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 a Pandas DF in which I want to convert column values to integer values. The information should be stored in meters but can be stored in kilometers as well, resulting in the following possible values:
23145 (correct)
23145.0 (.0 should be removed)
101.1 (should be multiplied *1000)
47,587 (should be multiplied *1000)
'No value known'
I tried different options with converting data types, but I always seem to break the existing integers and cannot check for them correctly because the type 'object' is the dtype. Sometimes faulty values or strings block conversion as well.
Any ideas how to check if the value currently is an integer and do nothing, remove .0 from applicable values and multiply where applicable.
I also have some other columns with integers (e.g. number 22321323) where randomly a .0 is assigned (e.g. number 22321323.0). How can I correctly convert these values to not include the .0?
If you use a .apply on the column, you should be able to very easily convert these values while casing on their type. For example:
import pandas as pd
def convert(x):
if isinstance(x, int):
return x
elif isinstance(x, float):
return int(x)
else:
# Defaults to 0 when not convertable
return 0
print(x)
df = pd.DataFrame({'col': [23145, 23145.0, 'No value known']})
df['col'] = df['col'].apply(convert)
I have a pandas DataFrame with multi level index. I want to sort by one of the index levels. It has float values, but occasionally few empty strings too which I want to be considered as nan.
df = pd.DataFrame(dict(x=[1,2,3,4]), index=[1,2,3,''])
df.index.name = 'i'
df.sort_values('i')
TypeError: '<' not supported between instances of 'str' and 'int'
One way to solve the problem is to replace the empty strings with nan, do the sort, and then replace nan with empty strings again.
I am wondering if there is any way we could tweek the sort_values to consider empty stings as nan.
Why there are empty strings in the first place?
In my application, actually the data read has missing values which is read as np.nan. But, np.nan values cause problem with groupby. So, they are replace to empty strings. I wish we had a constant like nan which is treated like empty string by groupby and like nan for numeric operations.
I am wondering if there is any way we could tweek the sort_values to consider empty stings as nan.
In pandas missing values are not empty values, only if save DataFrame with missing values then are replaced by empty strings.
Btw, main problem is mixed values - numeric with strings (empty values), best is convert all strings to numeric for avoid it.
You can replace empty values by missing values by rename:
df = pd.DataFrame(dict(x=[1,2,3,4]), index=[1,2,3,''])
df.index.name = 'i'
df = df.rename({'':np.nan})
df = df.sort_values('i')
print (df)
x
i
1.0 1
2.0 2
3.0 3
NaN 4
Possible solution if cannot be changed original data is get positions of sorted values by Index.argsort and change order by DataFrame.iloc:
df = df.iloc[df.rename({'':np.nan}).index.argsort()]
print (df)
x
i
1 1
2 2
3 3
4
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.
For example, the values in '/tmp/test.csv' (namely, 01, 02, 03) are meant to represent strings that happen to match /^\d+$/, as opposed to integers:
In [10]: print open('/tmp/test.csv').read()
A,B,C
01,02,03
By default, pandas.read_csv converts these values to integers:
In [11]: import pandas
In [12]: pandas.read_csv('/tmp/test.csv')
Out[12]:
A B C
0 1 2 3
I want to tell pandas.read_csv to leave all these values alone. I.e., perform no conversions whatsoever. Furthermore, I want this "please do nothing" directive to be applied across-the-board, without my having to specify any column names or numbers.
I tried this, which achieved nothing:
In [13]: import csv
In [14]: pandas.read_csv('/tmp/test.csv', quoting=csv.QUOTE_ALL)
Out[14]:
A B C
0 1 2 3
The only thing that worked was to define a big ol' ConstantDict class, and use an instance of it that always returns the identity function (lambda x: x) as the value for the converters parameter, and thereby trick pandas.read_csv into doing nothing:
In [15]: %cpaste
class ConstantDict(dict):
def __init__(self, value):
self.__value = value
def get(self, *args):
return self.__value
--
Pasting code; enter '--' alone on the line to stop or use Ctrl-D.
::::::
In [16]: pandas.read_csv('/tmp/test.csv', converters=ConstantDict(lambda x: x))
Out[16]:
A B C
0 01 02 03
That's a lot of gymnastics to get such a simple "please do nothing" request across. (It would be even more gymnastics if I were to make ConstantDict bullet-proof.)
Isn't there a simpler way to achieve this?
df = pd.read_csv('temp.csv', dtype=str)
From the docs:
dtype : Type name or dict of column -> type, default None
Data type for data or columns. E.g. {‘a’: np.float64, ‘b’: np.int32} (Unsupported with engine=’python’). Use str or object to preserve and not interpret dtype.