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Pandas vlookup like operation to a list

I am unable to properly explain my requirement, but I can show the expected result.
I have a dataframe that looks like so:
Series1
Series2
1370307
1370306
927092
927091
925392
925391
925390
925389
2344089
2344088
1827855
1827854
1715793
1715792
2356467
2356466
1463264
1463263
1712684
1712683
actual dataframe size: 902811 rows × 2 columns
then another dataframe of unique values of Series2. This I've done using value counts.
df2 = df['Series2'].value_counts().rename_axis('Series2').to_frame('counts').reset_index()
Then I need a list of matching Series1 values for each Series2 value:
The expected result is:
Series2
counts
Series1_List
2543113
6
[2543114, 2547568, 2559207, 2563778, 2564330, 2675803]
2557212
6
[2557213, 2557301, 2559192, 2576080, 2675693, 2712790]
2432032
5
[2432033, 2444169, 2490928, 2491392, 2528056]
2559269
5
[2559270, 2576222, 2588034, 2677710, 2713207]
2439554
5
[2439555, 2441882, 2442272, 2443590, 2443983]
2335180
5
[2335181, 2398282, 2527060, 2527321, 2565487]
2494111
4
[2494112, 2495321, 2526026, 2528492]
2559195
4
[2559196, 2570172, 2634537, 2675718]
2408775
4
[2408776, 2409117, 2563765, 2564320]
2408773
4
[2408774, 2409116, 2563764, 2564319]
I achieve this (although only for a subset of 50 rows) using the following code:
df2.loc[:50,'Series1_List'] = df2.loc[:50,'Series2'].apply(lambda x: df[df['Series2']==x]['Series1'].tolist())
If I do this for the whole dataframe it wouldn't complete even in 20 minutes.
So the question is whether there is a faster and efficient method of achieving the result?

IIUC, use:
df2 = (df.groupby('Series2', as_index=False)
.agg(counts=('Series1', 'count'), Series1_List=('Series1', list))
)

Pandas to modify values in csv file based on function

I have a CSV file that looks like below, this is same like my last question but this is by using Pandas.
Group Sam Dan Bori Son John Mave
A 0.00258844 0.983322 1.61479 1.2785 1.96963 10.6945
B 0.0026034 0.983305 1.61198 1.26239 1.9742 10.6838
C 0.0026174 0.983294 1.60913 1.24543 1.97877 10.6729
D 0.00263062 0.983289 1.60624 1.22758 1.98334 10.6618
E 0.00264304 0.98329 1.60332 1.20885 1.98791 10.6505
I have a function like below
def getnewno(value):
value = value + 30
if value > 40 :
value = value - 20
else:
value = value
return value
I want to send all these values to the getnewno function and get a newvalue and update the CSV file. How can this be accomplished in Pandas.
Expected output:
Group Sam Dan Bori Son John Mave
A 30.00258844 30.983322 31.61479 31.2785 31.96963 20.6945
B 30.0026034 30.983305 31.61198 31.26239 31.9742 20.6838
C 30.0026174 30.983294 31.60913 31.24543 31.97877 20.6729
D 30.00263062 30.983289 31.60624 31.22758 31.98334 20.6618
E 30.00264304 30.98329 31.60332 31.20885 31.98791 20.6505

The following should give you what you desire.
Applying a function
Your function can be simplified and here expressed as a lambda function.
It's then a matter of applying your function to all of the columns. There are a number of ways to do so. The first idea that comes to mind is to loop over df.columns. However, we can do better than this by using the applymap or transform methods:
import pandas as pd
# Read in the data from file
df = pd.read_csv('data.csv',
sep='\s+',
index_col=0)
# Simplified function with which to transform data
getnewno = lambda value: value + 10 if value > 10 else value + 30
# Looping over columns
#for col in df.columns:
# df[col] = df[col].apply(getnewno)
# Apply to all columns without loop
df = df.applymap(getnewno)
# Write out updated data
df.to_csv('data_updated.csv')
Using broadcasting
You can achieve your result using broadcasting and a little boolean logic. This avoids looping over any columns, and should ultimately prove faster and less memory intensive (although if your dataset is small any speed-up would be negligible):
import pandas as pd
df = pd.read_csv('data.csv',
sep='\s+',
index_col=0)
df += 30
make_smaller = df > 40
df[make_smaller] -= 20

First of all, your getnewno function looks too complicated... it can be simplified to e.g.:
def getnewno(value):
if value + 30 > 40:
return value - 20
else:
return value
you can even change value + 30 > 40 to value > 10.
Or even a oneliner if you want:
getnewno = lambda value: value-20 if value > 10 else value
Having the function you can apply it to specific values/columns. For example, if want you to create a column Mark_updated basing on Mark column, it should look like this (I assume your pandas DataFrame is called df):
df['Mark_updated'] = df['Mark'].apply(getnewno)

Use the mask function to do an if-else solution, before writing the data to csv
res = (df
.select_dtypes('number')
.add(30)
#the if-else comes in here
#if any entry in the dataframe is greater than 40, subtract 20 from it
#else leave as is
.mask(lambda x: x>40, lambda x: x.sub(20))
)
#insert the group column back
res.insert(0,'Group',df.Group.array)
write to csv
res.to_csv(filename)
Group Sam Dan Bori Son John Mave
0 A 30.002588 30.983322 31.61479 31.27850 31.96963 20.6945
1 B 30.002603 30.983305 31.61198 31.26239 31.97420 20.6838
2 C 30.002617 30.983294 31.60913 31.24543 31.97877 20.6729
3 D 30.002631 30.983289 31.60624 31.22758 31.98334 20.6618
4 E 30.002643 30.983290 31.60332 31.20885 31.98791 20.6505

Optimizing cartesian product between two Pandas Dataframe

I have two dataframes with the same columns:
Dataframe 1:
attr_1 attr_77 ... attr_8
userID
John 1.2501 2.4196 ... 1.7610
Charles 0.0000 1.0618 ... 1.4813
Genarito 2.7037 4.6707 ... 5.3583
Mark 9.2775 6.7638 ... 6.0071
Dataframe 2:
attr_1 attr_77 ... attr_8
petID
Firulais 1.2501 2.4196 ... 1.7610
Connie 0.0000 1.0618 ... 1.4813
PopCorn 2.7037 4.6707 ... 5.3583
I want to generate a correlation and p-value dataframe of all posible combinations, this would be the result:
userId petID Correlation p-value
0 John Firulais 0.091447 1.222927e-02
1 John Connie 0.101687 5.313359e-03
2 John PopCorn 0.178965 8.103919e-07
3 Charles Firulais -0.078460 3.167896e-02
The problem is that the cartesian product generates more than 3 million tuples. Taking minutes to finish. This is my code, I've written two alternatives:
First of all, initial DataFrames:
df1 = pd.DataFrame({
'userID': ['John', 'Charles', 'Genarito', 'Mark'],
'attr_1': [1.2501, 0.0, 2.7037, 9.2775],
'attr_77': [2.4196, 1.0618, 4.6707, 6.7638],
'attr_8': [1.7610, 1.4813, 5.3583, 6.0071]
}).set_index('userID')
df2 = pd.DataFrame({
'petID': ['Firulais', 'Connie', 'PopCorn'],
'attr_1': [1.2501, 0.0, 2.7037],
'attr_77': [2.4196, 1.0618, 4.6707],
'attr_8': [1.7610, 1.4813, 5.3583]
}).set_index('petID')
Option 1:
# Pre-allocate space
df1_keys = df1.index
res_row_count = len(df1_keys) * df2.values.shape[0]
genes = np.empty(res_row_count, dtype='object')
mature_mirnas = np.empty(res_row_count, dtype='object')
coff = np.empty(res_row_count)
p_value = np.empty(res_row_count)
i = 0
for df1_key in df1_keys:
df1_values = df1.loc[df1_key, :].values
for df2_key in df2.index:
df2_values = df2.loc[df2_key, :]
pearson_res = pearsonr(df1_values, df2_values)
users[i] = df1_key
pets[i] = df2_key
coff[i] = pearson_res[0]
p_value[i] = pearson_res[1]
i += 1
# After loop, creates the resulting Dataframe
return pd.DataFrame(data={
'userID': users,
'petID': pets,
'Correlation': coff,
'p-value': p_value
})
Option 2 (slower), from here:
# Makes a merge between all the tuples
def df_crossjoin(df1_file_path, df2_file_path):
df1, df2 = prepare_df(df1_file_path, df2_file_path)
df1['_tmpkey'] = 1
df2['_tmpkey'] = 1
res = pd.merge(df1, df2, on='_tmpkey').drop('_tmpkey', axis=1)
res.index = pd.MultiIndex.from_product((df1.index, df2.index))
df1.drop('_tmpkey', axis=1, inplace=True)
df2.drop('_tmpkey', axis=1, inplace=True)
return res
# Computes Pearson Coefficient for all the tuples
def compute_pearson(row):
values = np.split(row.values, 2)
return pearsonr(values[0], values[1])
result = df_crossjoin(mrna_file, mirna_file).apply(compute_pearson, axis=1)
Is there a faster way to solve such a problem with Pandas? Or I'll have no more option than parallelize the iterations?
Edit:
As the size of the dataframe increases the second option results in a better runtime, but It's still taking seconds to finish.
Thanks in advance

Of all the alternatives tested, the one that gave me the best results was the following:
An iteration product was made with
itertools.product().
All the iterations on both iterrows were performed on a Pool of
parallel processes (using a map function).
To give it a little more performance, the function compute_row_cython was compiled with Cython as it is advised in this section of the Pandas documentation:
In the cython_modules.pyx file:
from scipy.stats import pearsonr
import numpy as np
def compute_row_cython(row):
(df1_key, df1_values), (df2_key, df2_values) = row
cdef (double, double) pearsonr_res = pearsonr(df1_values.values, df2_values.values)
return df1_key, df2_key, pearsonr_res[0], pearsonr_res[1]
Then I set up the setup.py:
from distutils.core import setup
from Cython.Build import cythonize
setup(name='Compiled Pearson',
ext_modules=cythonize("cython_modules.pyx")
Finally I compiled it with: python setup.py build_ext --inplace
The final code was left, then:
import itertools
import multiprocessing
from cython_modules import compute_row_cython
NUM_CORES = multiprocessing.cpu_count() - 1
pool = multiprocessing.Pool(NUM_CORES)
# Calls to Cython function defined in cython_modules.pyx
res = zip(*pool.map(compute_row_cython, itertools.product(df1.iterrows(), df2.iterrows()))
pool.close()
end_values = list(res)
pool.join()
Neither Dask, nor the merge function with the apply used gave me better results. Not even optimizing the apply with Cython. In fact, this alternative with those two methods gave me memory error, when implementing the solution with Dask I had to generate several partitions, which degraded the performance as it had to perform many I/O operations.
The solution with Dask can be found in my other question.

Here's another method using same cross join but using the built in pandas method DataFrame.corrwith and scipy.stats.ttest_ind. Since we use less "loopy" implementation, this should perform better.
from scipy.stats import ttest_ind
mrg = df1.assign(key=1).merge(df2.assign(key=1), on='key').drop(columns='key')
x = mrg.filter(like='_x').rename(columns=lambda x: x.rsplit('_', 1)[0])
y = mrg.filter(like='_y').rename(columns=lambda x: x.rsplit('_', 1)[0])
df = mrg[['userID', 'petID']].join(x.corrwith(y, axis=1).rename('Correlation'))
df['p_value'] = ttest_ind(x, y, axis=1)[1]
userID petID Correlation p_value
0 John Firulais 1.000000 1.000000
1 John Connie 0.641240 0.158341
2 John PopCorn 0.661040 0.048041
3 Charles Firulais 0.641240 0.158341
4 Charles Connie 1.000000 1.000000
5 Charles PopCorn 0.999660 0.020211
6 Genarito Firulais 0.661040 0.048041
7 Genarito Connie 0.999660 0.020211
8 Genarito PopCorn 1.000000 1.000000
9 Mark Firulais -0.682794 0.006080
10 Mark Connie -0.998462 0.003865
11 Mark PopCorn -0.999569 0.070639

Applying function to every cell in a Dataframe based on index and col

I have a pandas dataframe with a format exactly like the one in this question and I'm trying to achieve the same result. In my case, I am calculating the fuzz-ratio between the row's index and it's corresponding col.
If I try this code (based on the answer to the linked question)
def get_similarities(x):
return x.index + x.name
test_df = test_df.apply(get_similarities)
the concatenation of the row index and col name happens cell-wise, just as intended. Running type(test_df) returns pandas.core.frame.DataFrame, as expected.
However, if I adapt the code to my scenario like so
def get_similarities(x):
return fuzz.partial_ratio(x.index, x.name)
test_df = test_df.apply(get_similarities)
it doesn't work. Instead of a dataframe, I get back a series (the return type of that function is an int)
I don't understand why the two samples would not behave the same nor how to fix my code so it returns a dataframe, with the fuzzy.ratio for each cell between the a row's index for that cell and the col name for that cell.

what about the following approach?
assuming that we have two sets of strings:
In [245]: set1
Out[245]: ['car', 'bike', 'sidewalk', 'eatery']
In [246]: set2
Out[246]: ['walking', 'caring', 'biking', 'eating']
Solution:
In [247]: from itertools import product
In [248]: res = np.array([fuzz.partial_ratio(*tup) for tup in product(set1, set2)])
In [249]: res = pd.DataFrame(res.reshape(len(set1), -1), index=set1, columns=set2)
In [250]: res
Out[250]:
walking caring biking eating
car 33 100 0 33
bike 25 25 75 25
sidewalk 73 20 22 36
eatery 17 33 0 50

There is a way to accomplish this via DataFrame.apply with some row manipulations.
Assuming the 'test_df` is as follows:
In [73]: test_df
Out[73]:
walking caring biking eating
car carwalking carcaring carbiking careating
bike bikewalking bikecaring bikebiking bikeeating
sidewalk sidewalkwalking sidewalkcaring sidewalkbiking sidewalkeating
eatery eaterywalking eaterycaring eaterybiking eateryeating
In [74]: def get_ratio(row):
...: return row.index.to_series().apply(lambda x: fuzz.partial_ratio(x,
...: row.name))
...:
In [75]: test_df.apply(get_ratio)
Out[75]:
walking caring biking eating
car 33 100 0 33
bike 25 25 75 25
sidewalk 73 20 22 36
eatery 17 33 0 50

It took some digging, but I figured it out. The problem comes from the fact that DataFrame.apply is either applied column-wise or row-wise, not cell by cell. So your get_similarities function is actually getting access to an entire row or column of data at a time! By default it gets the entire column -- so to solve your problem, you just have to make a get_similarities function that returns a list where you manually call fuzz.partial_ratio on each element, like this:
import pandas as pd
from fuzzywuzzy import fuzz
def get_similarities(x):
l = []
for rname in x.index:
print "Getting ratio for %s and %s" % (rname, x.name)
score = fuzz.partial_ratio(rname,x.name)
print "Score %s" % score
l.append(score)
print len(l)
print
return l
a = pd.DataFrame([[1,2],[3,4]],index=['apple','banana'], columns=['aple','banada'])
c = a.apply(get_similarities,axis=0)
print c
print type(c)
I left my print statements in their so you can see what the DataFrame.apply call is doing for yourself -- that's when it clicked for me.

Operating on pandas dataframes that may or may not be multiIndex

I have a few functions that make new columns in a pandas dataframe, as a function of existing columns in the dataframe. I have two different scenarios that occur here: (1) the dataframe is NOT multiIndex and has a set of columns, say [a,b] and (2) the dataframe is multiIndex and now has the same set of columns headers repeated N times, say [(a,1),(b,1),(a,2),(b,2)....(a,N),(n,N)].
I've been making the aforementioned functions in the style shown below:
def f(df):
if multiindex(df):
for s df[a].columns:
df[c,s] = someFunction(df[a,s], df[b,s])
else:
df[c] = someFunction(df[a], df[b])
Is there another way to do this, without having these if-multi-index/else statement everywhere and duplicating the someFunction code? I'd prefer NOT to split the multi indexed frame into N smaller dataframes (I often need to filter data or do things and keep the rows consistent across all the 1,2,...N frames, and keeping them together in one frame seems the to be the best way to do that).

you may still have to test if columns is a MultiIndex but this should be cleaner and more efficient. Caveat, this will not work if your function utilizes summary statistics on the column. For example, if someFunction divides by the the average of column 'a'.
Solution
def someFunction(a, b):
return a + b
def f(df):
df = df.copy()
ismi = isinstance(df.columns, pd.MultiIndex)
if ismi:
df = df.stack()
df['c'] = someFunction(df['a'], df['a'])
if ismi:
df = df.unstack()
return df
Setup
import pandas as pd
import numpy as np
setup_tuples = []
for c in ['a', 'b']:
for i in ['one', 'two', 'three']:
setup_tuples.append((c, i))
columns = pd.MultiIndex.from_tuples(setup_tuples)
rand_array = np.random.rand(10, len(setup_tuples))
df = pd.DataFrame(rand_array, columns=columns)
df looks like this
a b
one two three one two three
0 0.282834 0.490313 0.201300 0.140157 0.467710 0.352555
1 0.838527 0.707131 0.763369 0.265170 0.452397 0.968125
2 0.822786 0.785226 0.434637 0.146397 0.056220 0.003197
3 0.314795 0.414096 0.230474 0.595133 0.060608 0.900934
4 0.334733 0.118689 0.054299 0.237786 0.658538 0.057256
5 0.993753 0.552942 0.665615 0.336948 0.788817 0.320329
6 0.310809 0.199921 0.158675 0.059406 0.801491 0.134779
7 0.971043 0.183953 0.723950 0.909778 0.103679 0.695661
8 0.755384 0.728327 0.029720 0.408389 0.808295 0.677195
9 0.276158 0.978232 0.623972 0.897015 0.253178 0.093772
I constructed df to have MultiIndex columns. What I'd do is use the .stack() method to push the second level of the column index to be the second level of the row index.
df.stack() looks like this
a b
0 one 0.282834 0.140157
three 0.201300 0.352555
two 0.490313 0.467710
1 one 0.838527 0.265170
three 0.763369 0.968125
two 0.707131 0.452397
2 one 0.822786 0.146397
three 0.434637 0.003197
two 0.785226 0.056220
3 one 0.314795 0.595133
three 0.230474 0.900934
two 0.414096 0.060608
4 one 0.334733 0.237786
three 0.054299 0.057256
two 0.118689 0.658538
5 one 0.993753 0.336948
three 0.665615 0.320329
two 0.552942 0.788817
6 one 0.310809 0.059406
three 0.158675 0.134779
two 0.199921 0.801491
7 one 0.971043 0.909778
three 0.723950 0.695661
two 0.183953 0.103679
8 one 0.755384 0.408389
three 0.029720 0.677195
two 0.728327 0.808295
9 one 0.276158 0.897015
three 0.623972 0.093772
two 0.978232 0.253178
Now you can operate on df.stack() as if the columns were not a MultiIndex
Demonstration
print f(df)
will give you what you want
a b c \
one three two one three two one
0 0.282834 0.201300 0.490313 0.140157 0.352555 0.467710 0.565667
1 0.838527 0.763369 0.707131 0.265170 0.968125 0.452397 1.677055
2 0.822786 0.434637 0.785226 0.146397 0.003197 0.056220 1.645572
3 0.314795 0.230474 0.414096 0.595133 0.900934 0.060608 0.629591
4 0.334733 0.054299 0.118689 0.237786 0.057256 0.658538 0.669465
5 0.993753 0.665615 0.552942 0.336948 0.320329 0.788817 1.987507
6 0.310809 0.158675 0.199921 0.059406 0.134779 0.801491 0.621618
7 0.971043 0.723950 0.183953 0.909778 0.695661 0.103679 1.942086
8 0.755384 0.029720 0.728327 0.408389 0.677195 0.808295 1.510767
9 0.276158 0.623972 0.978232 0.897015 0.093772 0.253178 0.552317
three two
0 0.402600 0.980626
1 1.526739 1.414262
2 0.869273 1.570453
3 0.460948 0.828193
4 0.108599 0.237377
5 1.331230 1.105884
6 0.317349 0.399843
7 1.447900 0.367907
8 0.059439 1.456654
9 1.247944 1.956464