Develop Reference

Join an array to every row in the pandas dataframe

I have a data frame and an array as follows:
df = pd.DataFrame({'x': range(0,5), 'y' : range(1,6)})
s = np.array(['a', 'b', 'c'])
I would like to attach the array to every row of the data frame, such that I got a data frame as follows:
What would be the most efficient way to do this?

Just plain assignment:
# replace the first `s` with your desired column names
df[s] = [s]*len(df)

Try this:
for i in s:
df[i] = i
Output:
x y a b c
0 0 1 a b c
1 1 2 a b c
2 2 3 a b c
3 3 4 a b c
4 4 5 a b c

You could use pandas.concat:
pd.concat([df, pd.DataFrame(s).T], axis=1).ffill()
output:
x y 0 1 2
0 0 1 a b c
1 1 2 a b c
2 2 3 a b c
3 3 4 a b c
4 4 5 a b c

You can try using df.loc here.
df.loc[:, s] = s
print(df)
x y a b c
0 0 1 a b c
1 1 2 a b c
2 2 3 a b c
3 3 4 a b c
4 4 5 a b c

Pandas row replication python

So my dataframe has multiple columns, one of them is named "multiple" which contains boolean, only 1s and 0s. Now, I want to replicate all the rows 4 times only for all the df.loc[df.multiple==1]. How can I do that? (I don't want to replicate indexes)
example input:
df=
index strings multiple
0 A 0
1 B 1
2 C 1
3 D 0
4 E 1
Expected output:
index strings multiple
0 A 0
1 B 1
2 B 1
3 B 1
4 B 1
5 B 1
6 C 1
7 C 1
8 C 1
9 C 1
10 C 1
11 D 0
12 E 1
13 E 1
14 E 1
15 E 1
16 E 1

Here is another alternative, based on #Vinzent answer.
It is using the same approach to construct the repeats, but doesn't require to reconstruct the full dataframe. It is instead based on indexing. This solution is ~30% faster on the provided dataset and larger datasets.
df.loc[np.repeat(df.multiple, df.multiple.values*4+1).index].reset_index(drop=True)

This is what numpy.repeat is for:
import pandas as pd
import numpy as np
df = pd.DataFrame([['A', 0],
['B', 1],
['C', 1],
['D', 0],
['E', 1]],
columns=['strings', 'multiple'])
df = pd.DataFrame(np.repeat(df.values, df['multiple']*4+1, axis=0), columns=df.columns)
print(df)
# strings multiple
# 0 A 0
# 1 B 1
# 2 B 1
# 3 B 1
# 4 B 1
# 5 B 1
# 6 C 1
# 7 C 1
# 8 C 1
# 9 C 1
# 10 C 1
# 11 D 0
# 12 E 1
# 13 E 1
# 14 E 1
# 15 E 1
# 16 E 1

You can do it with pandas:
(df.groupby('multiple')
.apply(lambda x: pd.concat([x]*4) if x.name else x)
.droplevel(level=0)
.sort_index()
.reset_index(drop=True)
)

Create adjacency matrix from adjacency list

I have the next DF with two columns
A x
A y
A z
B x
B w
C x
C w
C i
I want to produce an adjacency matrix like this (count the intersection)
A B C
A 0 1 2
B 1 0 2
C 2 2 0
I have the next code but doesnt work:
import pandas as pd
df = pd.read_csv('lista.csv')
drugs = pd.read_csv('drugs.csv')
drugs = drugs['Drug'].tolist()
df = pd.crosstab(df.Drug, df.Gene)
df = df.reindex(index=drugs, columns=drugs)
How can i obtain the adjacency matrix?
Thanks

Try self merge on column 2 and then crosstab:
s = df.merge(df,on='col2').query('col1_x != col1_y')
pd.crosstab(s['col1_x'], s['col1_y'])
Output:
col1_y A B C
col1_x
A 0 1 1
B 1 0 2
C 1 2 0

Input:
>>> drugs
Drug Gene
0 A x
1 A y
2 A z
3 B x
4 B w
5 C x
6 C w
7 C i
Merge on gene before crosstab and fill diagonal with zeros
df = pd.merge(drugs, drugs, on="Gene")
df = pd.crosstab(df["Drug_x"], df["Drug_y"])
np.fill_diagonal(df.values, 0)
Output:
>>> df
Drug_y A B C
Drug_x
A 0 1 1
B 1 0 2
C 1 2 0

Convert N by N Dataframe to 3 Column Dataframe

I am using Python 2.7 with Pandas on a Windows 10 machine.
I have an n by n Dataframe where:
1) The index represents peoples names
2) The column headers are the same peoples names in the same order
3) Each cell of the Dataframeis the average number of times they email each other each day.
How would I transform that Dataframeinto a Dataframewith 3 columns, where:
1) Column 1 would be the index of the n by n Dataframe
2) Column 2 would be the row headers of the n by n Dataframe
3) Column 3 would be the cell value corresponding to those two names from the index, column header combination from the n by n Dataframe
Edit
Appologies for not providing an example of what I am looking for. I would like to take df1 and turn it into rel_df, from the code below.
import pandas as pd
from itertools import permutations
df1 = pd.DataFrame()
df1['index'] = ['a', 'b','c','d','e']
df1.set_index('index', inplace = True)
df1['a'] = [0,1,2,3,4]
df1['b'] = [1,0,2,3,4]
df1['c'] = [4,1,0,3,4]
df1['d'] = [5,1,2,0,4]
df1['e'] = [7,1,2,3,0]
##df of all relationships to build
flds = pd.Series(SO_df.fld1.unique())
flds = pd.Series(flds.append(pd.Series(SO_df.fld2.unique())).unique())
combos = []
for L in range(0, len(flds)+1):
for subset in permutations(flds, L):
if len(subset) == 2:
combos.append(subset)
if len(subset) > 2:
break
rel_df = pd.DataFrame.from_records(data = combos, columns = ['fld1','fld2'])
rel_df['value'] = [1,4,5,7,1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4]
print df1
>>> print df1
a b c d e
index
a 0 1 4 5 7
b 1 0 1 1 1
c 2 2 0 2 2
d 3 3 3 0 3
e 4 4 4 4 0
>>> print rel_df
fld1 fld2 value
0 a b 1
1 a c 4
2 a d 5
3 a e 7
4 b a 1
5 b c 1
6 b d 1
7 b e 1
8 c a 2
9 c b 2
10 c d 2
11 c e 2
12 d a 3
13 d b 3
14 d c 3
15 d e 3
16 e a 4
17 e b 4
18 e c 4
19 e d 4

Use melt:
df1 = df1.reset_index()
pd.melt(df1, id_vars='index', value_vars=df1.columns.tolist()[1:])
(If in your actual code you're explicitly setting the index as you do here, just skip that step rather than doing the reset_index; melt doesn't work on an index.)

# Flatten your dataframe.
df = df1.stack().reset_index()
# Remove duplicates (e.g. fld1 = 'a' and fld2 = 'a').
df = df.loc[df.iloc[:, 0] != df.iloc[:, 1]]
# Rename columns.
df.columns = ['fld1', 'fld2', 'value']
>>> df
fld1 fld2 value
1 a b 1
2 a c 4
3 a d 5
4 a e 7
5 b a 1
7 b c 1
8 b d 1
9 b e 1
10 c a 2
11 c b 2
13 c d 2
14 c e 2
15 d a 3
16 d b 3
17 d c 3
19 d e 3
20 e a 4
21 e b 4
22 e c 4
23 e d 4

Convert Two column data frame to occurrence matrix in pandas

Hi all I have a csv file which contains data as the format below
A a
A b
B f
B g
B e
B h
C d
C e
C f
The first column contains items second column contains available feature from feature vector=[a,b,c,d,e,f,g,h]
I want to convert this to occurence matrix look like below
a,b,c,d,e,f,g,h
A 1,1,0,0,0,0,0,0
B 0,0,0,0,1,1,1,1
C 0,0,0,1,1,1,0,0
Can anyone tell me how to do this using pandas?

Here is another way to do it using pd.get_dummies().
import pandas as pd
# your data
# =======================
df
col1 col2
0 A a
1 A b
2 B f
3 B g
4 B e
5 B h
6 C d
7 C e
8 C f
# processing
# ===================================
pd.get_dummies(df.col2).groupby(df.col1).apply(max)
a b d e f g h
col1
A 1 1 0 0 0 0 0
B 0 0 0 1 1 1 1
C 0 0 1 1 1 0 0

Unclear if your data has a typo or not but you can crosstab for this:
In [95]:
pd.crosstab(index=df['A'], columns = df['a'])
Out[95]:
a b d e f g h
A
A 1 0 0 0 0 0
B 0 0 1 1 1 1
C 0 1 1 1 0 0
In your sample data your second column has value a as the name of that column but in your expected output it's in the column as a value
EDIT
OK I fixed your input data so it generates the correct result:
In [98]:
import pandas as pd
import io
t="""A a
A b
B f
B g
B e
B h
C d
C e
C f"""
df = pd.read_csv(io.StringIO(t), sep='\s+', header=None, names=['A','a'])
df
Out[98]:
A a
0 A a
1 A b
2 B f
3 B g
4 B e
5 B h
6 C d
7 C e
8 C f
In [99]:
ct = pd.crosstab(index=df['A'], columns = df['a'])
ct
Out[99]:
a a b d e f g h
A
A 1 1 0 0 0 0 0
B 0 0 0 1 1 1 1
C 0 0 1 1 1 0 0

This approach yields the same result in a scipy sparse coo matrix much faster
from scipy import sparse
df['col1'] = df['col1'].astype("category")
df['col2'] = df['col2'].astype("category")
df['ones'] = 1
user_items = sparse.coo_matrix((df.ones.astype(float),
(df.col1.cat.codes,
df.col2.cat.codes)))