I have following data frame:
1 A a
1 A b
2 B c
1 A d
How do I append all the values of a row with same values to data frame:
1 A a,c,d
2 B c
You can use groupby and apply function join :
df.columns = ['a','b','c']
print (df)
a b c
0 1 A a
1 1 A b
2 2 B c
3 1 A d
print (df.groupby(['a', 'b'])['c'].apply(', '.join).reset_index())
a b c
0 1 A a, b, d
1 2 B c
Or if first column is index:
df.columns = ['a','b']
print (df)
a b
1 A a
1 A b
2 B c
1 A d
df1 = df.b.groupby([df.index, df.a]).apply(', '.join).reset_index(name='c')
df1.columns = ['a','b','c']
print (df1)
a b c
0 1 A a, b, d
1 2 B c
Related
I have the following example data set
A
B
C
D
foo
0
1
1
bar
0
0
1
baz
1
1
0
How could extract the column names of each 1 occurrence in a row and put that into another column E so that I get the following table:
A
B
C
D
E
foo
0
1
1
C, D
bar
0
0
1
D
baz
1
1
0
B, C
Note that there can be more than two 1s per row.
You can use DataFrame.dot.
df['E'] = df[['B', 'C', 'D']].dot(df.columns[1:] + ', ').str.rstrip(', ')
df
A B C D E
0 foo 0 1 1 C, D
1 bar 0 0 1 D
2 baz 1 1 0 B, C
Inspired by jezrael's answer in this post.
Another way is that you can convert each row to boolean and use it as a selection mask to filter the column names.
cols = pd.Index(['B', 'C', 'D'])
df['E'] = df[cols].astype('bool').apply(lambda row: ", ".join(cols[row]), axis=1)
df
A B C D E
0 foo 0 1 1 C, D
1 bar 0 0 1 D
2 baz 1 1 0 B, C
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
From Pandas Dataframe, how to get the index of all ones at the row level?
My data frame has around a hundred columns. here is an example:
a b c d
0 1 0 1 0
1 0 0 0 1
2 1 1 0 1
3 1 1 0 0
4 1 1 1 1
The expected result is
0 a,c
1 d
2 a,b,d
3 a,b
4 a,b,c,d
I found this question on stackoverflow
index of non "NaN" values in Pandas
but it works at the column level
Thanks in advance.
If there are only 1 and 0 values use DataFrame.dot for matrix multiplication with columns names and separator, last remove separator with Series.str.rstrip:
df['e'] = df.dot(df.columns + ', ').str.rstrip(', ')
#if exist another values like 0,1 and compare 1
#df['e'] = df.eq(1).dot(df.columns + ', ').str.rstrip(', ')
print (df)
a b c d e
0 1 0 1 0 a, c
1 0 0 0 1 d
2 1 1 0 1 a, b, d
3 1 1 0 0 a, b
4 1 1 1 1 a, b, c, d
Also for Series use:
s = df.dot(df.columns + ', ').str.rstrip(', ')
print (s)
0 a, c
1 d
2 a, b, d
3 a, b
4 a, b, c, d
dtype: object
Try:
df=df.stack()
df=df.loc[df.eq(1)].reset_index(level=1).groupby(level=0).agg(', '.join)
Outputs:
level_1
0 a, c
1 d
2 a, b, d
3 a, b
4 a, b, c, d
I have a pandas dataframe like this below:
A B C
a b c
d e f
where A B and C are column names. Now i have a list:
mylist = [1,2,3]
I want to replace the c in column C with list such as dataframe expands for all value of list, like below:
A B C
a b 1
a b 2
a b 3
d e f
Any help would be appreciated!
I tried this,
mylist = [1,2,3]
x=pd.DataFrame({'mylist':mylist})
x['C']='c'
res= pd.merge(df,x,on=['C'],how='left')
res['mylist']=res['mylist'].fillna(res['C'])
For further,
del res['C']
res.rename(columns={"mylist":"C"},inplace=True)
print res
Output:
A B C
0 a b 1
1 a b 2
2 a b 3
3 d e f
You can use:
print (df)
A B C
0 a b c
1 d e f
2 a b c
3 t e w
mylist = [1,2,3]
idx1 = df.index[df.C == 'c']
df = df.loc[idx1.repeat(len(mylist))].assign(C=mylist * len(idx1)).append(df[df.C != 'c'])
print (df)
A B C
0 a b 1
0 a b 2
0 a b 3
2 a b 1
2 a b 2
2 a b 3
1 d e f
3 t e w
Suppose I have a dataframe df such as
A B C
0 a 1
1 b 1
2 c 2
I would like to return B, and C when C==1, like so
B C
a 1
b 1
I've got as far as df.B[df.C==1], which returns
B
a
b
which is right (countwise) but wrong in the slice. How do I get C?
You can use loc or query:
print df.loc[df.C==1, ['B','C']]
B C
0 a 1
1 b 1
print df[['B','C']].query('C == 1')
B C
0 a 1
1 b 1
Or if you need only column C:
print df.loc[df.C==1, 'C']
0 1
1 1
Name: C, dtype: int64