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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 have two dataframes as:
df1.ix[1:3]
DateTime
2018-01-02 [-0.0031537018416199097, 0.006451397621428631,...
2018-01-03 [-0.0028882814454597745, -0.005829869983964528...
df2.ix[1:3]
DateTime
2018-01-02 [-0.03285881500135208, -0.027806145786217932, ...
2018-01-03 [-0.0001314381449719178, -0.006278235444742629...
len(df1.ix['2018-01-02'][0])
500
len(df2.ix['2018-01-02'][0])
500
When I do df1 + df2 I get:
len((df1 + df2).ix['2018-01-02'][0])
1000
So, the lists instead of being summation is being concatenated.
How do I add element wise the lists in the dataframes df1 and df2.
When an operation is applied between two dataframes, it gets broadcasted at element level. Element in your case is a list and when '+' operator is applied between two lists, it concatenates them. That's why resulting dataframe contains concatenated lists.
There can be multiple approaches for actually summing up elements of lists instead of concatenating.
One approach can be converting list elements into columns and then adding dataframes and then merging columns back to a single list.(which has been suggested in first answer but in a wrong way)
Step 1: Converting list elements to columns
df1=df1.apply(lambda row:pd.Series(row[0]), axis=1)
df2=df2.apply(lambda row:pd.Series(row[0]), axis=1)
We need to pass row[0] instead of row to get rid of column index associated with series.
Step 2: Add dataframes
df=df1+df2 #this dataframe will have 500 columns
Step 3: Merge columns back to lists
df=df.apply(lambda row:pd.Series({0:list(row)}),axis=1)
This is an interesting part. Why are we returning a series here? Why only returning list(row) doesn't work and keep retaining 500 columns?
Reason is - if length of list returned is same as length of columns in the beginning, then this list gets fit in columns and to us it seems nothing happened. Whereas if length of the list is not equal to number of columns, then it is returned as single list.
Let's look at an example.
Suppose I've a dataframe, having columns 0 ,1 and 2.
df=pd.DataFrame({0:[1,2,3],1:[4,5,6],2:[7,8,9]})
0 1 2
0 1 4 7
1 2 5 8
2 3 6 9
Number of columns in original dataframe are 3. If I try to return a list with two columns, it works and a series is returned,
df1=df.apply(lambda row:[row[0],row[1]],axis=1)
0 [1, 4]
1 [2, 5]
2 [3, 6]
dtype: object
Instead if try to return list of three numbers, it would get fit in columns.
df1=df.apply(list,axis=1)
0 1 2
0 1 4 7
1 2 5 8
2 3 6 9
So if we want to return list of same size as number of columns, we'll have to return it in form of Series where one row's value has been given as list.
Another approach can be, introduce one column of a dataframe into other and then add columns using apply function.
df1[1]=df2[0]
df=df1.apply(lambda r: list(np.array(r[0])+np.array(r[1])),axis=1)
We can take advantage of numpy arrays here. '+' operator on numpy arrays sums up corresponding values and gives a single numpy array.
Cast them to series so that they become columns, then add your dfs:
df1 = df1.apply(pd.Series, axis=1)
df2 = df2.apply(pd.Series, axis=1)
df1 + df2
Since .ix has been deprecated as of Pandas 0.20, I wonder what is the proper way to mix lable-based, boolean-based and position-based indexing in Pandas? I need to assign values to a slice of dataframe that can be best referenced with label or boolean on the index and position on the columns. For example (using .loc as placeholder for the desired slicing method):
df.loc[df['a'] == 'x', -12:-1] = 3
obviously this doesn't work, with which I get:
TypeError: cannot do slice indexing on <class 'pandas.core.indexes.base.Index'> with these indexers [-12] of <class 'int'>
If I use .iloc, I get:
NotImplementedError: iLocation based boolean indexing on an integer type is not available
So how do I do it, without chaining, obviously to avoid chained assignment problem.
Let's use .loc with the boolean indexing, and accessing the column labels via the dataframe column index with index slicing:
df.loc[df['a'] == 'x', df.columns[-12:-1]] = 3
maybe I should've explained clearer. I meant if your dataframe is indexed (with 0 to n), then you can use loc[] for a combination of number for rows and lable for column:
new_df = pd.DataFrame({'a':[1,2,3,4],'b':[5,6,7,8]})
new_df
Out[10]:
a b
0 1 5
1 2 6
2 3 7
3 4 8
new_df.loc[0,'a']
Out[11]:
1
Even though .ix has been removed, it looks like .loc does the same job now. You can make a mix reference using .loc.
I am new to pandas. I know how to use drop_duplicates and take the last observed row in a dataframe. Is there any way that I can use it to take only second last observed. Or any other way of doing it.
For example:
I would like to go from
df = pd.DataFrame(data={'A':[1,1,1,2,2,2],'B':[1,2,3,4,5,6]}) to
df1 = pd.DataFrame(data={'A':[1,2],'B':[2,5]})
The idea is that you'll group the data by the duplicate column , then check the length of group , if the length of group is greater than or equal 2 this mean that you can slice the second element of group , if the group has a length of one which mean that this value is not duplicated , then take index 0 which is the only element in the grouped data
df.groupby(df['A']).apply(lambda x : x.iloc[1] if len(x) >= 2 else x.iloc[0])
The first answer I think was on the right track, but possibly not quite right. I have extended your data to include 'A' groups with two observations, and an 'A' group with one observation, for the sake of completeness.
import pandas as pd
df = pd.DataFrame(data={'A':[1,1,1,2,2,2, 3, 3, 4],'B':[1,2,3,4,5,6, 7, 8, 9]})
def user_apply_func(x):
if len(x) == 2:
return x.iloc[0]
if len(x) > 2:
return x.iloc[-2]
return
df.groupby('A').apply(user_apply_func)
Out[7]:
A B
A
1 1 2
2 2 5
3 3 7
4 NaN NaN
For your reference the apply method automatically passes the data frame as the first argument.
Also, as you are always going to be reducing each group of data to a single observation you could also use the agg method (aggregate). apply is more flexible in terms of the length of the sequences that can be returned whereas agg must reduce the data to a single value.
df.groupby('A').agg(user_apply_func)
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.