I was experimenting with the kaggle.com Titanic data set (data on every person on the Titanic) and came up with a gender breakdown like this:
df = pd.DataFrame({'sex': ['male'] * 577 + ['female'] * 314})
gender = df.sex.value_counts()
gender
male 577
female 314
I would like to find out the percentage of each gender on the Titanic.
My approach is slightly less than ideal:
from __future__ import division
pcts = gender / gender.sum()
pcts
male 0.647587
female 0.352413
Is there a better (more idiomatic) way?
This function is implemented in pandas, actually even in value_counts(). No need to calculate :)
just type:
df.sex.value_counts(normalize=True)
which gives exactly the desired output.
Please note that value_counts() excludes NA values, so numbers might not add up to 1.
See here: http://pandas-docs.github.io/pandas-docs-travis/generated/pandas.Series.value_counts.html
(A column of a DataFrame is a Series)
In case you wish to show percentage one of the things that you might do is use value_counts(normalize=True) as answered by #fanfabbb.
With that said, for many purposes, you might want to show it in the percentage out of a hundred.
That can be achieved like so:
gender = df.sex.value_counts(normalize=True).mul(100).round(1).astype(str) + '%'
In this case, we multiply the results by hundred, round it to one decimal point and add the percentage sign.
If you want to merge counts with percentage, can use:
c = df.sex.value_counts(dropna=False)
p = df.sex.value_counts(dropna=False, normalize=True)
pd.concat([c,p], axis=1, keys=['counts', '%'])
I think I would probably do this in one go (without importing division):
1. * df.sex.value_counts() / len(df.sex)
or perhaps, remembering you want a percentage:
100. * df.sex.value_counts() / len(df.sex)
Much of a muchness really, your way looks fine too.
Related
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
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
Date
GoogleAnalytics_PVS
AdobeAnalytics_PVS
6-3-2020
4802
4922
6-4-2020
5939
5932
6-5-2020
5122
5298
I have a table structured like the one above where it returns the number of page views from two sources. Ideally, I would like another column that would return a discrepancy percentage.
Am I overthinking it or could I just do something like
df['Discrep_%'] = (df['GoogleAnalytics_PVS'] - df['AdobeAnalytics_PVS'] / df['GoogleAnalytics_PVS']) x 100
Is there a better method, please let me know, thanks!
complexity wise it's the same, but here is another way. hence there can be multiple ways but the one you are applying is also the better.
df_new = [df[df.columns.difference(['GoogleAnalytics_PVS', 'AdobeAnalytics_PVS'])]/df['GoogleAnalytics_PVS'] ]*100
df_new
I'm sorry, I know this is basic but I've tried to figure it out myself for 2 days by sifting through documentation to no avail.
My code:
import numpy as np
import pandas as pd
name = ["bob","bobby","bombastic"]
age = [10,20,30]
price = [111,222,333]
share = [3,6,9]
list = [name,age,price,share]
list2 = np.transpose(list)
dftest = pd.DataFrame(list2, columns = ["name","age","price","share"])
print(dftest)
name age price share
0 bob 10 111 3
1 bobby 20 222 6
2 bombastic 30 333 9
Want to divide all elements in 'price' column with all elements in 'share' column. I've tried:
print(dftest[['price']/['share']]) - Failed
dftest['price']/dftest['share'] - Failed, unsupported operand type
dftest.loc[:,'price']/dftest.loc[:,'share'] - Failed
Wondering if I could just change everything to int or float, I tried:
dftest.astype(float) - cant convert from str to float
Ive tried iter and items methods but could not understand the printouts...
My only suspicion is to use something called iterate, which I am unable to wrap my head around despite reading other old posts...
Please help me T_T
Apologies in advance for the somewhat protracted answer, but the question is somewhat unclear with regards to what exactly you're attempting to accomplish.
If you simply want price[0]/share[0], price[1]/share[1], etc. you can just do:
dftest['price_div_share'] = dftest['price'] / dftest['share']
The issue with the operand types can be solved by:
dftest['price_div_share'] = dftest['price'].astype(float) / dftest['share'].astype(float)
You're getting the cant convert from str to float error because you're trying to call astype(float) on the ENTIRE dataframe which contains string columns.
If you want to divide each item by each item, i.e. price[0] / share[0], price[1] / share[0], price[2] / share[0], price[0] / share[1], etc. You would need to iterate through each item and append the result to a new list. You can do that pretty easily with a for loop, although it may take some time if you're working with a large dataset. It would look something like this if you simply want the result:
new_list = []
for p in dftest['price'].astype(float):
for s in dftest['share'].astype(float):
new_list.append(p/s)
If you want to get this in a new dataframe you can simply save it to a new dataframe using pd.Dataframe() method:
new_df = pd.Dataframe(new_list, columns=[price_divided_by_share])
This new dataframe would only have one column (the result, as mentioned above). If you want the information from the original dataframe as well, then you would do something like the following:
new_list = []
for n, a, p in zip(dftest['name'], dftest['age'], dftest['price'].astype(float):
for s in dftest['share'].astype(float):
new_list.append([n, a, p, s, p/s])
new_df = pd.Dataframe(new_list, columns=[name, age, price, share, price_div_by_share])
If you check the data types of your dataframe, you will realise that they are all strings/object type :
dftest.dtypes
name object
age object
price object
share object
dtype: object
first step will be to change the relevant columns to numbers - this is one way:
dftest = dftest.set_index("name").astype(float)
dftest.dtypes
age float64
price float64
share float64
dtype: object
This way you make the names a useful index, and separate it from the numeric data. This is just a suggestion; you may have other reasons to leave names as a columns - in that case, you have to individually change the data types of each column.
Once that is done, you can safely execute your code :
dftest.div(dftest.share,axis=0)
age price share
name
bob 3.333333 37.0 1.0
bobby 3.333333 37.0 1.0
bombastic 3.333333 37.0 1.0
I assume this is what you expect as your outcome. If not, you can tweak it. Main part is get your data types as numbers before computation/division can occur.
Help with homework problem: "Let us define the "data science experience" of a given person as the person's largest score among Regression, Classification, and Clustering. Compute the average data science experience among all MSIS students."
Beginner to coding. I am trying to figure out how to check amongst columns and compare those columns to each other for the largest value. And then take the average of those found values.
I greatly appreciate your help in advance!
Picture of the sample data set: 1: https://i.stack.imgur.com/9OSjz.png
Provided Code:
import pandas as pd
df = pd.read_csv("cleaned_survey.csv", index_col=0)
df.drop(['ProgSkills','Languages','Expert'],axis=1,inplace=True)
Sample Data:
What I have tried so far:
df[data_science_experience]=df[["Regression","Classification","Clustering"]].values.max()
df['z']=df[['Regression','Classification','Clustering']].apply(np.max,axis=1)
df[data_science_experience]=df[["Regression","Classification","Clustering"]].apply(np.max,axis=1)
If you want to get the highest score of column 'hw1' you can get it with:
pd['hw1'].max(). this gives you a series of all the values in that column and max returns the maximum. for average use mean:
pd['hw1'].mean()
if you want to find the maximum of multiple columns, you can use:
maximum_list = list()
for col in pd.columns:
maximum_list.append(pd[col].max)
max = maximum_list.max()
avg = maximum_list.mean()
hope this helps.
First, you want to get only the rows with MSIS in the Program column. That can be done in the following way:
df[df['Program'] == 'MSIS']
Next, you want to get only the Regression, Classification and Clustering columns. The previous query filtered only rows; we can add to that, like this:
df.loc[df['Program'] == 'MSIS', ['Regression', 'Classification', 'Clustering']]
Now, for each row remaining, we want to take the maximum. That can be done by appending .max(axis=1) to the previous line (axis=1 because we want the maximum of each row, not each column).
At this point, we should have a DataFrame where each row represents the highest score of the three categories for each student. Now, all that's left to do is take the mean, which can be done with .mean(). The full code should therefore look like this:
df.loc[df['Program'] == 'MSIS', ['Regression', 'Classification', 'Clustering']].max(axis=1).mean()