I have a large pandas dataframe with many different types of observations that need different models applied to them. One column is which model to apply, and that can be mapped to a python function which accepts a dataframe and returns a dataframe. One approach would be just doing 3 steps:
split dataframe into n dataframes for n different models
run each dataframe through each function
concatenate output dataframes at the end
This just ends up not being super flexible particularly as models are added and removed. Looking at groupby it seems like I should be able to leverage that to make this look much cleaner code-wise, but I haven't been able to find a pattern that does what I'd like.
Also because of the size of this data, using apply isn't particularly useful as it would drastically slow down the runtime.
Quick example:
df = pd.DataFrame({"model":["a","b","a"],"a":[1,5,8],"b":[1,4,6]})
def model_a(df):
return df["a"] + df["b"]
def model_b(df):
return df["a"] - df["b"]
model_map = {"a":model_a,"b":model_b}
results = df.groupby("model")...
The expected result would look like [2,1,14]. Is there an easy way code-wise to do this? Note that the actual models are much more complicated and involve potentially hundreds of variables with lots of transformations, this is just a toy example.
Thanks!
You can use groupby/apply:
x.name contains the name of the group, here a and b
x contains the sub dataframe
df['r'] = df.groupby('model') \
.apply(lambda x: model_map[x.name](x)) \
.droplevel(level='model')
>>> df
model a b r
0 a 1 1 2
1 b 5 4 1
2 a 8 6 14
Or you can use np.select:
>>> np.select([df['model'] == 'a', df['model'] == 'b'],
[model_a(df), model_b(df)])
array([ 2, 1, 14])
I have what I'm sure is a fundamental lack of understanding about how dataframes work in Python. I am sure this is an easy question, but I have looked everywhere and can't find a good explanation. I am trying to understand why sometimes dataframe calculations seem to run on a row-by-row (or cell by cell) basis, and sometimes seem to run for an entire column... For example:
data = {'Name':['49-037-23094', '49-029-21476', '49-029-20812', '49-041-21318'], 'Depth':[20, 21, 7, 18]}
df = pd.DataFrame(data)
df
Which gives:
Name Depth
0 49-037-23094 20
1 49-029-21476 21
2 49-029-20812 7
3 49-041-21318 18
Now I know I can do:
df['DepthDouble']=df['Depth']*2
And get:
Name Depth DepthDouble
0 49-037-23094 20 40
1 49-029-21476 21 42
2 49-029-20812 7 14
3 49-041-21318 18 36
Which is what I would expect. But this doesn't always work, and I'm trying to understand why. For example, I am trying to run this code to modify the name:
df['newName']=''.join(re.findall('\d',str(df['Name'])))
which gives:
Name Depth DepthDouble \
0 49-037-23094 20 40
1 49-029-21476 21 42
2 49-029-20812 7 14
3 49-041-21318 18 36
newName
0 04903723094149029214762490292081234904121318
1 04903723094149029214762490292081234904121318
2 04903723094149029214762490292081234904121318
3 04903723094149029214762490292081234904121318
So it is taking all the values in my name column, removing the dashes, and concatenating them. Of course, I'd just like it to be a new name column exactly the same as the original "Name" column, but without the dashes.
So, can anyone help me understand what I am doing wrong here? I Don't understand why sometimes Dataframe calculations for one column are done row by row (e.g., the Depth Doubled column) and sometimes Python seems to take all values in the entire column and run the calculation (e.g., the newName column).
Surely the way to get around this isn't by making a loop for every index in the df to force it to run individually for each row for a given column?
If the output you're looking for is:
Name Depth newName
0 49-037-23094 20 4903723094
1 49-029-21476 21 4902921476
2 49-029-20812 7 4902920812
3 49-041-21318 18 4904121318
The way to get this is:
df['newName']=df['Name'].map(lambda name: ''.join(re.findall('\d', name)))
map is like apply but specifically for Series objects. Since you're applying to only the Name column you are operating on a Series.
If the lambda part is confusing, an equivalent way to write it is:
def find_digits(name):
return ''.join(re.findall('\d', name))
df['newName']=df['Name'].map(find_digits)
The equivalent operation in traditional for loops is:
newNameSeries = pd.Series(name='newName')
for name in df['Name']:
newNameSeries = newNameSeries.append(pd.Series(''.join(re.findall('\d', name))), ignore_index=True)
pd.concat([df, newNameSeries], axis=1).rename(columns={0:'newName'})
While there might be a slightly cleaner way to do the loop, you can see how much simpler the first approach is compared to trying to use for-loops. It's also faster. As you already have indicated you know, avoid for loops when using pandas.
The issue is that with str(df['Name']) you are converting the entire Name-column of your DataFrame to one single string. What you want to do instead is to use one of pandas' own methods for strings, which will be applied to every single element of the column.
For example, you could use pandas' replace method for strings:
import pandas as pd
data = {'Name':['49-037-23094', '49-029-21476', '49-029-20812', '49-041-21318'], 'Depth':[20, 21, 7, 18]}
df = pd.DataFrame(data)
df['newName'] = df['Name'].str.replace('-', '')
New to Python and Pandas, so please bear with me here.
I have created a dataframe with 10 rows, with a column called 'Distance' and I want to calculate a new column (TotalCost) with apply and a lambda funtion that I have created. Snippet below of the function
def TotalCost(Distance, m, c):
return m * df.Distance + c
where Distance is the column in the dataframe df, while m and c are just constants that I declare earlier in the main code.
I then try to apply it in the following manner:
df = df.apply(lambda row: TotalCost(row['Distance'], m, c), axis=1)
but when running this, I keep getting a dataframe as an output, instead of a single row.
EDIT: Adding in an example of input and desired output,
Input: df = {Distance: '1','2','3'}
if we assume m and c equal 10,
then the output of applying the function should be
df['TotalCost'] = 20,30,40
I will post the error below this, but what am I missing here? As far as I understand, my syntax is correct. Any assistance would be greatly appreciated :)
The error message:
ValueError: Wrong number of items passed 10, placement implies 1
Your lambda in apply should process only one row. BTW, apply return only calculated columns, not whole dataframe
def TotalCost(Distance,m,c): return m * Distance + c
df['TotalCost'] = df.apply(lambda row: TotalCost(row['Distance'],m,c),axis=1)
Your apply function will basically pass one row at a time to your lambda function and then returns a copy of your data frame with the edited or changed values
Finally it returns a modified copy of dataframe constructed with rows returned by lambda functions, instead of altering the original dataframe.
have a look at this link it should help you gain more insight
https://thispointer.com/pandas-apply-apply-a-function-to-each-row-column-in-dataframe/
import numpy as np
import pandas as pd
def star(x,m,c):
return x*m+c
vals=[(1,2,4),
(3,4,5),
(5,6,6) ]
df=pd.DataFrame(vals,columns=('one','two','three'))
res=df.apply(star,axis=0,args=[2,3])
Initial DataFrame
one two three
0 1 2 4
1 3 4 5
2 5 6 6
After applying the function you should get this stored in res
one two three
0 5 7 11
1 9 11 13
2 13 15 15
This is a more memory-efficient and cleaner way:
df.eval('total_cost = #m * Distance + #c', inplace=True)
Update: I also sometimes stick to assign,
df = df.assign(total_cost=lambda x: TotalCost(x['Distance'], m, c))
So, I've got a dataframe that looks like:
with 308 different ORIGIN_CITY_NAME and 12 different UNIQUE_CARRIER.
I am trying to remove the cities where the number of unique carrier airline is < 5 As such, I performed this function:
Now, I want i'd like to take this result and manipulate my original data, df in such a way that I can remove the rows where the ORIGIN_CITY_NAME corresponds to being TRUE.
I had an idea in mind which is to use the isin() function or the apply(lambda) function in Python but I'm not familiar how to go about it. Is there a more elegant way to go about this? Thank you!
filter was made for this
df.groubpy('ORIGIN_CITY_NAME').filter(
lambda d: d.UNIQUE_CARRIER.nunique() >= 5
)
However, to continue along the vein you were attempting to get results from...
I'd use map
mask = df.groubpy('ORIGIN_CITY_NAME').UNIQUE_CARRIER.nunique() >= 5
df[df.ORIGIN_CITY_NAME.map(mask)]
Or transform
mask = df.groupby('ORIGIN_CITY_NAME').UNIQUE_CARRIER.transform(
lambda x: x.nunique() >= 5
)
df[mask]
I have a pandas dataframe with two columns. I need to change the values of the first column without affecting the second one and get back the whole dataframe with just first column values changed. How can I do that using apply() in pandas?
Given a sample dataframe df as:
a b
0 1 2
1 2 3
2 3 4
3 4 5
what you want is:
df['a'] = df['a'].apply(lambda x: x + 1)
that returns:
a b
0 2 2
1 3 3
2 4 4
3 5 5
For a single column better to use map(), like this:
df = pd.DataFrame([{'a': 15, 'b': 15, 'c': 5}, {'a': 20, 'b': 10, 'c': 7}, {'a': 25, 'b': 30, 'c': 9}])
a b c
0 15 15 5
1 20 10 7
2 25 30 9
df['a'] = df['a'].map(lambda a: a / 2.)
a b c
0 7.5 15 5
1 10.0 10 7
2 12.5 30 9
Given the following dataframe df and the function complex_function,
import pandas as pd
def complex_function(x, y=0):
if x > 5 and x > y:
return 1
else:
return 2
df = pd.DataFrame(data={'col1': [1, 4, 6, 2, 7], 'col2': [6, 7, 1, 2, 8]})
col1 col2
0 1 6
1 4 7
2 6 1
3 2 2
4 7 8
there are several solutions to use apply() on only one column. In the following I will explain them in detail.
I. Simple solution
The straightforward solution is the one from #Fabio Lamanna:
df['col1'] = df['col1'].apply(complex_function)
Output:
col1 col2
0 2 6
1 2 7
2 1 1
3 2 2
4 1 8
Only the first column is modified, the second column is unchanged. The solution is beautiful. It is just one line of code and it reads almost like english: "Take 'col1' and apply the function complex_function to it."
However, if you need data from another column, e.g. 'col2', it won't work. If you want to pass the values of 'col2' to variable y of the complex_function, you need something else.
II. Solution using the whole dataframe
Alternatively, you could use the whole dataframe as described in this SO post or this one:
df['col1'] = df.apply(lambda x: complex_function(x['col1']), axis=1)
or if you prefer (like me) a solution without a lambda function:
def apply_complex_function(x):
return complex_function(x['col1'])
df['col1'] = df.apply(apply_complex_function, axis=1)
There is a lot going on in this solution that needs to be explained. The apply() function works on pd.Series and pd.DataFrame. But you cannot use df['col1'] = df.apply(complex_function).loc[:, 'col1'], because it would throw a ValueError.
Hence, you need to give the information which column to use. To complicate things, the apply() function does only accept callables. To solve this, you need to define a (lambda) function with the column x['col1'] as argument; i.e. we wrap the column information in another function.
Unfortunately, the default value of the axis parameter is zero (axis=0), which means it will try executing column-wise and not row-wise. This wasn't a problem in the first solution, because we gave apply() a pd.Series. But now the input is a dataframe and we must be explicit (axis=1). (I marvel how often I forget this.)
Whether you prefer the version with the lambda function or without is subjective. In my opinion the line of code is complicated enough to read even without a lambda function thrown in. You only need the (lambda) function as a wrapper. It is just boilerplate code. A reader should not be bothered with it.
Now, you can modify this solution easily to take the second column into account:
def apply_complex_function(x):
return complex_function(x['col1'], x['col2'])
df['col1'] = df.apply(apply_complex_function, axis=1)
Output:
col1 col2
0 2 6
1 2 7
2 1 1
3 2 2
4 2 8
At index 4 the value has changed from 1 to 2, because the first condition 7 > 5 is true but the second condition 7 > 8 is false.
Note that you only needed to change the first line of code (i.e. the function) and not the second line.
Side note
Never put the column information into your function.
def bad_idea(x):
return x['col1'] ** 2
By doing this, you make a general function dependent on a column name! This is a bad idea, because the next time you want to use this function, you cannot. Worse: Maybe you rename a column in a different dataframe just to make it work with your existing function. (Been there, done that. It is a slippery slope!)
III. Alternative solutions without using apply()
Although the OP specifically asked for a solution with apply(), alternative solutions were suggested. For example, the answer of #George Petrov suggested to use map(); the answer of #Thibaut Dubernet proposed assign().
I fully agree that apply() is seldom the best solution, because apply() is not vectorized. It is an element-wise operation with expensive function calling and overhead from pd.Series.
One reason to use apply() is that you want to use an existing function and performance is not an issue. Or your function is so complex that no vectorized version exists.
Another reason to use apply() is in combination with groupby(). Please note that DataFrame.apply() and GroupBy.apply() are different functions.
So it does make sense to consider some alternatives:
map() only works on pd.Series, but accepts dict and pd.Series as input. Using map() with a function is almost interchangeable with using apply(). It can be faster than apply(). See this SO post for more details.
df['col1'] = df['col1'].map(complex_function)
applymap() is almost identical for dataframes. It does not support pd.Series and it will always return a dataframe. However, it can be faster. The documentation states: "In the current implementation applymap calls func twice on the first column/row to decide whether it can take a fast or slow code path.". But if performance really counts you should seek an alternative route.
df['col1'] = df.applymap(complex_function).loc[:, 'col1']
assign() is not a feasible replacement for apply(). It has a similar behaviour in only the most basic use cases. It does not work with the complex_function. You still need apply() as you can see in the example below. The main use case for assign() is method chaining, because it gives back the dataframe without changing the original dataframe.
df['col1'] = df.assign(col1=df.col1.apply(complex_function))
Annex: How to speed up apply()?
I only mention it here because it was suggested by other answers, e.g. #durjoy. The list is not exhaustive:
Do not use apply(). This is no joke. For most numeric operations, a vectorized method exists in pandas. If/else blocks can often be refactored with a combination of boolean indexing and .loc. My example complex_function could be refactored in this way.
Refactor to Cython. If you have a complex equation and the parameters of the equation are in your dataframe, this might be a good idea. Check out the official pandas user guide for more information.
Use raw=True parameter. Theoretically, this should improve the performance of apply() if you are just applying a NumPy reduction function, because the overhead of pd.Series is removed. Of course, your function has to accept an ndarray. You have to refactor your function to NumPy. By doing this, you will have a huge performance boost.
Use 3rd party packages. The first thing you should try is Numba. I do not know swifter mentioned by #durjoy; and probably many other packages are worth mentioning here.
Try/Fail/Repeat. As mentioned above, map() and applymap() can be faster - depending on the use case. Just time the different versions and choose the fastest. This approach is the most tedious one with the least performance increase.
You don't need a function at all. You can work on a whole column directly.
Example data:
>>> df = pd.DataFrame({'a': [100, 1000], 'b': [200, 2000], 'c': [300, 3000]})
>>> df
a b c
0 100 200 300
1 1000 2000 3000
Half all the values in column a:
>>> df.a = df.a / 2
>>> df
a b c
0 50 200 300
1 500 2000 3000
Although the given responses are correct, they modify the initial data frame, which is not always desirable (and, given the OP asked for examples "using apply", it might be they wanted a version that returns a new data frame, as apply does).
This is possible using assign: it is valid to assign to existing columns, as the documentation states (emphasis is mine):
Assign new columns to a DataFrame.
Returns a new object with all original columns in addition to new ones. Existing columns that are re-assigned will be overwritten.
In short:
In [1]: import pandas as pd
In [2]: df = pd.DataFrame([{'a': 15, 'b': 15, 'c': 5}, {'a': 20, 'b': 10, 'c': 7}, {'a': 25, 'b': 30, 'c': 9}])
In [3]: df.assign(a=lambda df: df.a / 2)
Out[3]:
a b c
0 7.5 15 5
1 10.0 10 7
2 12.5 30 9
In [4]: df
Out[4]:
a b c
0 15 15 5
1 20 10 7
2 25 30 9
Note that the function will be passed the whole dataframe, not only the column you want to modify, so you will need to make sure you select the right column in your lambda.
If you are really concerned about the execution speed of your apply function and you have a huge dataset to work on, you could use swifter to make faster execution, here is an example for swifter on pandas dataframe:
import pandas as pd
import swifter
def fnc(m):
return m*3+4
df = pd.DataFrame({"m": [1,2,3,4,5,6], "c": [1,1,1,1,1,1], "x":[5,3,6,2,6,1]})
# apply a self created function to a single column in pandas
df["y"] = df.m.swifter.apply(fnc)
This will enable your all CPU cores to compute the result hence it will be much faster than normal apply functions. Try and let me know if it become useful for you.
Let me try a complex computation using datetime and considering nulls or empty spaces. I am reducing 30 years on a datetime column and using apply method as well as lambda and converting datetime format. Line if x != '' else x will take care of all empty spaces or nulls accordingly.
df['Date'] = df['Date'].fillna('')
df['Date'] = df['Date'].apply(lambda x : ((datetime.datetime.strptime(str(x), '%m/%d/%Y') - datetime.timedelta(days=30*365)).strftime('%Y%m%d')) if x != '' else x)
Make a copy of your dataframe first if you need to modify a column
Many answers here suggest modifying some column and assign the new values to the old column. It is common to get the SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame. warning. This happens when your dataframe was created from another dataframe but is not a proper copy.
To silence this warning, make a copy and assign back.
df = df.copy()
df['a'] = df['a'].apply('add', other=1)
apply() only needs the name of the function
You can invoke a function by simply passing its name to apply() (no need for lambda). If your function needs additional arguments, you can pass them either as keyword arguments or pass the positional arguments as args=. For example, suppose you have file paths in your dataframe and you need to read files in these paths.
def read_data(path, sep=',', usecols=[0]):
return pd.read_csv(path, sep=sep, usecols=usecols)
df = pd.DataFrame({'paths': ['../x/yz.txt', '../u/vw.txt']})
df['paths'].apply(read_data) # you don't need lambda
df['paths'].apply(read_data, args=(',', [0, 1])) # pass the positional arguments to `args=`
df['paths'].apply(read_data, sep=',', usecols=[0, 1]) # pass as keyword arguments
Don't apply a function, call the appropriate method directly
It's almost never ideal to apply a custom function on a column via apply(). Because apply() is a syntactic sugar for a Python loop with a pandas overhead, it's often slower than calling the same function in a list comprehension, never mind, calling optimized pandas methods. Almost all numeric operators can be directly applied on the column and there are corresponding methods for all of them.
# add 1 to every element in column `a`
df['a'] += 1
# for every row, subtract column `a` value from column `b` value
df['c'] = df['b'] - df['a']
If you want to apply a function that has if-else blocks, then you should probably be using numpy.where() or numpy.select() instead. It is much, much faster. If you have anything larger than 10k rows of data, you'll notice the difference right away.
For example, if you have a custom function similar to func() below, then instead of applying it on the column, you could operate directly on the columns and return values using numpy.select().
def func(row):
if row == 'a':
return 1
elif row == 'b':
return 2
else:
return -999
# instead of applying a `func` to each row of a column, use `numpy.select` as below
import numpy as np
conditions = [df['col'] == 'a', df['col'] == 'b']
choices = [1, 2]
df['new'] = np.select(conditions, choices, default=-999)
As you can see, numpy.select() has very minimal syntax difference from an if-else ladder; only need to separate conditions and choices into separate lists. For other options, check out this answer.