Suppose I have a dataframe like this
Create sample dataframe:
import pandas as pd
import numpy as np
data = {
'gender': np.random.choice(['m', 'f'], size=100),
'vaccinated': np.random.choice([0, 1], size=100),
'got sick': np.random.choice([0, 1], size=100)
}
df = pd.DataFrame(data)
and I want to see, by gender, what proportion of vaccinated people got sick.
I've tries something like this:
df.groupby('gender').agg(lambda group: sum(group['vaccinated']==1 & group['sick']==1)
/sum(group['sick']==1))
but this doesn't work because agg works on the series level. Same applies for transform. apply doesn't work either, but I'm not as clear why or how apply functions on groupby objects.
Any ideas how to accomplish this with a single line of code?
You could first filter for the vaccinated people and then group by gender and calculate the proportion of people that got sick..
df[df.vaccinated == 1].groupby("gender").agg({"got sick":"mean"})
Output:
got sick
gender
f 0.548387
m 0.535714
In this case the proportion is calculated based on a sample data that I've created
The docs for GroupBy.apply state that the function is applied "group-wise". This means that the function is called on each group separately as a data frame.
That is, df.groupby(c).apply(f) is conceptually equivalent to:
results = {}
for val in df[c]:
group = df.loc[df[c] == val]
result = f(group)
results[val] = result
pd.concat(results)
We can use this understanding to apply your custom aggregation function, using a top-level def just to make the code easier to read:
def calc_vax_sick_frac(group):
vaccinated = group['vaccinated'] == 1
sick = group['sick'] == 1
return (vaccinated & sick).sum() / sick.sum()
(
df
.groupby('gender')
.apply(calc_vax_sick_frac)
)
I'm trying to remove certain values with that code, however pandas does not give me to, instead outputs
ValueError: Unable to coerce to Series, length must be 10: given 2
Here is my code:
import pandas as pd
df = pd.read_csv("/Volumes/SSD/IT/DataSets/Automobile_data.csv")
print(df.shape)
columns_df = ['index', 'company', 'body-style', 'wheel-base', 'length', 'engine-type',
'num-of-cylinders', 'horsepower', 'average-mileage', 'price']
prohibited_symbols = ['?','Nan''n.a']
df = df[df[columns_df] != prohibited_symbols]
print(df)
Try:
df = df[~df[columns_df].str.contains('|'.join(prohibited_symbols))]
The regex operator '|' helps remove records that contain any of your prohibited symbols.
Because what you are trying is not doing what you imagine it should.
df = df[df[columns_df] != prohibited_symbols]
Above line will always return False values for everything. You can't iterate over a list of prohibited symbols like that. != will do only a simple inequality check and none of your cells will be equal to the list of prohibited symbols probably. Also using that syntax will not delete those values from your cells.
You'll have to use a for loop and clean every column like this.
for column in columns_df:
df[column] = df[column].str.replace('|'.join(prohibited_symbols), '', regex=True)
You can as well specify the values you consider as null with the na_values argument when reading the data and then use dropna from pandas.
Example:
import pandas as pd
df = pd.read_csv("/Volumes/SSD/IT/DataSets/Automobile_data.csv", na_values=['?','Nan''n.a'])
df = df.dropna()
I need to create a derived field in a pandas dataframe based on the value in another field of the same dataframe. Like so:
def newfield(row):
If row.col1 == ‘x’
return ‘value is x’
Elif row.col1 == ‘y’
Return ‘value is y’
Then i call it:
df.newfield = df.apply(lambda row:newfield(row),axis=1)
Is there a way to do it without’apply’? Also would like to make it less verbose. Np.where only allows 2 conditions, but i have more than 2.
Yes, you can use np.select:
df['newfield'] = np.select([df['col1'] == 'x', df['col1'=='y'],
['value is x', 'value is y'],
np.nan)
I'm trying to replace the values in one column of a dataframe. The column ('female') only contains the values 'female' and 'male'.
I have tried the following:
w['female']['female']='1'
w['female']['male']='0'
But receive the exact same copy of the previous results.
I would ideally like to get some output which resembles the following loop element-wise.
if w['female'] =='female':
w['female'] = '1';
else:
w['female'] = '0';
I've looked through the gotchas documentation (http://pandas.pydata.org/pandas-docs/stable/gotchas.html) but cannot figure out why nothing happens.
Any help will be appreciated.
If I understand right, you want something like this:
w['female'] = w['female'].map({'female': 1, 'male': 0})
(Here I convert the values to numbers instead of strings containing numbers. You can convert them to "1" and "0", if you really want, but I'm not sure why you'd want that.)
The reason your code doesn't work is because using ['female'] on a column (the second 'female' in your w['female']['female']) doesn't mean "select rows where the value is 'female'". It means to select rows where the index is 'female', of which there may not be any in your DataFrame.
You can edit a subset of a dataframe by using loc:
df.loc[<row selection>, <column selection>]
In this case:
w.loc[w.female != 'female', 'female'] = 0
w.loc[w.female == 'female', 'female'] = 1
w.female.replace(to_replace=dict(female=1, male=0), inplace=True)
See pandas.DataFrame.replace() docs.
Slight variation:
w.female.replace(['male', 'female'], [1, 0], inplace=True)
This should also work:
w.female[w.female == 'female'] = 1
w.female[w.female == 'male'] = 0
This is very compact:
w['female'][w['female'] == 'female']=1
w['female'][w['female'] == 'male']=0
Another good one:
w['female'] = w['female'].replace(regex='female', value=1)
w['female'] = w['female'].replace(regex='male', value=0)
You can also use apply with .get i.e.
w['female'] = w['female'].apply({'male':0, 'female':1}.get):
w = pd.DataFrame({'female':['female','male','female']})
print(w)
Dataframe w:
female
0 female
1 male
2 female
Using apply to replace values from the dictionary:
w['female'] = w['female'].apply({'male':0, 'female':1}.get)
print(w)
Result:
female
0 1
1 0
2 1
Note: apply with dictionary should be used if all the possible values of the columns in the dataframe are defined in the dictionary else, it will have empty for those not defined in dictionary.
Using Series.map with Series.fillna
If your column contains more strings than only female and male, Series.map will fail in this case since it will return NaN for other values.
That's why we have to chain it with fillna:
Example why .map fails:
df = pd.DataFrame({'female':['male', 'female', 'female', 'male', 'other', 'other']})
female
0 male
1 female
2 female
3 male
4 other
5 other
df['female'].map({'female': '1', 'male': '0'})
0 0
1 1
2 1
3 0
4 NaN
5 NaN
Name: female, dtype: object
For the correct method, we chain map with fillna, so we fill the NaN with values from the original column:
df['female'].map({'female': '1', 'male': '0'}).fillna(df['female'])
0 0
1 1
2 1
3 0
4 other
5 other
Name: female, dtype: object
Alternatively there is the built-in function pd.get_dummies for these kinds of assignments:
w['female'] = pd.get_dummies(w['female'],drop_first = True)
This gives you a data frame with two columns, one for each value that occurs in w['female'], of which you drop the first (because you can infer it from the one that is left). The new column is automatically named as the string that you replaced.
This is especially useful if you have categorical variables with more than two possible values. This function creates as many dummy variables needed to distinguish between all cases. Be careful then that you don't assign the entire data frame to a single column, but instead, if w['female'] could be 'male', 'female' or 'neutral', do something like this:
w = pd.concat([w, pd.get_dummies(w['female'], drop_first = True)], axis = 1])
w.drop('female', axis = 1, inplace = True)
Then you are left with two new columns giving you the dummy coding of 'female' and you got rid of the column with the strings.
w.replace({'female':{'female':1, 'male':0}}, inplace = True)
The above code will replace 'female' with 1 and 'male' with 0, only in the column 'female'
There is also a function in pandas called factorize which you can use to automatically do this type of work. It converts labels to numbers: ['male', 'female', 'male'] -> [0, 1, 0]. See this answer for more information.
w.female = np.where(w.female=='female', 1, 0)
if someone is looking for a numpy solution. This is useful to replace values based on a condition. Both if and else conditions are inherent in np.where(). The solutions that use df.replace() may not be feasible if the column included many unique values in addition to 'male', all of which should be replaced with 0.
Another solution is to use df.where() and df.mask() in succession. This is because neither of them implements an else condition.
w.female.where(w.female=='female', 0, inplace=True) # replace where condition is False
w.female.mask(w.female=='female', 1, inplace=True) # replace where condition is True
dic = {'female':1, 'male':0}
w['female'] = w['female'].replace(dic)
.replace has as argument a dictionary in which you may change and do whatever you want or need.
I think that in answer should be pointed which type of object do you get in all methods suggested above: is it Series or DataFrame.
When you get column by w.female. or w[[2]] (where, suppose, 2 is number of your column) you'll get back DataFrame.
So in this case you can use DataFrame methods like .replace.
When you use .loc or iloc you get back Series, and Series don't have .replace method, so you should use methods like apply, map and so on.
To answer the question more generically so it applies to more use cases than just what the OP asked, consider this solution. I used jfs's solution solution to help me. Here, we create two functions that help feed each other and can be used whether you know the exact replacements or not.
import numpy as np
import pandas as pd
class Utility:
#staticmethod
def rename_values_in_column(column: pd.Series, name_changes: dict = None) -> pd.Series:
"""
Renames the distinct names in a column. If no dictionary is provided for the exact name changes, it will default
to <column_name>_count. Ex. female_1, female_2, etc.
:param column: The column in your dataframe you would like to alter.
:param name_changes: A dictionary of the old values to the new values you would like to change.
Ex. {1234: "User A"} This would change all occurrences of 1234 to the string "User A" and leave the other values as they were.
By default, this is an empty dictionary.
:return: The same column with the replaced values
"""
name_changes = name_changes if name_changes else {}
new_column = column.replace(to_replace=name_changes)
return new_column
#staticmethod
def create_unique_values_for_column(column: pd.Series, except_values: list = None) -> dict:
"""
Creates a dictionary where the key is the existing column item and the value is the new item to replace it.
The returned dictionary can then be passed the pandas rename function to rename all the distinct values in a
column.
Ex. column ["statement"]["I", "am", "old"] would return
{"I": "statement_1", "am": "statement_2", "old": "statement_3"}
If you would like a value to remain the same, enter the values you would like to stay in the except_values.
Ex. except_values = ["I", "am"]
column ["statement"]["I", "am", "old"] would return
{"old", "statement_3"}
:param column: A pandas Series for the column with the values to replace.
:param except_values: A list of values you do not want to have changed.
:return: A dictionary that maps the old values their respective new values.
"""
except_values = except_values if except_values else []
column_name = column.name
distinct_values = np.unique(column)
name_mappings = {}
count = 1
for value in distinct_values:
if value not in except_values:
name_mappings[value] = f"{column_name}_{count}"
count += 1
return name_mappings
For the OP's use case, it is simple enough to just use
w["female"] = Utility.rename_values_in_column(w["female"], name_changes = {"female": 0, "male":1}
However, it is not always so easy to know all of the different unique values within a data frame that you may want to rename. In my case, the string values for a column are hashed values so they hurt the readability. What I do instead is replace those hashed values with more readable strings thanks to the create_unique_values_for_column function.
df["user"] = Utility.rename_values_in_column(
df["user"],
Utility.create_unique_values_for_column(df["user"])
)
This will changed my user column values from ["1a2b3c", "a12b3c","1a2b3c"] to ["user_1", "user_2", "user_1]. Much easier to compare, right?
If you have only two classes you can use equality operator. For example:
df = pd.DataFrame({'col1':['a', 'a', 'a', 'b']})
df['col1'].eq('a').astype(int)
# (df['col1'] == 'a').astype(int)
Output:
0 1
1 1
2 1
3 0
Name: col1, dtype: int64
I would like to extract certain section of a URL, residing in a column of a Pandas Dataframe and make that a new column. This
ref = df['REFERRERURL']
ref.str.findall("\\d\\d\\/(.*?)(;|\\?)",flags=re.IGNORECASE)
returns me a Series with tuples in it. How can I take out only one part of that tuple before the Series is created, so I can simply turn that into a column? Sample data for referrerurl is
http://wap.blah.com/xxx/id/11/someproduct_step2;jsessionid=....
In this example I am interested in creating a column that only has 'someproduct_step2' in it.
Thanks,
In [25]: df = DataFrame([['http://wap.blah.com/xxx/id/11/someproduct_step2;jsessionid=....']],columns=['A'])
In [26]: df['A'].str.findall("\\d\\d\\/(.*?)(;|\\?)",flags=re.IGNORECASE).apply(lambda x: Series(x[0][0],index=['first']))
Out[26]:
first
0 someproduct_step2
in 0.11.1 here is a neat way of doing this as well
In [34]: df.replace({ 'A' : "http:.+\d\d\/(.*?)(;|\\?).*$"}, { 'A' : r'\1'} ,regex=True)
Out[34]:
A
0 someproduct_step2
This also worked
def extract(x):
res = re.findall("\\d\\d\\/(.*?)(;|\\?)",x)
if res: return res[0][0]
session['RU_2'] = session['REFERRERURL'].apply(extract)