Not sure if this is a good idea after all, but having a dictionary with arrays as values, such as
DF = {'z_eu': array([127.45064758, 150.4478288 , 150.74781189, -98.3227338 , -98.25155681, -98.24993753]),
'Process': array(['initStep', 'Transportation', 'Transportation', 'Transportation', 'Transportation', 'phot']),
'Creator': array(['SynRad', 'SynRad', 'SynRad', 'SynRad', 'SynRad', 'SynRad']) }
I need to do a selection of the numeric data (z_eu) based on values of the other two keys.
One workaround I came up with so far, was to extract the arrays and iterate through them, thereby creating another array which contains the valid data.
proc = DF['Process']; z= DF['z_eu']; creat = DF['Creator']
data = [z for z,p,c in zip(z, proc,creat) if (p == 'initStep') and c=='SynRad' ]
But somehow this seems to me as effort which can be completely avoided by dealing more intelligently with the dictionary in the first place? Also, the zip() takes a long time as well.
I know that dataframes are a valid alternative but unfortunately, since I'm dealing with strings, pandas appears to be too slow.
Any hints are most welcome!
A bit simpler, using conditional slicing you could write
data = DF['z_eu'][(DF['Process'] == 'initStep') & (DF['Creator'] == 'SynRad')]
...or still using zip, you could simplify to
data = [z for z, p, c in zip(*DF.values()) if p == 'initStep' and c == 'SynRad']
Basically also conditional slicing, using a pandas DataFrame:
df = pd.DataFrame(DF)
data = df.loc[(df['Process'] == 'initStep') & (df['Creator'] == 'SynRad'), 'z_eu']
print(data)
# 0 127.450648
# Name: z_eu, dtype: float64
In principle I'd say there's nothing wrong with handling numpy arrays in a dict. You'll have a lot of flexibility and sometimes operations are more efficient if you do them straight in numpy (you could even utilize numba for purely numerical, expensive calculations) - but if that is not needed and you're fine with basically a n*m table, pandas dfs are nice and convenient.
If your dataset is large and you want to perform many look-ups as the one shown, you might not want to perform those on strings. To improve performance, you could e.g. come up with unique IDs (integers) for each 'Process' or 'Creator' from the example. You'll just need to be able to map those back to the original strings, so keep that data as well.
You can loop through one array and via the index get the right element
z_eu = DF['z_eu']
process = DF['Process']
creator = DF['Creator']
result = []
for i in range(len(z_eu)):
if process[i] == 'initStep' and creator[i] == 'SynRad':
result.append(z_eu[i])
print(result)
I have a dataframe containing a column type of categorical data, and I have a table (dictionary) of parameter values for each possible type, each entry of which looks like
type1: [x1,x2,x3]
I have working code looking like this:
def foo(df):
[x1,x2,x3] = parameters[df.type]
return (* formula depending on x1,x2,x3,df.A,df.B *)
df['new_variable'] = df.apply(lambda x: foo(x), axis = 1)
Iterating through the rows like this (.apply(..., axis=1)) is of course very slow, and I'd like an efficient solution, but I don't know how to do the table-lookup in a neat manner. For instance, I can't just do
df['new_variable'] = (* formula depending on parameters[df.type][0:3],df.A,df.B *)
as that throws a TypeError: 'Series' objects are mutable, thus they cannot be hashed (I'm naively trying to use a Series as a key, which doesn't work).
I suppose I could make new columns for the parameter values, but that seems inelegant somehow, and I'm sure there is a better way. What's the best way to do this?
EDIT: I just realised I can get a column with the lists of parameters via
df.type.map(parameters)
but I can't access the entries of those lists, as the usual index-conventions don't seem to work. E.g. df.type.map(parameters).loc[:,2] gives an IndexingError: Too many indexers; basically pandas gets confused when having too many dimensions without sticking it all in a MultiIndex. Is there a way to get around this?
EDIT2: a minimal example:
df = pd.DataFrame([['dog',4],['dog',6],['cat',1],['cat',4]],columns = ['type','A'])
parameters = {'dog': [1,2], 'cat': [3,-1]}
def foo(x):
[a,b]=parameters[x.type]
return a * x.A + b
df['new'] = df.apply(foo,axis=1)
produces the desired output
type A new
0 dog 4 6
1 dog 6 8
2 cat 1 2
3 cat 4 11
For a vectorised solution you should split your series of lists, which is what df['type'].map(parameters) gives, into separate columns. You can then leverage efficient NumPy operations:
params = pd.DataFrame(df['type'].map(parameters).values.tolist(),
columns=['a', 'b'])
df['new'] = params['a'] * df['A'] + params['b']
As you note, pd.DataFrame.apply is a thinly veiled, and generally inefficient, loop. It should be avoided wherever possible.
Suppose we have an input array with some (but not all) nan values from which we want to write into a nan-initialized output array. After writing non-nan data into the output array there are still nan values and I don't understand at all why:
# minimal example just for testing purposes
import numpy as np
# fix state of seed
np.random.seed(1000)
# create input array and nan-filled output array
a = np.random.rand(6,3,5)
b = np.zeros((6,3,5)) * np.nan
x = [np.arange(6),1,2]
# select data in one dimension with others fixed
y_temp = a[x]
# set arbitrary index to nan
y_temp[1] = np.nan
ind_valid = ~np.isnan(y_temp)
# select non-nan values
y = y_temp[ind_valid]
# write input to output at corresponding indices
b[x][ind_valid] = y
print b[x][ind_valid]
# surprise, surprise :(
# [ nan nan nan nan nan nan]
# workaround (that will of course cost computation time, even if not much)
c = np.zeros(len(y_temp)) * np.nan
c[ind_valid] = y
b[x] = c
print b[x][ind_valid]
# and this is what we want to have
# [ 0.39719446 nan 0.39820488 0.68190824 0.86534558 0.69910395]
I thought the array b would reserve some block in memory and by indexing with x it "knows" those indices. Then it should also know them when selecting only some of them with ind_valid and be able to write into exactly those bit addresses in memory. No idea, but maybe it's sth. similar as python nested list unexpected behaviour? Please explain and maybe also provide a nice solution instead of the proposed workaround! Thanks!
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 an array of size: (50, 50). Within this array there is a slice of size (20,10).
Only this slice contains data, the remainder is all set to nan.
How do I cut this slice out of my large array?
You can get this using fancy indexing to collect the items that are not NaN:
a = a[ np.logical_not( np.isnan(a) ) ].reshape(20,10)
or, alternatively, as suggested by Joe Kington:
a = a[ ~np.isnan(a) ]
Do you know where the NaNs are? If so, something like this should work:
newarray = np.copy(oldarray[xstart:xend,ystart:yend])
where xstart and xend are the beginning and end of the slice you want in the x dimension and similarly for y. You can then delete the old array to free up memory if you don't need it anymore.
If you don't know where the NaNs are, this should do the trick:
# in this example, the starting array is A, numpy is imported as np
boolA = np.isnan(A) #get a boolean array of where the nans are
nonnanidxs = zip(*np.where(boolA == False)) #all the indices which are non NaN
#slice out the nans
corner1 = nonnanidxs[0]
corner2 = nonnanidxs[-1]
xdist = corner2[0] - corner1[0] + 1
ydist = corner2[1] - corner1[1] + 1
B = copy(A[corner1[0]:corner1[0]+xdist,corner1[1]:corner1[1]+ydist])
#B is now the array you want
Note that this would be pretty slow for large arrays because np.where looks through the whole thing. There's an open issue in the number bug tracker for a method that finds the first index equal to some value and then stops. There might be a more elegant way to do this, this is just the first thing that came to my head.
EDIT: ignore, sgpc's answer is much better.