Merge multiple binary encoded rows into one in pandas dataframe - python-3.x

I have a pandas.DataFrame that looks like this:
A B C D E F
0 0 1 0 0 0
1 1 0 0 0 0
2 0 1 0 0 0
3 0 0 0 1 0
4 0 0 1 0 0
There are several rows that share a 1 in their columns and in each row there is only one 1 present. I want to merge the rows with each other so the resulting dataFrame would onyl consist of one row, that combines all the 1s of the dataframe, like this:
A B C D E F
0 1 1 1 1 0
Is there a smart, easy way to do this with pandas?


Use DataFrame.sum, then compare for greater or equal by Series.ge and last convert to 0,1 by Series.view:
s = df.sum().ge(1).view('i1')
Another idea if 0,1 values only is use DataFrame.any with convert mask to 0,1:
s = df.any().view('i1')
print (s)
A 1
B 1
C 1
D 1
E 1
F 0
dtype: int8


We can do
df.sum().ge(1).astype(int)
Out[316]:
A 1
B 1
C 1
D 1
E 1
F 0
dtype: int32

Related

How to return all rows that have equal number of values of 0 and 1?

I have dataframe that has 50 columns each column have either 0 or 1. How do I return all rows that have an equal (tie) in the number of 0 and 1 (25 "0" and 25 "1").
An example on a 4 columns:
A B C D
1 1 0 0
1 1 1 0
1 0 1 0
0 0 0 0
based on the above example it should return the first and the third row.
A B C D
1 1 0 0
1 0 1 0

Because you have four columns, we assume you must have atleast two sets of 1 in a row. So, please try
df[df.mean(1).eq(0.5)]

Dataframe conversion

I am trying to convert a data frame into a 1,0 matrix format
data = pd.DataFrame({'Val1':['A','B','B'],
'Val2':['C','A','D'],
'Val3':['E','F','C'],
'Comb':['Comb1','Comb2','Comb3']})
data:
Val1 Val2 Val3 Comb
0 A C E Comb1
1 B A F Comb2
2 B D C Comb3
What I need is to convert to below data frame
Comb A C D E B F
0 Comb1 1 1 0 1 0 0
1 Comb2 1 0 0 0 1 1
2 Comb3 0 1 1 0 1 0
I was able to do it with a FOR loop but as my dataframe increases, the processing time increases. Is there a better way to do it?
header = set(data[['Val1','Val2','Val3']].values.ravel())
matrix = pd.DataFrame(columns=header)
for i in range(data.shape[0]):
temp_dict = {data["Val1"].iloc[i]:1, data["Val2"].iloc[i]:1, data["Val3"].iloc[i]:1}
matrix = matrix.append(temp_dict, ignore_index=True)
matrix = matrix.loc[:, matrix.columns.notnull()]
matrix = matrix.fillna(0)
matrix = pd.merge(data[["Comb"]],matrix, left_index=True, right_index=True, how= 'outer')
Thanks!

There may be a better solution, but this is what came to my mind: convert each raw to a dictionary of "present" letters, build a Series from the dictionary, and combine the Series into a dataframe.
data.loc[:, 'Val1':'Val3'].apply(lambda row:
pd.Series({letter: 1 for letter in row}), axis=1)\
.fillna(0).astype(int).join(data.Comb)
# A B C D E F Comb
#0 1 0 1 0 1 0 Comb1
#1 1 1 0 0 0 1 Comb2
#2 0 1 1 1 0 0 Comb3

There are propably multiple ways to solve this, I used pd.crosstab for it:
import pandas as pd
data = pd.DataFrame({'Val1':['A','B','B'],
'Val2':['C','A','D'],
'Val3':['E','F','C'],
'Comb':['Comb1','Comb2','Comb3']})
data["lst"] = data[['Val1', 'Val2', 'Val3']].values.tolist()
data = data.explode("lst")
print(pd.crosstab(data["Comb"], data["lst"]))
Out[20]:
lst A B C D E F
Comb
Comb1 1 0 1 0 1 0
Comb2 1 1 0 0 0 1
Comb3 0 1 1 1 0 0

I guess this will work. Please let me know if it works
pd.get_dummies(data, columns=['Val1','Val2','Val3'],prefix="",prefix_sep="").groupby(axis=1,level=0).sum()

Here's another way:
data.melt('Comb').set_index('Comb')['value'].str.get_dummies().sum(level=0).reset_index()
Output:
Comb A B C D E F
0 Comb1 1 0 1 0 1 0
1 Comb2 1 1 0 0 0 1
2 Comb3 0 1 1 1 0 0

List column name having value greater than zero

I have following dataframe
A | B | C | D
1 0 2 1
0 1 1 0
0 0 0 1
I want to add the new column have any value of row in the column greater than zero along with column name
A | B | C | D | New
1 0 2 1 A-1, C-2, D-1
0 1 1 0 B-1, C-1
0 0 0 1 D-1

We can use mask and stack
s=df.mask(df==0).stack().\
astype(int).astype(str).\
reset_index(level=1).apply('-'.join,1).add(',').sum(level=0).str[:-1]
df['New']=s
df
Out[170]:
A B C D New
0 1 0 2 1 A-1,C-2,D-1
1 0 1 1 0 B-1,C-1
2 0 0 0 1 D-1

Combine the column names with the df values that are not zero and then filter out the None values.
df = pd.read_clipboard()
arrays = np.where(df!=0, df.columns.values + '-' + df.values.astype('str'), None)
new = []
for array in arrays:
new.append(list(filter(None, array)))
df['New'] = new
df
Out[1]:
A B C D New
0 1 0 2 1 [A-1, C-2, D-1]
1 0 1 1 0 [B-1, C-1]
2 0 0 0 1 [D-1]

Excel check for value in different cells per row

Having this kind of data:
A B C D E
1 1 0 1 0 0
2 0 1 1 0 1
3 1 0 1 1 0
4 0 1 0 1 0
I would like to show true/false in column F if column A, C and E has the value of 1.
So not looking for a value in range - but different columns.

You can use the AND function, something like:
=IF(AND(A1=1,C1=1,E1=1),"TRUE","FALSE")

=IF(A2;IF(C2;IF(E2;TRUE;FALSE);FALSE);FALSE)
This will display TRUE if ALL three cells are 1, else FALSE.

Transform a Matrix into an Edge List in EXCEL

I have an excel matrix that looks like this:
ID A B C D E F G
XYZ 0 1 0 1 0 1 0
ZVC 1 0 1 0 1 0 1
...
ABC 0 1 0 1 0 1 1
I would like to transform this matrix into three columns:
XYZ A 0
XYZ B 1
XYZ C 0
...
ABC F 1
ABC G 1
What would be an efficient way to do that (possibly without macros)?

I actually found a simple solution to my problem: https://www.youtube.com/watch?v=N3wWQjRWkJc

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