By default, the mean() method should ignore the nan value, but for my case, it didn't work. It still takes the nan value.
a = np.array([1,9])
b = np.array([3,nan])
c = np.array([7,8])
d = {'value': [a,b,a,c], 'group': [3,3,4,4], 'garbage':['asd','acas','asdasdc','ghfas']}
df = pd.DataFrame(data=d)
df
OUTPUT:
value group garbage
0 [1, 9] 3 asd
1 [3.0, nan] 3 acas
2 [1, 9] 4 asdasdc
3 [7, 8] 4 ghfas
for i,j in df.groupby('group')['value']:
print(j.mean())
print("=========")
OUTPUT:
[ 2. nan]
=========
[4. 8.5]
=========
I am not sure what you are trying to do here, but Ill take a stab at it.
Firstly, the values column is a column of numpy arrays, so it is two dimensional. Then when you run groupby, j becomes a pd.Series of numpy arrays. Thus, when you call mean you are taking the mean by aligning the axes of the numpy arrays. This is pretty unadvisable because these objects can change shape which will cause an error.
I think what you are trying to do is take the mean across all the arrays in each group. You can do that with.
for i,j in df.groupby('group')['value']:
print(np.nanmean(np.concatenate(j.values)))
Whatever you are trying to do, it is going to be way easier to interact with once you combine the values in your loop.
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'm looking for a way to quickly and effectively filter through a dataframe column and remove values that don't meet a condition.
Say, I have a column with the numbers 4, 5 and 10. I want to filter the column and replace any numbers above 7 with 0. How would I go about this?
You're talking about two separate things - filtering and value replacement. They both have uses and end up being similar in nature but for filtering I'll point to this great answer.
Let's say our data frame is called df and looks like
A B
1 4 10
2 4 2
3 10 1
4 5 9
5 10 3
Column A fits your statement of a column only having values 4, 5, 10. If you wanted to replace numbers above 7 with 0, this would do it:
df["A"] = [0 if x > 7 else x for x in df["A"]]
If you read through the right-hand side it cleanly explains what it is doing. It helps to include parentheses to separate out the "what to do" with the "what you're doing it over":
df["A"] = [(0 if x > 7 else x) for x in df["A"]]
If you want to do a manipulation over multiple columns, then utilizing zip allows you to do it easily. For example, if you want the sum of columns A and B then:
df["sum"] = [x[0] + x[1] for x in zip(df["A"], df["B"])]
Take care when you overwrite data - this removes information. It's a good practice to have the transformed data in other columns so you can trace back when something inevitably goes wonky.
There is many options. One possibility for if then... is np.where
import pandas as pd
import numpy as np
df = pd.DataFrame({'x': [1, 200, 4, 5, 6, 11],
'y': [4, 5, 10, 24, 4 , 3]})
df['y'] = np.where(df['y'] > 7, 0, df['y'])
I am able to read a file 'data.csv' that looks something like this:
Time, Upper Layer Temperature,...,
0, 450,...,
...
Using pandas:
data=pd.read_table('data.csv',header=0,sep=',')
To plot, I can't seem to get syntax right:
pl.plot(data.Time,data.'Upper Layer Temperature',label="ULT")
What is the correct way to be able to plot by column name when the column name has multiple words separated by spaces?
You can use a dictionary-style access instead of attribute-style, namely data["Upper Layer Temperature"]. For example:
>>> df = pd.DataFrame({"A": [1,2,3], "B has a space": [10,20,30]})
>>> df
A B has a space
0 1 10
1 2 20
2 3 30
[3 rows x 2 columns]
>>> df["B has a space"]
0 10
1 20
2 30
Name: B has a space, dtype: int64
>>> plt.plot(df["A"], df["B has a space"], label="fred")
[<matplotlib.lines.Line2D object at 0xb75106c>]
But note, the way you're reading in your file, you're probably going to have an extra space in the column names:
>>> !cat data.csv
Time, Upper Layer Temperature,...,
0, 450,...,
>>> data=pd.read_table('data.csv',header=0,sep=',')
>>> data.columns
Index([u'Time', u' Upper Layer Temperature', u'...', u'Unnamed: 3'], dtype='object')
You might be able to use skipinitialspace=True to avoid this:
>>> data=pd.read_table('data.csv',header=0,sep=',',skipinitialspace=True)
>>> data.columns
Index([u'Time', u'Upper Layer Temperature', u'...', u'Unnamed: 3'], dtype='object')
There is a method to plot Series histograms, but is there a function to retrieve the histogram counts to do further calculations on top of it?
I keep using numpy's functions to do this and converting the result to a DataFrame or Series when I need this. It would be nice to stay with pandas objects the whole time.
If your Series was discrete you could use value_counts:
In [11]: s = pd.Series([1, 1, 2, 1, 2, 2, 3])
In [12]: s.value_counts()
Out[12]:
2 3
1 3
3 1
dtype: int64
You can see that s.hist() is essentially equivalent to s.value_counts().plot().
If it was of floats an awful hacky solution could be to use groupby:
s.groupby(lambda i: np.floor(2*s[i]) / 2).count()
Since hist and value_counts don't use the Series' index, you may as well treat the Series like an ordinary array and use np.histogram directly. Then build a Series from the result.
In [4]: s = Series(randn(100))
In [5]: counts, bins = np.histogram(s)
In [6]: Series(counts, index=bins[:-1])
Out[6]:
-2.968575 1
-2.355032 4
-1.741488 5
-1.127944 26
-0.514401 23
0.099143 23
0.712686 12
1.326230 5
1.939773 0
2.553317 1
dtype: int32
This is a really convenient way to organize the result of a histogram for subsequent computation.
To index by the center of each bin instead of the left edge, you could use bins[:-1] + np.diff(bins)/2.
If you know the number of bins you want, you can use pandas' cut function, which is now accessible via value_counts. Using the same random example:
s = pd.Series(np.random.randn(100))
s.value_counts(bins=5)
Out[55]:
(-0.512, 0.311] 40
(0.311, 1.133] 25
(-1.335, -0.512] 14
(1.133, 1.956] 13
(-2.161, -1.335] 8
Based on this answer from a related question you can get the bin edges and histogram counts as follows:
s = pd.Series(np.random.randn(100))
ax = s.hist()
for rect in dd.patches:
((x0, y0), (x1, y1)) = rect.get_bbox().get_points()
print(((x0, y0), (x1, y1)))