So I have two pandas dataframes, A and B.
A is 1000 rows x 500 columns, filled with binary values indicating either presence or absence.
B is 1024 rows x 10 columns, and is a full iteration of 0's and 1's, hence having 1024 rows.
I am trying to find which rows in A, at a particular 10 columns of A, correspond with a given row in B. I need the whole row to match up, rather than element by element.
For example, I would want
A[(A.ix[:,(1,2,3,4,5,6,7,8,9,10)==(1,0,1,0,1,0,0,1,0,0)).all(axis=1)]
To return something that rows (3,5,8,11,15) in A match up with that (1,0,1,0,1,0,0,1,0,0) row of B at those particular columns (1,2,3,4,5,6,7,8,9,10)
And I want to do this over every row in B.
The best way I could figure out to do this was:
import numpy as np
for i in B:
B_array = np.array(i)
Matching_Rows = A[(A.ix[:,(1,2,3,4,5,6,7,8,9,10)] == B_array).all(axis=1)]
Matching_Rows_Index = Matching_Rows.index
This isn't terrible for one instance, but I use it in a while loop that runs around 20,000 times; therefore, it slows it down quite a bit.
I have been messing around with DataFrame.apply to no avail. Could map work better?
I was just hoping someone saw something obviously more efficient as I am fairly new to python.
Thanks and best regards!
We can abuse the fact that both dataframes have binary values 0 or 1 by collapsing the relevant columns from A and all columns from B into 1D arrays each, when considering each row as a sequence of binary numbers that could be converted to decimal number equivalents. This should reduce the problem set considerably, which would help with performance. Now, after getting those 1D arrays, we can use np.in1d to look for matches from B in A and finally np.where on it to get the matching indices.
Thus, we would have an implementation like so -
# Setup 1D arrays corresponding to selected cols from A and entire B
S = 2**np.arange(10)
A_ID = np.dot(A[range(1,11)],S)
B_ID = np.dot(B,S)
# Look for matches that exist from B_ID in A_ID, whose indices
# would be desired row indices that have matched from B
out_row_idx = np.where(np.in1d(A_ID,B_ID))[0]
Sample run -
In [157]: # Setup dataframes A and B with rows 0, 4 in A having matches from B
...: A_arr = np.random.randint(0,2,(10,14))
...: B_arr = np.random.randint(0,2,(7,10))
...:
...: B_arr[2] = A_arr[4,1:11]
...: B_arr[4] = A_arr[4,1:11]
...: B_arr[5] = A_arr[0,1:11]
...:
...: A = pd.DataFrame(A_arr)
...: B = pd.DataFrame(B_arr)
...:
In [158]: S = 2**np.arange(10)
...: A_ID = np.dot(A[range(1,11)],S)
...: B_ID = np.dot(B,S)
...: out_row_idx = np.where(np.in1d(A_ID,B_ID))[0]
...:
In [159]: out_row_idx
Out[159]: array([0, 4])
You can use merge with reset_index - output are indexes of B which are equal in A in custom columns:
A = pd.DataFrame({'A':[1,0,1,1],
'B':[0,0,1,1],
'C':[1,0,1,1],
'D':[1,1,1,0],
'E':[1,1,0,1]})
print (A)
A B C D E
0 1 0 1 1 1
1 0 0 0 1 1
2 1 1 1 1 0
3 1 1 1 0 1
B = pd.DataFrame({'0':[1,0,1],
'1':[1,0,1],
'2':[1,0,0]})
print (B)
0 1 2
0 1 1 1
1 0 0 0
2 1 1 0
print (pd.merge(B.reset_index(),
A.reset_index(),
left_on=B.columns.tolist(),
right_on=A.columns[[0,1,2]].tolist(),
suffixes=('_B','_A')))
index_B 0 1 2 index_A A B C D E
0 0 1 1 1 2 1 1 1 1 0
1 0 1 1 1 3 1 1 1 0 1
2 1 0 0 0 1 0 0 0 1 1
print (pd.merge(B.reset_index(),
A.reset_index(),
left_on=B.columns.tolist(),
right_on=A.columns[[0,1,2]].tolist(),
suffixes=('_B','_A'))[['index_B','index_A']])
index_B index_A
0 0 2
1 0 3
2 1 1
You can do it in pandas by using loc or ix and telling it to find the rows where the ten columns are all equal. Like this:
A.loc[(A[1]==B[1]) & (A[2]==B[2]) & (A[3]==B[3]) & A[4]==B[4]) & (A[5]==B[5]) & (A[6]==B[6]) & (A[7]==B[7]) & (A[8]==B[8]) & (A[9]==B[9]) & (A[10]==B[10])]
This is quite ugly in my opinion but it will work and gets rid of the loop so it should be significantly faster. I wouldn't be surprised if someone could come up with a more elegant way of coding the same operation.
In this special case, your rows of 10 zeros and ones can be interpreted as 10 digit binaries. If B is in order, then it can be interpreted as a range from 0 to 1023. In this case, all we need to do is take A's rows in 10 column chunks and calculate what its binary equivalent is.
I'll start by defining a range of powers of two so I can do matrix multiplication with it.
twos = pd.Series(np.power(2, np.arange(10)))
Next, I'll relabel A's columns into a MultiIndex and stack to get my chunks of 10.
A = pd.DataFrame(np.random.binomial(1, .5, (1000, 500)))
A.columns = pd.MultiIndex.from_tuples(zip((A.columns / 10).tolist(), (A.columns % 10).tolist()))
A_ = A.stack(0)
A_.head()
Finally, I'll multiply A_ with twos to get integer representation of each row and unstack.
A_.dot(twos).unstack()
This is now a 1000 x 50 DataFrame where each cell represents which of B's rows we matched for that particular 10 column chunk for that particular row of A. There isn't even a need for B.
Related
I want to count the number of common elements between rows (each row has 6 elements/ columns)
My dataframe (df) looks something like this:
>>> df
Customer Number Most Frequent Called 1 Most Frequent Called 2 Most Frequent Called 3 Most Frequent Called 4 Most Frequent Called 5
0 552711620 161359852 611336215 884140437 804548991 135953430
1 561712520 186359312 666336115 855140357 899548041 134953530
2 331112180 316659812 436926115 545220357 117748041 984213530
3 873212120 196357673 331112180 565777359 174348053 554212940
4 113219540 733352993 975632166 569117345 175888077 364212923
...
I have tried this code:
connection_df = pd.DataFrame()
for i in range(len(df)):
connection_list = []
for j in range(len(df)):
intersection = set(df.iloc[i]).intersection(df.iloc[j])
connection_list.append(len(intersection))
connection_df.insert(loc=i, column = str(i), value = connection_list)
This will give me a dataframe of a form of a matrix like this:
>>> connection_df
0 1 2 3 4
0 6 0 0 0 0
1 0 6 0 0 0
2 0 0 6 1 0
3 0 0 1 6 0
4 0 0 0 0 6
This piece of code does what I want, but as I'm using loops, they are very inefficient. Potentially there will be millions of rows so I want to ask for any suggestions on optimizing these codes. Thanks.
An efficient solution consist in performing all the operation with Numpy (by converting the whole dataframe to a Numpy matrix), computing only the upper part of the matrix as the intersection between two sets is symmetric, and pre-computing all the sets.
def fastConnectionDf(df):
size = len(df)
connection_mat = np.zeros((size, size), dtype=np.int)
df_mat = df.to_numpy()
uniqueSets = [np.unique(df_mat[i]) for i in range(size)] # Precompute all the sets
for i in range(size):
connection_mat[i,i] = len(uniqueSets[i])
for j in range(i+1, size):
intersection = np.intersect1d(uniqueSets[i], uniqueSets[j], assume_unique=True)
connection_mat[i,j] = len(intersection)
connection_mat = np.maximum(connection_mat, connection_mat.T)
connection_df = pd.DataFrame(connection_mat)
return connection_df
On my machine, this solution is 28 times faster on the example dataframe (and up to 50 times faster on bigger dataframes).
Note that it is possible to improve the algorithm by:
just counting the number of intersecting elements rather than creating an array with all the items
using a more clever implementation can sort the arrays before to speed up the set intersections (see np.searchsorted)
using Numba to speed up the computation on big dataframes
The two first improvements are hard (impossible?) to perform efficiently only with Numpy, but possible with Numba although this is a bit complex to do.
I have a task that is completely driving me mad. Lets suppose we have this df:
import pandas as pd
k = {'random_col':{0:'a',1:'b',2:'c'},'isin':{0:'ES0140074008', 1:'ES0140074008ES0140074010', 2:'ES0140074008ES0140074016ES0140074024'},'n_isins':{0:1,1:2,2:3}}
k = pd.DataFrame(k)
What I want to do is to double or triple a row a number of times goberned by col n_isins which is a number obtained by dividing the lentgh of col isin didived by 12, as isins are always strings of 12 characters.
So, I need 1 time row 0, 2 times row 1 and 3 times row 2. My real numbers are up-limited by 6 so it is a hard task. I began by using booleans and slicing the col isin but that does not take me to nothing. Hopefully my explanation is good enough. Also I need the col isin sliced like this [0:11] + ' ' + [12:23]... splitting by the 'E' but I think I know how to do that, I just post it cause is the criteria that rules the number of times I have to copy each row. Thanks in advance!
I think you need numpy.repeat with loc, last remove duplicates in index by reset_index. Last for new column use custom splitting function with numpy.concatenate:
n = np.repeat(k.index, k['n_isins'])
k = k.loc[n].reset_index(drop=True)
print (k)
isin n_isins random_col
0 ES0140074008 1 a
1 ES0140074008ES0140074010 2 b
2 ES0140074008ES0140074010 2 b
3 ES0140074008ES0140074016ES0140074024 3 c
4 ES0140074008ES0140074016ES0140074024 3 c
5 ES0140074008ES0140074016ES0140074024 3 c
#https://stackoverflow.com/a/7111143/2901002
def chunks(s, n):
"""Produce `n`-character chunks from `s`."""
for start in range(0, len(s), n):
yield s[start:start+n]
s = np.concatenate(k['isin'].apply(lambda x: list(chunks(x, 12))))
df['new'] = pd.Series(s, index = df.index)
print (df)
isin n_isins random_col new
0 ES0140074008 1 a ES0140074008
1 ES0140074008ES0140074010 2 b ES0140074008
2 ES0140074008ES0140074010 2 b ES0140074010
3 ES0140074008ES0140074016ES0140074024 3 c ES0140074008
4 ES0140074008ES0140074016ES0140074024 3 c ES0140074016
5 ES0140074008ES0140074016ES0140074024 3 c ES0140074024
I have a very simple query.
I have a csv that looks like this:
ID X Y
1 10 3
2 20 23
3 21 34
And I want to add a new column called Z which is equal to 1 if X is equal to or bigger than Y, or 0 otherwise.
My code so far is:
import pandas as pd
data = pd.read_csv("XYZ.csv")
for x in data["X"]:
if x >= data["Y"]:
Data["Z"] = 1
else:
Data["Z"] = 0
You can do this without using a loop by using ge which means greater than or equal to and cast the boolean array to int using astype:
In [119]:
df['Z'] = (df['X'].ge(df['Y'])).astype(int)
df
Out[119]:
ID X Y Z
0 1 10 3 1
1 2 20 23 0
2 3 21 34 0
Regarding your attempt:
for x in data["X"]:
if x >= data["Y"]:
Data["Z"] = 1
else:
Data["Z"] = 0
it wouldn't work, firstly you're using Data not data, even with that fixed you'd be comparing a scalar against an array so this would raise a warning as it's ambiguous to compare a scalar with an array, thirdly you're assigning the entire column so overwriting the column.
You need to access the index label which your loop didn't you can use iteritems to do this:
In [125]:
for idx, x in df["X"].iteritems():
if x >= df['Y'].loc[idx]:
df.loc[idx, 'Z'] = 1
else:
df.loc[idx, 'Z'] = 0
df
Out[125]:
ID X Y Z
0 1 10 3 1
1 2 20 23 0
2 3 21 34 0
But really this is unnecessary as there is a vectorised method here
Firstly, your code is just fine. You simply capitalized your dataframe name as 'Data' instead of making it 'data'.
However, for efficient code, EdChum has a great answer above. Or another method similar to the for loop in efficiency but easier code to remember:
import numpy as np
data['Z'] = np.where(data.X >= data.Y, 1, 0)
This is a continuation of my question. Fastest way to compare rows of two pandas dataframes?
I have two dataframes A and B:
A is 1000 rows x 500 columns, filled with binary values indicating either presence or absence.
For a condensed example:
A B C D E
0 0 0 0 1 0
1 1 1 1 1 0
2 1 0 0 1 1
3 0 1 1 1 0
B is 1024 rows x 10 columns, and is a full iteration from 0 to 1023 in binary form.
Example:
0 1 2
0 0 0 0
1 0 0 1
2 0 1 0
3 0 1 1
4 1 0 0
5 1 0 1
6 1 1 0
7 1 1 1
I am trying to find which rows in A, at a particular 10 columns of A, correspond with each row of B.
Each row of A[My_Columns_List] is guaranteed to be somewhere in B, but not every row of B will match up with a row in A[My_Columns_List]
For example, I want to show that for columns [B,D,E] of A,
rows [1,3] of A match up with row [6] of B,
row [0] of A matches up with row [2] of B,
row [2] of A matches up with row [3] of B.
I have tried using:
pd.merge(B.reset_index(), A.reset_index(),
left_on = B.columns.tolist(),
right_on =A.columns[My_Columns_List].tolist(),
suffixes = ('_B','_A')))
This works, but I was hoping that this method would be faster:
S = 2**np.arange(10)
A_ID = np.dot(A[My_Columns_List],S)
B_ID = np.dot(B,S)
out_row_idx = np.where(np.in1d(A_ID,B_ID))[0]
But when I do this, out_row_idx returns an array containing all the indices of A, which doesn't tell me anything.
I think this method will be faster, but I don't know why it returns an array from 0 to 999.
Any input would be appreciated!
Also, credit goes to #jezrael and #Divakar for these methods.
I'll stick by my initial answer but maybe explain better.
You are asking to compare 2 pandas dataframes. Because of that, I'm going to build dataframes. I may use numpy, but my inputs and outputs will be dataframes.
Setup
You said we have a a 1000 x 500 array of ones and zeros. Let's build that.
A_init = pd.DataFrame(np.random.binomial(1, .5, (1000, 500)))
A_init.columns = pd.MultiIndex.from_product([range(A_init.shape[1]/10), range(10)])
A = A_init
In addition, I gave A a MultiIndex to easily group by columns of 10.
Solution
This is very similar to #Divakar's answer with one minor difference that I'll point out.
For one group of 10 ones and zeros, we can treat it as a bit array of length 8. We can then calculate what it's integer value is by taking the dot product with an array of powers of 2.
twos = 2 ** np.arange(10)
I can execute this for every group of 10 ones and zeros in one go like this
AtB = A.stack(0).dot(twos).unstack()
I stack to get a row of 50 groups of 10 into columns in order to do the dot product more elegantly. I then brought it back with the unstack.
I now have a 1000 x 50 dataframe of numbers that range from 0-1023.
Assume B is a dataframe with each row one of 1024 unique combinations of ones and zeros. B should be sorted like B = B.sort_values().reset_index(drop=True).
This is the part I think I failed at explaining last time. Look at
AtB.loc[:2, :2]
That value in the (0, 0) position, 951 means that the first group of 10 ones and zeros in the first row of A matches the row in B with the index 951. That's what you want!!! Funny thing is, I never looked at B. You know why, B is irrelevant!!! It's just a goofy way of representing the numbers from 0 to 1023. This is the difference with my answer, I'm ignoring B. Ignoring this useless step should save time.
These are all functions that take two dataframes A and B and returns a dataframe of indices where A matches B. Spoiler alert, I'll ignore B completely.
def FindAinB(A, B):
assert A.shape[1] % 10 == 0, 'Number of columns in A is not a multiple of 10'
rng = np.arange(A.shape[1])
A.columns = pd.MultiIndex.from_product([range(A.shape[1]/10), range(10)])
twos = 2 ** np.arange(10)
return A.stack(0).dot(twos).unstack()
def FindAinB2(A, B):
assert A.shape[1] % 10 == 0, 'Number of columns in A is not a multiple of 10'
rng = np.arange(A.shape[1])
A.columns = pd.MultiIndex.from_product([range(A.shape[1]/10), range(10)])
# use clever bit shifting instead of dot product with powers
# questionable improvement
return (A.stack(0) << np.arange(10)).sum(1).unstack()
I'm channelling my inner #Divakar (read, this is stuff I've learned from Divakar)
def FindAinB3(A, B):
assert A.shape[1] % 10 == 0, 'Number of columns in A is not a multiple of 10'
a = A.values.reshape(-1, 10)
a = np.einsum('ij->i', a << np.arange(10))
return pd.DataFrame(a.reshape(A.shape[0], -1), A.index)
Minimalist One Liner
f = lambda A: pd.DataFrame(np.einsum('ij->i', A.values.reshape(-1, 10) << np.arange(10)).reshape(A.shape[0], -1), A.index)
Use it like
f(A)
Timing
FindAinB3 is an order of magnitude faster
I have a dataframe of the below structure. I want to get the column numbers which has the same value (for a specific value) when i compare two rows.
1 1 0 1 1
0 1 0 1 0
0 1 0 0 1
1 0 0 0 1
0 0 0 0 0
1 0 0 0 1
So for example when I use the above sample df to compare two rows to get the columns which has 1 in it, I should get col(1) and col(3) when I compare row(0) and row(1). Similarly, when I compare row(1) and row(2), I should get col(1). I want to know if there is a more efficient solution in python.
NB: I want only the matching column numbers and also I will specify the rows to compare.
Consider the following dataframe:
import numpy as np
df = pd.DataFrame(np.random.binomial(1, 0.2, (2, 10000)))
It will be a binary matrix of size 2x10000.
np.where((df.iloc[0] * df.iloc[1]))
Or,
np.where((df.iloc[0]) & (df.iloc[1]))
returns the columns that have 1s in both rows. Multiplication seems to be faster:
%timeit np.where((df.iloc[0]) & (df.iloc[1]))
1000 loops, best of 3: 400 µs per loop
%timeit np.where((df.iloc[0] * df.iloc[1]))
1000 loops, best of 3: 269 µs per loop
Here's a simple function. You can modify it as needed, depending on how you represent your data. I'm assuming a list of lists:
df = [[1,1,0,1,1],
[0,1,0,1,0],
[0,1,0,0,1],
[1,0,0,0,1],
[0,0,0,0,0],
[1,0,0,0,1]]
def compare_rows(df,row1,row2):
"""Returns the column numbers in which both rows contain 1's"""
column_numbers = []
for i,_ in enumerate(df[0]):
if (df[row1][i] == 1) and (df[row2][i] ==1):
column_numbers.append(i)
return column_numbers
compare_rows(df,0,1) produces the output:
[1,3]