let say I have a dataframe that looks like this:
df = pd.DataFrame(index=list('abcde'), data={'A': range(5), 'B': range(5)})
df
Out[92]:
A B
a 0 0
b 1 1
c 2 2
d 3 3
e 4 4
Asumming that this dataframe already exist, how can I simply add a level 'C' to the column index so I get this:
df
Out[92]:
A B
C C
a 0 0
b 1 1
c 2 2
d 3 3
e 4 4
I saw SO anwser like this python/pandas: how to combine two dataframes into one with hierarchical column index? but this concat different dataframe instead of adding a column level to an already existing dataframe.
-
As suggested by #StevenG himself, a better answer:
df.columns = pd.MultiIndex.from_product([df.columns, ['C']])
print(df)
# A B
# C C
# a 0 0
# b 1 1
# c 2 2
# d 3 3
# e 4 4
option 1
set_index and T
df.T.set_index(np.repeat('C', df.shape[1]), append=True).T
option 2
pd.concat, keys, and swaplevel
pd.concat([df], axis=1, keys=['C']).swaplevel(0, 1, 1)
A solution which adds a name to the new level and is easier on the eyes than other answers already presented:
df['newlevel'] = 'C'
df = df.set_index('newlevel', append=True).unstack('newlevel')
print(df)
# A B
# newlevel C C
# a 0 0
# b 1 1
# c 2 2
# d 3 3
# e 4 4
You could just assign the columns like:
>>> df.columns = [df.columns, ['C', 'C']]
>>> df
A B
C C
a 0 0
b 1 1
c 2 2
d 3 3
e 4 4
>>>
Or for unknown length of columns:
>>> df.columns = [df.columns.get_level_values(0), np.repeat('C', df.shape[1])]
>>> df
A B
C C
a 0 0
b 1 1
c 2 2
d 3 3
e 4 4
>>>
Another way for MultiIndex (appanding 'E'):
df.columns = pd.MultiIndex.from_tuples(map(lambda x: (x[0], 'E', x[1]), df.columns))
A B
E E
C D
a 0 0
b 1 1
c 2 2
d 3 3
e 4 4
I like it explicit (using MultiIndex) and chain-friendly (.set_axis):
df.set_axis(pd.MultiIndex.from_product([df.columns, ['C']]), axis=1)
This is particularly convenient when merging DataFrames with different column level numbers, where Pandas (1.4.2) raises a FutureWarning (FutureWarning: merging between different levels is deprecated and will be removed ... ):
import pandas as pd
df1 = pd.DataFrame(index=list('abcde'), data={'A': range(5), 'B': range(5)})
df2 = pd.DataFrame(index=list('abcde'), data=range(10, 15), columns=pd.MultiIndex.from_tuples([("C", "x")]))
# df1:
A B
a 0 0
b 1 1
# df2:
C
x
a 10
b 11
# merge while giving df1 another column level:
pd.merge(df1.set_axis(pd.MultiIndex.from_product([df1.columns, ['']]), axis=1),
df2,
left_index=True, right_index=True)
# result:
A B C
x
a 0 0 10
b 1 1 11
Another method, but using a list comprehension of tuples as the arg to pandas.MultiIndex.from_tuples():
df.columns = pd.MultiIndex.from_tuples([(col, 'C') for col in df.columns])
df
A B
C C
a 0 0
b 1 1
c 2 2
d 3 3
e 4 4
Related
I have been trying to melt this columns
d = {'key': [1,2,3,4,5], 'a': ['None','a', 'None','None','None'], 'b': ['None','None','b','None','None'],'c':['None','None','None','c','c']}
df = pd.DataFrame(d)
I need to look like this
key
letter
1
None
2
a
3
b
4
c
5
c
I tried:
df = pd.melt(df,id_vars=['key'], var_name = 'letters')
but i got:
key
letters
value
1
a
None
2
a
a
3
a
None
4
a
None
5
a
None
1
b
None
2
b
None
3
b
b
4
b
None
5
b
None
1
c
None
2
c
None
3
c
None
4
c
c
5
c
c
If need get first non None value per rows after key column use DataFrame.set_index with replace possible None strings with back filling missing values and selected first column by position, last use Series.reset_index:
df = (df.set_index('key')
.replace('None', np.nan)
.bfill(axis=1)
.iloc[:, 0]
.reset_index(name='letter')))
print (df)
key letter
0 1 NaN
1 2 a
2 3 b
3 4 c
4 5 c
If possible multiple non None value per rows use:
d = {'key': [1,2,3,4,5],
'a': ['None','a', 'None','None','None'],
'b': ['None','b','b','None','None'],
'c':['None','None','None','c','c']}
df = pd.DataFrame(d)
df = (df[['key']].join(df.set_index('key')
.replace('None', np.nan)
.stack()
.groupby(level=0)
.agg(','.join)
.rename('letter'), on='key'))
print (df)
key letter
0 1 NaN
1 2 a,b
2 3 b
3 4 c
4 5 c
Or:
df = (df.set_index('key')
.replace('None', np.nan)
.apply(lambda x: ','.join(x.dropna()), axis=1)
.replace('', np.nan)
.reset_index(name='letter'))
print (df)
key letter
0 1 NaN
1 2 a,b
2 3 b
3 4 c
4 5 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
I have a DataFrame which looks like this:
df:-
A B
1 a
1 a
1 b
2 c
3 d
Now using this dataFrame i want to get the following new_df:
new_df:-
item val_not_present
1 c #1 doesn't have values c and d(values not part of group 1)
1 d
2 a #2 doesn't have values a,b and d(values not part of group 2)
2 b
2 d
3 a #3 doesn't have values a,b and c(values not part of group 3)
3 b
3 c
or an individual DataFrame for each items like:
df1:
item val_not_present
1 c
1 d
df2:-
item val_not_present
2 a
2 b
2 d
df3:-
item val_not_present
3 a
3 b
3 c
I want to get all the values which are not part of that group.
You can use np.setdiff and explode:
values_b = df.B.unique()
pd.DataFrame(df.groupby("A")["B"].unique().apply(lambda x: np.setdiff1d(values_b,x)).rename("val_not_present").explode())
Output:
val_not_present
A
1 c
1 d
2 a
2 b
2 d
3 a
3 b
3 c
Another approach is using crosstab/pivot_table to get counts and then filter on where count is 0 and transform to dataframe:
m = pd.crosstab(df['A'],df['B'])
pd.DataFrame(m.where(m.eq(0)).stack().index.tolist(),columns=['A','val_not_present'])
A val_not_present
0 1 c
1 1 d
2 2 a
3 2 b
4 2 d
5 3 a
6 3 b
7 3 c
You could convert B to a categorical datatype and then compute the value counts. Categorical variables will show categories that have frequency counts of zero so you could do something like this:
df['B'] = df['B'].astype('category')
new_df = (
df.groupby('A')
.apply(lambda x: x['B'].value_counts())
.reset_index()
.query('B == 0')
.drop(labels='B', axis=1)
.rename(columns={'level_1':'val_not_present',
'A':'item'})
)
What I want to achieve is the following in Pandas:
a = [1,2,3,4]
b = ['a', 'b']
Can I create a DataFrame like:
column1 column2
'a' 1
'a' 2
'a' 3
'a' 4
'b' 1
'b' 2
'b' 3
'b' 4
Use itertools.product with DataFrame constructor:
a = [1, 2, 3, 4]
b = ['a', 'b']
from itertools import product
# pandas 0.24.0+
df = pd.DataFrame(product(b, a), columns=['column1', 'column2'])
# pandas below
# df = pd.DataFrame(list(product(b, a)), columns=['column1', 'column2'])
print (df)
column1 column2
0 a 1
1 a 2
2 a 3
3 a 4
4 b 1
5 b 2
6 b 3
7 b 4
I will put here another method, just in case someone prefers it.
full mockup below:
import pandas as pd
a = [1,2,3,4]
b = ['a', 'b']
df=pd.DataFrame([(y, x) for x in a for y in b], columns=['column1','column2'])
df
result below:
column1 column2
0 a 1
1 b 1
2 a 2
3 b 2
4 a 3
5 b 3
6 a 4
7 b 4
I have a list l=['a', 'b' ,'c']
and a dataframe with columns d,e,f and values are all numbers
How can I insert list l in my dataframe just below the columns.
Setup
df = pd.DataFrame(np.ones((2, 3), dtype=int), columns=list('def'))
l = list('abc')
df
d e f
0 1 1 1
1 1 1 1
Option 1
I'd accomplish this task by adding a level to the columns object
df.columns = pd.MultiIndex.from_tuples(list(zip(df.columns, l)))
df
d e f
a b c
0 1 1 1
1 1 1 1
Option 2
Use a dictionary comprehension passed to the dataframe constructor
pd.DataFrame({(i, j): df[i] for i, j in zip(df, l)})
d e f
a b c
0 1 1 1
1 1 1 1
But if you insist on putting it in the dataframe proper... (keep in mind, this turns the dataframe into dtype object and we lose significant computational efficiencies.)
Alternative 1
pd.DataFrame([l], columns=df.columns).append(df, ignore_index=True)
d e f
0 a b c
1 1 1 1
2 1 1 1
Alternative 2
pd.DataFrame([l] + df.values.tolist(), columns=df.columns)
d e f
0 a b c
1 1 1 1
2 1 1 1
Use pd.concat
In [1112]: df
Out[1112]:
d e f
0 0.517243 0.731847 0.259034
1 0.318821 0.551298 0.773115
2 0.194192 0.707525 0.804102
3 0.945842 0.614033 0.757389
In [1113]: pd.concat([pd.DataFrame([l], columns=df.columns), df], ignore_index=True)
Out[1113]:
d e f
0 a b c
1 0.517243 0.731847 0.259034
2 0.318821 0.551298 0.773115
3 0.194192 0.707525 0.804102
4 0.945842 0.614033 0.757389
Are you looking for append i.e
df = pd.DataFrame([[1,2,3]],columns=list('def'))
I = ['a','b','c']
ndf = df.append(pd.Series(I,index=df.columns.tolist()),ignore_index=True)
Output:
d e f
0 1 2 3
1 a b c
If you want add list to columns for MultiIndex:
df.columns = [df.columns, l]
print (df)
d e f
a b c
0 4 7 1
1 5 8 3
2 4 9 5
3 5 4 7
4 5 2 1
5 4 3 0
print (df.columns)
MultiIndex(levels=[['d', 'e', 'f'], ['a', 'b', 'c']],
labels=[[0, 1, 2], [0, 1, 2]])
If you want add list to specific position pos:
pos = 0
df1 = pd.DataFrame([l], columns=df.columns)
print (df1)
d e f
0 a b c
df = pd.concat([df.iloc[:pos], df1, df.iloc[pos:]], ignore_index=True)
print (df)
d e f
0 a b c
1 4 7 1
2 5 8 3
3 4 9 5
4 5 4 7
5 5 2 1
6 4 3 0
But if append this list to numeric dataframe, get mixed types - numeric with strings, so some pandas functions should failed.
Setup:
df = pd.DataFrame({'d':[4,5,4,5,5,4],
'e':[7,8,9,4,2,3],
'f':[1,3,5,7,1,0]})
print (df)