So i have dataframe which looks like this one:
>>>df = pd.DataFrame({
'id': [i for i in range(5)],
'1': ['a', 'b', 'c', 'd', 'e'],
'2': ['f', 'g', 'h', 'i', 'g']
})
>>>df
id 1 2
0 0 a f
1 1 b g
2 2 c h
3 3 d i
4 4 e g
I want to convert this dataframe to following dataframe
>>>df_concatenated
id val
1 0 a
1 1 b
1 2 c
1 3 d
1 4 e
2 0 f
2 1 g
2 2 h
2 3 i
2 4 g
One way is to pd.melt
pd.melt(df, id_vars=['id'], value_vars=['1','2']).set_index('variable',append=True)
The other is by splitting by .loc accessor and concatenating. Long but it works
res1=df.iloc[:,[0,2]]
res1.columns=['id','val']
res=df.iloc[:,:2]
res.columns=['id','val']
res2=pd.concat([res1,res])
res2
variable id value
0 1 0 a
1 1 1 b
2 1 2 c
3 1 3 d
4 1 4 e
5 2 0 f
6 2 1 g
7 2 2 h
8 2 3 i
9 2 4 g
You can try this:
df = df.rename({"1":"val"},axis=1)
df_temp = df[["id","2"]]
df_temp = df_temp.rename({"2":"val"},axis=1)
df.drop("2",axis=1,inplace=True)
out_df = pd.concat([df,df_temp],axis=0).reset_index(drop=True)
print(out_df)
output:
id val
0 0 a
1 1 b
2 2 c
3 3 d
4 4 e
5 0 f
6 1 g
7 2 h
8 3 i
9 4 g
Related
I have this simple dataframe df:
df = pd.DataFrame({'c':[1,1,1,2,2,2,2],'type':['m','n','o','m','m','n','n']})
my goal is to count values of type for each c, and then add a column with the size of c. So starting with:
In [27]: g = df.groupby('c')['type'].value_counts().reset_index(name='t')
In [28]: g
Out[28]:
c type t
0 1 m 1
1 1 n 1
2 1 o 1
3 2 m 2
4 2 n 2
the first problem is solved. Then I can also:
In [29]: a = df.groupby('c').size().reset_index(name='size')
In [30]: a
Out[30]:
c size
0 1 3
1 2 4
How can I add the size column directly to the first dataframe? So far I used map as:
In [31]: a.index = a['c']
In [32]: g['size'] = g['c'].map(a['size'])
In [33]: g
Out[33]:
c type t size
0 1 m 1 3
1 1 n 1 3
2 1 o 1 3
3 2 m 2 4
4 2 n 2 4
which works, but is there a more straightforward way to do this?
Use transform to add a column back to the orig df from a groupby aggregation, transform returns a Series with its index aligned to the orig df:
In [123]:
g = df.groupby('c')['type'].value_counts().reset_index(name='t')
g['size'] = df.groupby('c')['type'].transform('size')
g
Out[123]:
c type t size
0 1 m 1 3
1 1 n 1 3
2 1 o 1 3
3 2 m 2 4
4 2 n 2 4
Another solution with transform len:
df['size'] = df.groupby('c')['type'].transform(len)
print df
c type size
0 1 m 3
1 1 n 3
2 1 o 3
3 2 m 4
4 2 m 4
5 2 n 4
6 2 n 4
Another solution with Series.map and Series.value_counts:
df['size'] = df['c'].map(df['c'].value_counts())
print (df)
c type size
0 1 m 3
1 1 n 3
2 1 o 3
3 2 m 4
4 2 m 4
5 2 n 4
6 2 n 4
You can calculate the groupby object and use it multiple times:
g = df.groupby('c')['type']
df = g.value_counts().reset_index(name='counts')
df['size'] = g.transform('size')
or
g.value_counts().reset_index(name='counts').assign(size=g.transform('size'))
Output:
c type counts size
0 1 m 1 3
1 1 n 1 3
2 1 o 1 3
3 2 m 2 4
4 2 n 2 4
I have 2 data frames like :
df_out:
a b c d
1 1 2 1
2 1 2 3
3 1 3 5
df_fin:
a e f g
1 0 2 1
2 5 2 3
3 1 3 5
5 2 4 6
7 3 2 5
I want to get result as :
a b c d a e f g
1 1 2 1 1 0 2 1
2 1 2 3 2 5 2 3
3 1 3 5 3 1 3 5
in the other word I have two diffrent data frames that are common in one column(a), I want two compare this two columns(df_fin.a and df_out.a) and select the rows from df_fin that have the same value in column a and create new dataframe that has selected rows from df_fin and added columns from df_out ?
I think you need merge with left join:
df = pd.merge(df_out, df_fin, on='a', how='left')
print (df)
a b c d e f g
0 1 1 2 1 0 2 1
1 2 1 2 3 5 2 3
2 3 1 3 5 1 3 5
EDIT:
df1 = df_fin[df_fin['a'].isin(df_out['a'])]
df2 = df_out.join(df1.set_index('a'), on='a')
print (df2)
a b c d e f g
0 1 1 2 1 0 2 1
1 2 1 2 3 5 2 3
2 3 1 3 5 1 3 5
Given the dataframe:
df = pd.DataFrame([['foo', 123, 4, 5, 0, 1], ['foo', 123, 4, 0, 9, 1], ['bar', 33, 0, 0, 3, 5]], columns=list('ABCDEF'))
[out]:
A B C D E F
0 foo 123 4 5 0 1
1 foo 123 4 0 9 1
2 bar 33 0 0 3 5
The goal is to sum certain columns ('C', 'D', 'E', F') using other columns ('A' and 'B') as keys to achieve:
A B C D E F
0 foo 123 8 5 9 2
2 bar 33 0 0 3 5
I've tried:
df.groupby(['A', 'B']).sum()
[out]:
C D E F
A B
bar 33 0 0 3 5
foo 123 8 5 9 2
How do I change it back to the non-indexed matrix? i.e.
A B C D E F
0 foo 123 8 5 9 2
2 bar 33 0 0 3 5
You need to add .reset_index().
df.groupby(['A','B']).sum().reset_index()
A B C D E F
0 bar 33 0 0 3 5
1 foo 123 8 5 9 2
or
df.set_index(['A','B']).sum(level=[0,1]).reset_index()
A B C D E F
0 bar 33 0 0 3 5
1 foo 123 8 5 9 2
You can use parameter as_index=False for return df:
df1 = df.groupby(['A', 'B'], as_index=False).sum()
print (df1)
A B C D E F
0 bar 33 0 0 3 5
1 foo 123 8 5 9 2
I have a pandas dataframe df with columns [a, b, c, d, e, f]. I want to perform a group by on df. I can best describe what it's supposed to do in SQL:
SELECT a, b, min(c), min(d), max(e), sum(f)
FROM df
GROUP BY a, b
How do I do this group by using pandas on my dataframe df?
consider df:
a b c d e f
1 1 2 5 9 3
1 1 3 3 4 5
2 2 4 7 4 4
2 2 5 3 8 8
I expect the result to be:
a b c d e f
1 1 2 3 9 8
2 2 4 3 8 12
use agg
df = pd.DataFrame(
dict(
a=list('aaaabbbb'),
b=list('ccddccdd'),
c=np.arange(8),
d=np.arange(8),
e=np.arange(8),
f=np.arange(8),
)
)
funcs = dict(c='min', d='min', e='max', f='sum')
df.groupby(['a', 'b']).agg(funcs).reset_index()
a b c e f d
0 a c 0 1 1 0
1 a d 2 3 5 2
2 b c 4 5 9 4
3 b d 6 7 13 6
with your data
a b c e f d
0 1 1 2 9 8 3
1 2 2 4 8 12 3
I have a DataFrame with columns [A, B, C, D, E, F, G, H].
An index has been made with columns [D, G, H]:
>>> print(dgh_columns)
Index(['D', 'G', 'H'], dtype='object')
How can I retrieve the original DataFrame without the columns D, G, H ?
Is there an index subset operation?
Ideally, this would be:
df[df.index - dgh_columns]
But this doesn't seem to work
I think you can use Index.difference:
df[df.columns.difference(dgh_columns)]
Sample:
df = pd.DataFrame({'A':[1,2,3],
'B':[4,5,6],
'C':[7,8,9],
'D':[1,3,5],
'E':[7,8,9],
'F':[1,3,5],
'G':[5,3,6],
'H':[7,4,3]})
print (df)
A B C D E F G H
0 1 4 7 1 7 1 5 7
1 2 5 8 3 8 3 3 4
2 3 6 9 5 9 5 6 3
dgh_columns = pd.Index(['D', 'G', 'H'])
print (df[df.columns.difference(dgh_columns)])
A B C E F
0 1 4 7 7 1
1 2 5 8 8 3
2 3 6 9 9 5
Numpy solution with numpy.setxor1d or numpy.setdiff1d:
dgh_columns = pd.Index(['D', 'G', 'H'])
print (df[np.setxor1d(df.columns, dgh_columns)])
A B C E F
0 1 4 7 7 1
1 2 5 8 8 3
2 3 6 9 9 5
dgh_columns = pd.Index(['D', 'G', 'H'])
print (df[np.setdiff1d(df.columns, dgh_columns)])
A B C E F
0 1 4 7 7 1
1 2 5 8 8 3
2 3 6 9 9 5
use drop
df.drop(list('DGH'), axis=1)
df = pd.DataFrame({'A':[1,2,3],
'B':[4,5,6],
'C':[7,8,9],
'D':[1,3,5],
'E':[7,8,9],
'F':[1,3,5],
'G':[5,3,6],
'H':[7,4,3]})
df.drop(list('DGH'), 1)