and populate it
df.loc[df.gt(2).any(1), "difference"] = 'Difference'
Col_ F_1 F_2 difference
0 1 5 0 Difference
In addition to David's reponse you may also try this:
if ((df > 2).astype(int)).sum(axis=1).values[0] == 1:
df['difference']='Difference'

Dataframe non-null values differ from value_counts() values

There is an inconsistency with dataframes that I cant explain. In the following, I'm not looking for a workaround (already found one) but an explanation of what is going on under the hood and how it explains the output.
One of my colleagues which I talked into using python and pandas, has a dataframe "data" with 12,000 rows.
"data" has a column "length" that contains numbers from 0 to 20. she wants to divided the dateframe into groups by length range: 0 to 9 in group 1, 9 to 14 in group 2, 15 and more in group 3. her solution was to add another column, "group", and fill it with the appropriate values. she wrote the following code:
data['group'] = np.nan
mask = data['length'] < 10;
data['group'][mask] = 1;
mask2 = (data['length'] > 9) & (data['phraseLength'] < 15);
data['group'][mask2] = 2;
mask3 = data['length'] > 14;
data['group'][mask3] = 3;
This code is not good, of course. the reason it is not good is because you dont know in run time whether data['group'][mask3], for example, will be a view and thus actually change the dataframe, or it will be a copy and thus the dataframe would remain unchanged. It took me quit sometime to explain it to her, since she argued correctly that she is doing an assignment, not a selection, so the operation should always return a view.
But that was not the strange part. the part the even I couldn't understand is this:
After performing this set of operation, we verified that the assignment took place in two different ways:
By typing data in the console and examining the dataframe summary. It told us we had a few thousand of null values. The number of null values was the same as the size of mask3 so we assumed the last assignment was made on a copy and not on a view.
By typing data.group.value_counts(). That returned 3 values: 1,2 and 3 (surprise) we then typed data.group.value_counts.sum() and it summed up to 12,000!
So by method 2, the group column contained no null values and all the values we wanted it to have. But by method 1 - it didnt!
Can anyone explain this?
see docs here.
You dont' want to set values this way for exactly the reason you pointed; since you don't know if its a view, you don't know that you are actually changing the data. 0.13 will raise/warn that you are attempting to do this, but easiest/best to just access like:
data.loc[mask3,'group'] = 3
which will guarantee you inplace setitem