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 the following sample dataframe:
Market Value
0 282024800.37
1 317460884.85
2 1260854026.24
3 320556927.27
4 42305412.79
I am trying to round the values in this dataframe to the nearest whole number. Desired output:
Market Value
282024800
317460885
1260854026
320556927
42305413
I tried:
df.values.round()
and the result was
Market Value
282025000.00
317461000.00
1260850000.00
320557000.00
42305400.00
What am I doing wrong?
Thanks
This might be more appropriate posted as a comment, but put here for proper format.
I can't produce your result. With numpy 1.18.1 and Pandas 1.1.0,
df.round().astype('int')
gives me:
Market Value
0 282024800
1 317460885
2 1260854026
3 320556927
4 42305413
The only thing I can think of is that you may have a 32 bit system, where
df.astype('float32').round().astype('int')
gives me
Market Value
0 282024800
1 317460896
2 1260854016
3 320556928
4 42305412
The following will keep your data information intact as a float put will have it display/print to the nearest int.
Big caveat: it is only possible to have this apply to ALL dataframes at once (it is a pandas wide option) rather than just a single dataframe.
pd.set_option("display.precision", 0)
If you like #noah's solution but don't want to have to change the variables back if you output something, you can use the following helper function:
import pandas as pd
from contextlib import contextmanager
#contextmanager
def temp_pandas_options(options):
seen_options = set()
old_values = {}
if isinstance(options, dict):
options_pairs = list(options.items())
else:
options_pairs = options
for option, value in options_pairs:
assert not option in seen_options, f"Already saw option {option}"
old_values[option] = pd.get_option(option)
pd.set_option(option, value)
yield
for option, old_value in old_values.items():
pd.set_option(option, old_value)
Then you can run
with temp_pandas_options({'display.float_format': '{:.0f}'.format}):
print(market_value_df)
and get
Market value
0 282024800
1 317460885
2 1260854026
3 320556927
4 42305413
This question is similar to this one, but in my case I need to apply a function that returns a Series rather than a single value for each group — that question is about aggregating with sum, but I need to use rank (so the difference is like that between agg and transform).
I have data on firms over time. This generates some dummy data that looks like my use case:
import numpy as np
import pandas as pd
dates = pd.date_range('1926', '2020', freq='M')
ndates = len(dates)
nfirms = 5000
cols = list('ABCDE')
df = pd.DataFrame(np.random.randn(nfirms*ndates,len(cols)),
index=np.tile(dates,nfirms),
columns=cols)
df.insert(0, 'id', np.repeat(np.arange(nfirms), ndates))
I need to calculate ranks of column E within each date (the index), but keeping column id.
If I just use groupby and .rank I get this:
df.groupby(level=0)['E'].rank()
1926-01-31 3226.0
1926-02-28 1042.0
1926-03-31 1611.0
1926-04-30 2591.0
1926-05-31 30.0
...
2019-08-31 1973.0
2019-09-30 227.0
2019-10-31 4381.0
2019-11-30 1654.0
2019-12-31 1572.0
Name: E, Length: 5640000, dtype: float64
This has the same dimension as df but I'm not sure it's safe to merge on the index — I really need to join on the id column also. Can I assume that the order remains the same?
If the order in the output is the same as in the output, I think I can do this:
df['ranks'] = df.groupby(level=0)['E'].rank()
But something about this seems strange, and I assume there is a way to include additional columns in the groupby output.
(I'm also not clear if calling .rank() is equivalent to .transform('rank').)
I need to extract dataframes from json data stored in every row of initial dataframe and concat them all together. Currently it works for me over iteration and takes ages.
Input data is dataframe, containing JSON dictionaries:
print(json_table)
json_responce timestamp request
27487 {'explore_tabs.. 2019-07-02 02:05:25 Lisboa, Portugal
27488 {'explore_tabs.. 2019-07-02 02:05:27 Ribeira, Portugal
The json_responce field is being unwraped to dataframe:
from pandas.io.json import json_normalize
from ast import literal_eval
json = literal_eval(json_table.loc[0,'json_responce'])
df_normalized = json_normalize(json['explore_tabs'][0]['sections'][0]
['listings'])
which gives a nice unwrapped dataframe for each row of the initial df
Having 27000 rows of json containing df, I iterate over initial df, which creates new df at every step and concat's to the final_df, to concat all the data together:
def unwrap_json_and_concat(json_table):
final_df = pd.DataFrame()
for i in json_table.index:
row = literal_eval(json_table.loc[i,'json_responce'])
df = json_normalize(row['explore_tabs'][0]['sections']
[0]['listings'])
final_df = pd.concat([final_df,df])
return final_df
As expected, that takes ages to iterate over, with significant slowing towards the end of calculation due to the increasing size of the final_df.
I know how to create functions for apply, but I believe it will not give much perfomance either, due to the fact, that new dataframe is being created every row anyways.
How to vectorize this calculation?
Thank you!
I have a big array of data:
>>> len(b)
6636849
>>> print(b)
[['60D19E9E-4E2C-11E2-AA9A-52540027E502' '100015361']
['60D19EB6-4E2C-11E2-AA9A-52540027E502' '100015385']
['60D19ECE-4E2C-11E2-AA9A-52540027E502' '100015409']
...,
['8CC90633-447E-11E6-B010-005056A76B49' '106636785']
['F8C74244-447E-11E6-B010-005056A76B49' '106636809']
['F8C7425C-447E-11E6-B010-005056A76B49' '106636833']]
I need to get the filtered dataset, i.e, everything containing (or starting with) '106' in the string). Something like the following code with substring operation instead of math operation:
>>> len(b[b[:,1] > '10660600'])
30850
I don't think numpy is well suited for this type of operation. You can do it simply using basic python operations. Here it is with some sample data a:
import random # for the test data
a = []
for i in range(10000):
a.append(["".join(random.sample('abcdefg',3)), "".join(random.sample('01234567890',8))])
answer = [i for i in a if i[1].find('106') != -1]
Keep in mind that startswith is going to be a lot faster than find, because find has to look for matching substrings in all positions.
It's not too clear why you need do this with such a large list/array in the first place, and there might be a better solution when it comes to not including these values in the list in the first place.
Here's a simple pandas solution
import pandas as pd
df = pd.DataFrame(b, columns=['1st String', '2nd String'])
df_filtered = df[df['2nd String'].str.contains('106')]
This gives you
In [29]: df_filtered
Out[29]:
1st String 2nd String
3 8CC90633-447E-11E6-B010-005056A76B49 106636785
4 F8C74244-447E-11E6-B010-005056A76B49 106636809
5 F8C7425C-447E-11E6-B010-005056A76B49 106636833
Update: Timing Results
Using Benjamin's list a as the test sample:
In [20]: %timeit [i for i in a if i[1].find('106') != -1]
100 loops, best of 3: 2.2 ms per loop
In [21]: %timeit df[df['2nd String'].str.contains('106')]
100 loops, best of 3: 5.94 ms per loop
So it looks like Benjamin's answer is actually about 3x faster. This surprises me since I was under the impression that the operation in pandas is vectorized. Moreover, the speed ratio does not change when a is 100 times longer.
Look at the functions in the np.char submodule:
data = [['60D19E9E-4E2C-11E2-AA9A-52540027E502', '100015361'],
['60D19EB6-4E2C-11E2-AA9A-52540027E502', '100015385'],
['60D19ECE-4E2C-11E2-AA9A-52540027E502', '100015409'],
['8CC90633-447E-11E6-B010-005056A76B49', '106636785'],
['F8C74244-447E-11E6-B010-005056A76B49', '106636809'],
['F8C7425C-447E-11E6-B010-005056A76B49', '106636833']]
data = np.array([r[1] for r in data], np.str)
idx = np.char.startswith(data, '106')
print(idx)