I have a dataframe:
a b c
0 1 2 3
1 1 1 1
2 3 7 NaN
3 2 3 5
...
I want to fill column "three" inplace (update the values) where the values are NaN using a machine learning algorithm.
I don't know how to do it inplace. Sample code:
import pandas as pd
import numpy as np
from sklearn.linear_model import LinearRegression
df=pd.DataFrame([range(3), [1, 5, np.NaN], [2, 2, np.NaN], [4,5,9], [2,5,7]],columns=['a','b','c'])
x=[]
y=[]
for row in df.iterrows():
index,data = row
if(not pd.isnull(data['c'])):
x.append(data[['a','b']].tolist())
y.append(data['c'])
model = LinearRegression()
model.fit(x,y)
#this line does not do it in place.
df[~df.c.notnull()].assign(c = lambda x:model.predict(x[['a','b']]))
But this gives me a copy of the dataframe. Only option I have left is using a for loop however, I don't want to do that. I think there should be more pythonic way of doing it using pandas. Can someone please help? Or is there any other way of doing this?
You'll have to do something like :
df.loc[pd.isnull(df['three']), 'three'] = _result of model_
This modifies directly dataframe df
This way you first filter the dataframe to keep the slice you want to modify (pd.isnull(df['three'])), then from that slice you select the column you want to modify (three).
On the right hand side of the equal, it expects to get an array / list / series with the same number of lines than the filtered dataframe ( in your example, one line)
You may have to adjust depending on what your model returns exactly
EDIT
You probably need to do stg like this
pred = model.predict(df[['a', 'b']])
df['pred'] = model.predict(df[['a', 'b']])
df.loc[pd.isnull(df['c']), 'c'] = df.loc[pd.isnull(df['c']), 'pred']
Note that a significant part of the issue comes from the way you are using scikit learn in your example. You need to pass the whole dataset to the model when you predict.
The simplest way is yo transpose first, then forward fill/backward fill at your convenience.
df.T.ffill().bfill().T
Related
I am trying to import a .csv of EMG data as a Dataframe and filter each column of data using a list comprehension. Below is a dummy dataframe.
from scipy.signal import butter, filtfilt
test_array = pd.DataFrame(np.random.normal(0,2,size=(1000,6)),columns = ['time','RF','VM','TA','GM','BF'])
b,a = butter(4,[0.05,0.9],'bandpass',analog=False)
columns = ['RF','VM','TA','GM','BF']
filtered_df = pd.DataFrame([filtfilt(b,a,test_array.loc[:,i] for i in test_array[columns]])
The code above gives a version of the expected output, but instead of returning filtered_df as a (1000,5) dataframe, it is returning a (5,1000) dataframe.
I've tried using df.transpose() on the back end to fix the orientation, but it seems like there should be a more straightforward solution to preventing the transposing in the first place. Is there a way to get the desired output?
This issue is related to how you building the new dataframe. Just passing in a list from:
[filtfilt(b,a,test_array.loc[:,i]) for i in test_array[columns]]
pandas will read that in as a dataframe with four rows and column names representing the indices of the numpy array. If you build your dataframe using a dictionary mapped to each column name like:
results = [filtfilt(b,a,test_array.loc[:,i]) for i in test_array[columns]]
filtered_df = pd.DataFrame(data = dict(zip(columns, results)))
you get your desired result
RF VM TA GM BF
0 -0.072520 0.025846 0.111571 0.043277 0.024290
1 -2.674829 3.139997 0.285869 -0.162487 3.759851
2 -0.521439 3.481993 0.427854 -1.411966 5.422871
3 -2.719175 5.162347 2.195120 -0.535819 -1.721818
4 0.451544 1.730292 0.930652 -2.017700 -0.926594
.. ... ... ... ... ...
995 -5.240183 -0.625118 2.176452 2.065998 1.561615
996 -3.084039 -0.017626 -0.377022 -1.996366 2.041706
997 -5.122489 1.476979 -3.219335 1.609466 -3.707151
998 -2.072177 -0.870773 0.546386 0.031297 0.247766
999 0.141538 -0.048204 -0.601213 0.499631 0.246530
[1000 rows x 5 columns]
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 series called df1['geometry.coordinates'] of coordinate values in the following format:
geometry.coordinates
0 [150.792711, -34.210868]
1 [151.551228, -33.023339]
2 [148.92149870748742, -34.767207772932835]
3 [151.033742, -33.919998]
4 [150.953963043732, -32.3935017885229]
... ...
432 [114.8927165, -28.902492300000002]
433 [115.34601918477634, -30.041742290803096]
434 [115.4632611, -30.8581035]
435 [121.42151909999998, -30.7804027]
436 [115.69424934340425, -30.680970908597665]
I want to plot each point on a graph, probably through using a scatter plot.
I tried: df1['geometry.coordinates'].plot.scatter() but it gets confused because it only reads it as one list value rather than two and therefore I always get the following error:
TypeError: scatter() missing 2 required positional arguments: 'x' and 'y'
Anyone know how I can solve this?
You need to separate the column containing the list so that you can specify x and y in the plot call.
You can split a column containing a list by constructing a data frame from a list.
pd.DataFrame(df2["geometry.coordinates"].to_list(), columns=['x', 'y']).plot.scatter(x=“x”, y=“y”)
Step 1: Split array into multiple columns
df1[['x','y']] = pd.DataFrame(df1['geometry.coordinates'].tolist(), index= df1.index)
Step 2: Plot
df1.plot.scatter(x = 'x', y = 'y', s = 30) #s is size of dots
You are not giving the parameters to scatter(), so the error is quite logical. Something among the lines of df.scatter.plot(df[0],df[1]) should work.
Also, as you are working working with column vectors, you need to transpose your data for it to be viewed as rows: df.scatter.plot(df.T[0],df.T[1])
I did it this way.
import matplotlib.pyplot as plt
geometry = pd.Series([
[150.792711, -34.210868],
[151.551228, -33.023339],
[148.92149870748742, -34.767207772932835],
[151.033742, -33.919998],
[150.953963043732, -32.3935017885229]])
df = pd.DataFrame(geometry.to_list(), columns = ['x','y'])
plt.scatter(x = df['x'], y = df['y'],
edgecolor ='black')
plt.grid(alpha=.15)
you can try
import pandas as pd
geometry_coordinates=[[150.792711, -34.210868],
[151.551228, -33.023339],
[148.92149870748742, -34.767207772932835],
[151.033742, -33.919998],
[150.953963043732, -32.3935017885229],
[114.8927165, -28.902492300000002],
[115.34601918477634, -30.041742290803096],
[115.4632611, -30.8581035],
[121.42151909999998, -30.7804027],
[115.69424934340425, -30.680970908597665]]
geometry_coordinates=pd.DataFrame(geometry_coordinates,columns=['lat','long'])
geometry_coordinates.plot.scatter(x='lat',y='long')
I've got a column in a Pandas dataframe comprised of variable-length lists and I'm trying to find an efficient way of extracting elements conditional on list length. Consider this minimal reproducible example:
t = pd.DataFrame({'a':[['1234','abc','444'],
['5678'],
['2468','def']]})
Say I want to extract the 2nd element (where relevant) into a new column, and use NaN otherwise. I was able to get it in a very inefficient way:
_ = []
for index,row in t.iterrows():
if (len(row['a']) > 1):
_.append(row['a'][1])
else:
_.append(np.nan)
t['element_two'] = _
And I gave an attempt using np.where(), but I'm not specifying the 'if' argument correctly:
np.where(t['a'].str.len() > 1, lambda x: x['a'][1], np.nan)
Corrections and tips to other solutions would be greatly appreciated! I'm coming from R where I take vectorization for granted.
I'm on pandas 0.25.3 and numpy 1.18.1.
Use str accesor :
n = 2
t['second'] = t['a'].str[n-1]
print(t)
a second
0 [1234, abc, 444] abc
1 [5678] NaN
2 [2468, def] def
While not incredibly efficient, apply is at least clean:
t['a'].apply(lambda _: np.nan if len(_)<2 else _[1])
I have a pandas DataFrame, which contains 610 rows, and every row contains a nested list of coordinate pairs, it looks like that:
[1377778.4800000004, 6682395.377599999] is one coordinate pair.
I want to unnest every row, so instead of one row containing a list of coordinates I will have one row for every coordinate pair, i.e.:
I've tried s.apply(pd.Series).stack() from this question Split nested array values from Pandas Dataframe cell over multiple rows but unfortunately that didn't work.
Please any ideas? Many thanks in advance!
Here my new answer to your problem. I used "reduce" to flatten your nested array and then I used "itertools chain" to turn everything into a 1d list. After that I reshaped the list into a 2d array which allows you to convert it to the dataframe that you need. I tried to be as generic as possible. Please let me know if there are any problems.
#libraries
import operator
from functools import reduce
from itertools import chain
#flatten lists of lists using reduce. Then turn everything into a 1d list using
#itertools chain.
reduced_coordinates = list(chain.from_iterable(reduce(operator.concat,
geometry_list)))
#reshape the coordinates 1d list to a 2d and convert it to a dataframe
df = pd.DataFrame(np.reshape(reduced_coordinates, (-1, 2)))
df.columns = ['X', 'Y']
One thing you can do is use numpy. It allows you to perform a lot of list/ array operations in a fast and efficient way. This includes "unnesting" (reshaping) lists. Then you only have to convert to pandas dataframe.
For example,
import numpy as np
#your list
coordinate_list = [[[1377778.4800000004, 6682395.377599999],[6582395.377599999, 2577778.4800000004], [6582395.377599999, 2577778.4800000004]]]
#convert list to array
coordinate_array = numpy.array(coordinate_list)
#print shape of array
coordinate_array.shape
#reshape array into pairs of
reshaped_array = np.reshape(coordinate_array, (3, 2))
df = pd.DataFrame(reshaped_array)
df.columns = ['X', 'Y']
The output will look like this. Let me know if there is something I am missing.
import pandas as pd
import numpy as np
data = np.arange(500).reshape([250, 2])
cols = ['coord']
new_data = []
for item in data:
new_data.append([item])
df = pd.DataFrame(data=new_data, columns=cols)
print(df.head())
def expand(row):
row['x'] = row.coord[0]
row['y'] = row.coord[1]
return row
df = df.apply(expand, axis=1)
df.drop(columns='coord', inplace=True)
print(df.head())
RESULT
coord
0 [0, 1]
1 [2, 3]
2 [4, 5]
3 [6, 7]
4 [8, 9]
x y
0 0 1
1 2 3
2 4 5
3 6 7
4 8 9