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 am not certain if this question is appropriate here, and apologies in advance if it is not.
I am a pandas maintainer, and recently I've been working on fixing bugs in pandas groupby when used with dropna=True and transform for the 1.5 release. For example, in pandas 1.4.2,
import pandas as pd
df = pd.DataFrame({'a': [1, 1, np.nan], 'b': [2, 3, 4]})
print(df.groupby('a', dropna=True).transform('sum'))
produces the incorrect (in particular, the last row) output
b
0 5
1 5
2 5
While working on this, I've been wondering how useful the dropna argument is in groupby. For aggregations (e.g. df.groupby('a').sum()) and filters (e.g. df.groupby('a').head(2)), it seems to me it's always possible to drop the offending rows prior to the groupby. In addition to this, in my use of pandas if I have null values in the groupers, then I want them in the groupby result. For transformations, where the resulting index should match that of the input, the value is instead filled with null. For the above code block, the output should be
b
0 5.0
1 5.0
2 NaN
But I can't imagine this result ever being useful. In case it is, it also is not too difficult to accomplish:
result = df.groupby('a', dropna=False).transform('sum')
result.loc[df['a'].isnull()] = np.nan
If we were able to deprecate and then remove the dropna argument to groupby (i.e. groupby always behaves as if dropna=False), then this would help simplify a good part of the groupby code.
So I'd like to ask if there are examples where dropna=True and the operation might be otherwise hard to accomplish.
Thanks!
I want to have a function that can operate on either a row or a column of a 2D ndarray. Assume the array has C order. The function changes values in the 2D data.
Inside the function I want to have identical index syntax whether it is called with a row or column. A row slice is [n,:] and column slice [:,n] so they have different shapes. Inside the function this requires different indexing expressions.
Is there a way to do this that does not require moving or allocating memory? I am under the impression that using reshape will force a copy to make the data to make it contiguous. Is there a way to use nditer in the function?
Do you mean like this:
In [74]: def foo(arr, n):
...: arr += n
...:
In [75]: arr = np.ones((2,3),int)
In [76]: foo(arr[0,:],1)
In [77]: arr
Out[77]:
array([[2, 2, 2],
[1, 1, 1]])
In [78]: foo(arr[:,1],[100,200])
In [79]: arr
Out[79]:
array([[ 2, 102, 2],
[ 1, 201, 1]])
In the first case I'm adding 1 to one row of the array, ie. a row slice. In the second case I'm add a array (list) to a column. In that case n has to have the right length.
Usually we don't worry about whether the values are C contiguous. Striding takes care of access either way.
I am wondering about the best way to change values in a subset of rows in a dataframe.
Let's say I want to double the values in column value in rows where selected is true.
In [1]: import pandas as pd
In [2]: df = pd.DataFrame({'value': [1, 2, 3, 4], 'selected': [False, False, True, True]})
In [3]: df
Out[3]:
selected value
0 False 1
1 False 2
2 True 3
3 True 4
There are several ways to do this:
# 1. Subsetting with .loc on left and right hand side:
df.loc[df['selected'], 'value'] = df.loc[df['selected'], 'value'] * 2
# 2. Subsetting with .loc on left hand side:
df.loc[df['selected'], 'value'] = df['value'] * 2
# 3. Using where()
df['value'] = (df['value'] * 2).where(df['selected'], df['value'])
If I only subset on the left hand side (option 2), would Pandas actually make the calculation for all rows and then discard the result for all but the selected rows?
In terms of evaluation, is there any difference between using loc and where?
Your #2 option is the most standard and recommended way to do this. Your #1 option is fine also, but the extra code is unnecessary because ix/loc/iloc are designed to pass the boolean selection through and do the necessary alignment to make sure it applies only to your desired subset.
# 2. Subsetting with .loc on left hand side:
df.loc[df['selected'], 'value'] = df['value'] * 2
If you don't use ix/loc/iloc on the left hand side, problems can arise that we don't want to get into in a simple answer. Hence, using ix/loc/iloc is generally the safest and most recommened way to go. There is nothing wrong with your option #3, but it is the least readable of the three.
One faster and acceptable alternative you should know about is numpy's where() function:
df['value'] = np.where( df['selected'], df['value'] * 2, df['value'] )
The first argument is the selection or mask, the second is the value to assign if True, and third is the value to assign if false. It's especially useful if you want to also create or change the value if the selection is False.
My first question here!
I'm having some trouble figuring out what I'm doing wrong here, trying to append columns to an existing pd.DataFrame object. Specifically, my original dataframe has n-many columns, and I want to use apply to append an additional 2n-many columns to it. The problem seems to be that doing this via apply() doesn't work, in that if I try to append more than n-many columns, it falls over. This doesn't make sense to me, and I was hoping somebody could either shed some light on to why I'm seeing this behaviour, or suggest a better approach.
For example,
df = pd.DataFrame(np.random.rand(10,2))
def this_works(x):
return 5 * x
def this_fails(x):
return np.append(5 * x, 5 * x)
df.apply(this_works, 1) # Two columns of output, as expected
df.apply(this_fails, 1) # Unexpected failure...
Any ideas? I know there are other ways to create the data columns, this approach just seemed very natural to me and I'm quite confused by the behaviour.
SOLVED! CT Zhu's solution below takes care of this, my error arises from not properly returning a pd.Series object in the above.
Are you trying to do a few different calculations on your df and put the resulting vectors together in one larger DataFrame, like in this example?:
In [39]:
print df
0 1
0 0.718003 0.241216
1 0.580015 0.981128
2 0.477645 0.463892
3 0.948728 0.653823
4 0.056659 0.366104
5 0.273700 0.062131
6 0.151237 0.479318
7 0.425353 0.076771
8 0.317731 0.029182
9 0.543537 0.589783
In [40]:
print df.apply(lambda x: pd.Series(np.hstack((x*5, x*6))), axis=1)
0 1 2 3
0 3.590014 1.206081 4.308017 1.447297
1 2.900074 4.905639 3.480088 5.886767
2 2.388223 2.319461 2.865867 2.783353
3 4.743640 3.269114 5.692369 3.922937
4 0.283293 1.830520 0.339951 2.196624
5 1.368502 0.310656 1.642203 0.372787
6 0.756187 2.396592 0.907424 2.875910
7 2.126764 0.383853 2.552117 0.460624
8 1.588656 0.145909 1.906387 0.175091
9 2.717685 2.948917 3.261222 3.538701
FYI in this trivial case you can do 5 * df !
I think the issue here is that np.append flattens the Series:
In [11]: np.append(df[0], df[0])
Out[11]:
array([ 0.33145275, 0.14964056, 0.86268119, 0.17311983, 0.29618537,
0.48831228, 0.64937305, 0.03353709, 0.42883925, 0.99592229,
0.33145275, 0.14964056, 0.86268119, 0.17311983, 0.29618537,
0.48831228, 0.64937305, 0.03353709, 0.42883925, 0.99592229])
what you want is it to create four columns (isn't it?). The axis=1 means that you are doing this row-wise (i.e. x is the row which is a Series)...
In general you want apply to return either:
a single value, or
a Series (with unique index).
Saying that I kinda thought the following may work (to get four columns):
In [21]: df.apply((lambda x: pd.concat([x[0] * 5, x[0] * 5], axis=1)), axis=1)
TypeError: ('cannot concatenate a non-NDFrame object', u'occurred at index 0')
In [22]: df.apply(lambda x: np.array([1, 2, 3, 4]), axis=1)
ValueError: Shape of passed values is (10,), indices imply (10, 2)
In [23]: df.apply(lambda x: pd.Series([1, 2, 3, 4]), axis=1) # works
Maybe I expected the first to raise about non-unique index... but I was surprised that the second failed.