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.
I'm trying to remove outliers using IQR method. However, the shape of my df remains the same.
Here is the code:
def IQR_outliers(df):
Q1=df.quantile(0.25)
Q3=df.quantile(0.75)
IQR=Q3-Q1
df=df[~((df<(Q1-1.5*IQR)) | (df>(Q3+1.5*IQR)))]
return df
IQR_outliers(df['Distance'])
IQR_outliers(df['Price'])
Your function considers the whole object that is passed, but you're only passing a single series each time you use it. You're also not capturing the output. All of these things stack on top of each to make your problem pretty complex.
So here's what I would do:
add a column argument to your function
modifying the function to only consider that column when selecting rows from the entire dataframe
pipe the dataframe to that function a couple of times
So that's:
def IQR_outliers(df, column):
Q1 = df[column].quantile(0.25)
Q3 = df[column].quantile(0.75)
IQR = Q3 - Q1
df = df.loc[lambda df: ~((df[column] < (Q1 - 1.5 * IQR)) | (df[column] > (Q3 + 1.5 * IQR)))]
return df
revised_df = df.pipe(IQR_outliers, 'Distance').pipe(IQR_outliers, 'Price')
Note that the way you've demonstrated this, you'll very likely drop rows where Distance is an outlier even if Price is not. If you don't want to do that, you'll need to stack your dataframe, apply this function to a groupby operation, and then optionally unstack the dataframe
Let's say if I have a Pandas df called df_1 where one of the rows looks like this:
id
rank_url_agg
url_list
2223
['gtech.com','gm.com', 'ford.com']
['google.com','gtech.com','autoblog.com','gm.com', 'ford.com']
I want to create a new column called url_list_agg which does the following things for each row:
Iterate through the URLs in url_list
If URL doesn't exist in rank_url_agg in the same row, assign a value of 0.
If URL exists in rank_url_agg, then assign the value that corresponds to the difference between the length of the rank_url_agg list and the index of that URL in rank_url_agg.
Once done iterating through all URLs in url_list, wrap the results into a list.
So at the end, the first row in the new url_list_agg column will become [0,3,0,2,1].
I've tried running the following script (only to test the 1st row and not entire dataframe):
for item in agg_report['url_list'][0]:
if item in agg_report['rank_url_agg'][0]:
item=len(rank_url_agg[0]) - agg_report['rank_url_agg'][0].index(item)
else:
item=0
But when I checked agg_report['url_list'][0], it still returned just this list: ['google.com','gtech.com','autoblog.com','gm.com', 'ford.com']. So my code didn't work.
Any advice on how to achieve this goal for every row in the dataframe will be greatly appreciated!
You're not assigning back to the actual dataframe.
def idx(a, b):
return [len(a) - a.index(x) if x in a else 0 for x in b]
df_1 = df_1.assign(url_list_agg=[*map(idx, df_1.rank_url_agg, df_1.url_list)])
I'm attempting to determine the number of widget failures from a test population.
Each widget can fail in 0, 1, or multiple ways. I'd like to calculate the number of failures of for each failure method, but once a widget is known to have failed, it should be excluded from future sums. In other words, the failure modes are known and ordered. If a widget fails via mode 1 and mode 3, I don't care about mode 3: I just want to count mode 1.
I have a dataframe with one row per item, and one column per failure mode. If the widget fails in that mode, the column value is 1, else it is 0.
d = {"item_1":
{"failure_1":0, "failure_2":0},
"item_2":
{"failure_1":1, "failure_2":0},
"item_3":
{"failure_1":0, "failure_2":1},
"item_4":
{"failure_1":1, "failure_2":1}}
df = pd.DataFrame(d).T
display(df)
Output:
failure_1 failure_2
item_1 0 0
item_2 1 0
item_3 0 1
item_4 1 1
If I just want to sum the columns, that's easy: df.sum(). And if I want to calculate percentage failures, easy too: df.sum()/len(df). But this counts widgets that fail in multiple ways, multiple times. For the problem stated, the best I can come up with is this:
# create empty df to store results
df2 = pd.DataFrame(columns=["total_failures"])
for col in df.columns:
# create a row, named after the column, and assign it the value of the sum
df2.loc[col] = df[col].sum()
# drop rows in the df column that are equal to 1
df = df.loc[df[col] != 1]
display(df2)
Output:
total_failures
failure_1 2
failure_2 1
This requires creating another dataframe (that's fine), but also requires iterating over the existing dataframe columns and deleting it a couple of rows at a time. If the dataframe takes a while to generate, or is needed for future calculations, this is not workable. I can deal with iterating over the columns.
Is there a way to do this without deleting the original df, or making a temporary copy? (Not workable with large data sets.)
You can do a cumsum on axis=1 and wherever the value is greater than 1 , mask it as 0 and then take sum:
out = df.mask(df.cumsum(axis=1).gt(1), 0).sum().to_frame('total_failures')
print(out)
total_failures
failure_1 2
failure_2 1
This way the original df is retained too.
I have a large dataframe that I want to sample based on values on the target column value, which is binary : 0/1
I want to extract equal number of rows that have 0's and 1's in the "target" column. I was thinking of using the pandas sampling function but not sure how to declare the equal number of samples I want from both classes for the dataframe based on the target column.
I was thinking of using something like this:
df.sample(n=10000, weights='target', random_state=1)
Not sure how to edit it to get 10k records with 5k 1's and 5k 0's in the target column. Any help is appreciated!
You can group the data by target and then sample,
df = pd.DataFrame({'col':np.random.randn(12000), 'target':np.random.randint(low = 0, high = 2, size=12000)})
new_df = df.groupby('target').apply(lambda x: x.sample(n=5000)).reset_index(drop = True)
new_df.target.value_counts()
1 5000
0 5000
Edit: Use DataFrame.sample
You get similar results using DataFrame.sample
new_df = df.groupby('target').sample(n=5000)
You can use DataFrameGroupBy.sample method as follwing:
sample_df = df.groupby("target").sample(n=5000, random_state=1)
Also found this to be a good method:
df['weights'] = np.where(df['target'] == 1, .5, .5)
sample_df = df.sample(frac=.1, random_state=111, weights='weights')
Change the value of frac depending on the percent of data you want back from the original dataframe.
You will have to run a df0.sample(n=5000) and df1.sample(n=5000) and then combine df0 and df1 into a dfsample dataframe. You can create df0 and df1 by df.filter() with some logic. If you provide sample data I can help you construct that logic.