Develop Reference

Spread single value in group across all other NaN values in group

In this example, we attempt to apply value in group and column to all other NaNs, that are in the same group and column.
import pandas as pd
df = pd.DataFrame({'id':[1,1,2,2,3,4,5], 'Year':[2000,2000, 2001, 2001, 2000, 2000, 2000], 'Values': [1, 3, 2, 3, 4, 5,6]})
df['pct'] = df.groupby(['id', 'Year'])['Values'].apply(lambda x: x/x.shift() - 1)
print(df)
id Year Values pct
0 1 2000 1 NaN
1 1 2000 3 2.0
2 2 2001 2 NaN
3 2 2001 3 0.5
4 3 2000 4 NaN
5 4 2000 5 NaN
6 5 2000 6 NaN
I have tried to use .ffill() to fill the NaN's within each group that contains a value. For example, the code is trying to make it so that the NaN associated with index 0, to be 2.0, and the NaN associated to index 2 to be 0.5.
df['pct'] = df.groupby(['id', 'Year'])['pct'].ffill()
print(df)
id Year Values pct
0 1 2000 1 NaN
1 1 2000 3 2.0
2 2 2001 2 NaN
3 2 2001 3 0.5
4 3 2000 4 NaN
5 4 2000 5 NaN
6 5 2000 6 NaN

It should be bfill
df['pct'] = df.groupby(['id', 'Year'])['pct'].bfill()
df
Out[109]:
id Year Values pct
0 1 2000 1 2.0
1 1 2000 3 2.0
2 2 2001 2 0.5
3 2 2001 3 0.5
4 3 2000 4 NaN
5 4 2000 5 NaN
6 5 2000 6 NaN

Create dataframe with hierarchical indices and extra columns from non-hierarchically indexed dataframe

Consider a simple dataframe:
import numpy as np
import pandas as pd
x = pd.DataFrame(np.arange(10).reshape(5,2))
print(x)
0 1
0 0 1
1 2 3
2 4 5
3 6 7
4 8 9
I would like to create a hierarchically indexed dataframe of the form:
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN
where the 'a' columns correspond to the original dataframe columns and the 'b' columns are blank (or nan).
I can certainly create a hierarchically indexed dataframe with all NaNs and loop over the columns of the original dataframe, writing them into
the new dataframe. Is there something more compact than that?

you can do with MultiIndex.from_product
extra_level = ['a', 'b']
new_cols = pd.MultiIndex.from_product([x.columns, extra_level])
x.columns = new_cols[::len(x.columns)] # select all the first element of extra_level
x = x.reindex(columns=new_cols)
print(x)
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN

Very much like #Ben.T I am using MultiIndex.from_product:
x.assign(l='a')
.set_index('l', append=True)
.unstack()
.reindex(pd.MultiIndex.from_product([x.columns.tolist(), ['a','b']]), axis=1)
Output:
0 1
a b a b
0 0 NaN 1 NaN
1 2 NaN 3 NaN
2 4 NaN 5 NaN
3 6 NaN 7 NaN
4 8 NaN 9 NaN

Retain NaN values after concatenating

I have a df where I want to apply a function. How can I retain the NaN values even after concatenating two columns? I want to avoid np.where since the real function has more elif conditions
df fruit year price vol signifiance
0 apple 2010 1 5 NaN
1 apple 2011 2 4 NaN
2 apple 2012 3 3 NaN
3 NaN 2013 3 3 NaN
4 NaN NaN NaN 3 NaN
5 apple 2015 3 3 important
df = df.fillna('')
def func(row):
if (pd.notna(row['year'])):
return row['fruit'] + row['significance'] +row['price']+ '_test'
else:
return np.NaN
df['final'] = row.apply(func, axis=1)
Expected Output
df fruit year price vol significance final
0 apple 2010 1 5 NaN apple1_test
1 apple 2011 2 4 NaN apple2_test
2 apple 2012 3 3 NaN apple3_test
3 NaN 2013 3 3 NaN 3_test
4 NaN 2014 NaN 3 NaN NaN
5 apple 2015 3 3 important appleimportant3_test

df = df.fillna('')
def func(row):
a = f"{row['fruit']}{row['significance']}{row['price']}"
if a:
return a + '_test'
return np.NaN

First remove df = df.fillna('') and then use your solution with added elif for test if missing values in both columns:
def func(row):
if (pd.notna(row['fruit'])) & (pd.notna(row['signifiance'])):
return row['fruit'] +'_' + row['signifiance']
elif (pd.isna(row['fruit'])) & (pd.isna(row['signifiance'])):
return 'apple'
else:
return row['fruit']
df['final'] = df.apply(func, axis=1)
print (df)
df fruit year price vol signifiance final
0 0 apple 2010 1 5 NaN apple
1 1 apple 2011 2 4 NaN apple
2 2 apple 2012 3 3 NaN apple
3 3 apple 2013 3 3 NaN apple
4 4 NaN 2014 3 3 NaN apple
5 5 apple 2015 3 3 important apple_important

pd.wide_to_long() lost data

I'm very new to Python. I've tried to reshape a data set using pd.wide_to_long. The original dataframe looks like this:
chk1 chk2 chk3 ... chf1 chf2 chf3 id var1 var2
0 3 4 2 ... nan nan nan 1 1 0
1 4 4 4 ... nan nan nan 2 1 0
2 2 nan nan ... 3 4 3 3 0 1
3 3 3 3 ... 3 2 2 4 1 0
I used the following code:
df2 = pd.wide_to_long(df,
stubnames=['chk', 'chf'],
i=['id', 'var1', 'var2'],
j='type')
When checking the data after these codes, it looks like this
chk chf
id var1 var2 egenskap
1 1 0 1 3 nan
2 4 nan
3 2 nan
4 nan nan
5 4 nan
6 nan nan
7 4 nan
8 4 nan
2 1 0 1 4 nan
2 4 nan
3 4 nan
4 5 nan
But when I check the columns in the new data set, it seems that all columns except 'chk' and 'chf' are gone!
df2.columns
Out[47]: Index(['chk', 'chf'], dtype='object')
df2.columns
for col in df2.columns:
print(col)
chk
chf
From the dataview it looks like 'id', 'var1', 'var2' have been merged into one common index:
Screenprint dataview here
Can someone please help me? :)

Unmelt Pandas DataFrame

I have a pandas dataframe with two id variables:
df = pd.DataFrame({'id': [1,1,1,2,2,3],
'num': [10,10,12,13,14,15],
'q': ['a', 'b', 'd', 'a', 'b', 'z'],
'v': [2,4,6,8,10,12]})
id num q v
0 1 10 a 2
1 1 10 b 4
2 1 12 d 6
3 2 13 a 8
4 2 14 b 10
5 3 15 z 12
I can pivot the table with:
df.pivot('id','q','v')
And end up with something close:
q a b d z
id
1 2 4 6 NaN
2 8 10 NaN NaN
3 NaN NaN NaN 12
However, what I really want is (the original unmelted form):
id num a b d z
1 10 2 4 NaN NaN
1 12 NaN NaN 6 NaN
2 13 8 NaN NaN NaN
2 14 NaN 10 NaN NaN
3 15 NaN NaN NaN 12
In other words:
'id' and 'num' my indices (normally, I've only seen either 'id' or 'num' being the index but I need both since I'm trying to retrieve the original unmelted form)
'q' are my columns
'v' are my values in the table
Update
I found a close solution from Wes McKinney's blog:
df.pivot_table(index=['id','num'], columns='q')
v
q a b d z
id num
1 10 2 4 NaN NaN
12 NaN NaN 6 NaN
2 13 8 NaN NaN NaN
14 NaN 10 NaN NaN
3 15 NaN NaN NaN 12
However, the format is not quite the same as what I want above.

You could use set_index and unstack
In [18]: df.set_index(['id', 'num', 'q'])['v'].unstack().reset_index()
Out[18]:
q id num a b d z
0 1 10 2.0 4.0 NaN NaN
1 1 12 NaN NaN 6.0 NaN
2 2 13 8.0 NaN NaN NaN
3 2 14 NaN 10.0 NaN NaN
4 3 15 NaN NaN NaN 12.0

You're really close slaw. Just rename your column index to None and you've got what you want.
df2 = df.pivot_table(index=['id','num'], columns='q')
df2.columns = df2.columns.droplevel().rename(None)
df2.reset_index().fillna("null").to_csv("test.csv", sep="\t", index=None)
Note that the the 'v' column is expected to be numeric by default so that it can be aggregated. Otherwise, Pandas will error out with:
DataError: No numeric types to aggregate
To resolve this, you can specify your own aggregation function by using a custom lambda function:
df2 = df.pivot_table(index=['id','num'], columns='q', aggfunc= lambda x: x)

you can remove name q.
df1.columns=df1.columns.tolist()
Zero's answer + remove q =
df1 = df.set_index(['id', 'num', 'q'])['v'].unstack().reset_index()
df1.columns=df1.columns.tolist()
id num a b d z
0 1 10 2.0 4.0 NaN NaN
1 1 12 NaN NaN 6.0 NaN
2 2 13 8.0 NaN NaN NaN
3 2 14 NaN 10.0 NaN NaN
4 3 15 NaN NaN NaN 12.0

This might work just fine:
Pivot
df2 = (df.pivot_table(index=['id', 'num'], columns='q', values='v')).reset_index())
Concatinate the 1st level column names with the 2nd
df2.columns =[s1 + str(s2) for (s1,s2) in df2.columns.tolist()]

Came up with a close solution
df2 = df.pivot_table(index=['id','num'], columns='q')
df2.columns = df2.columns.droplevel()
df2.reset_index().fillna("null").to_csv("test.csv", sep="\t", index=None)
Still can't figure out how to drop 'q' from the dataframe

It can be done in three steps:
#1: Prepare auxilary column 'id_num':
df['id_num'] = df[['id', 'num']].apply(tuple, axis=1)
df = df.drop(columns=['id', 'num'])
#2: 'pivot' is almost an inverse of melt:
df, df.columns.name = df.pivot(index='id_num', columns='q', values='v').reset_index(), ''
#3: Bring back 'id' and 'num' columns:
df['id'], df['num'] = zip(*df['id_num'])
df = df.drop(columns=['id_num'])
This is a result, but with different order of columns:
a b d z id num
0 2.0 4.0 NaN NaN 1 10
1 NaN NaN 6.0 NaN 1 12
2 8.0 NaN NaN NaN 2 13
3 NaN 10.0 NaN NaN 2 14
4 NaN NaN NaN 12.0 3 15
Alternatively with proper order:
def multiindex_pivot(df, columns=None, values=None):
#inspired by: https://github.com/pandas-dev/pandas/issues/23955
names = list(df.index.names)
df = df.reset_index()
list_index = df[names].values
tuples_index = [tuple(i) for i in list_index] # hashable
df = df.assign(tuples_index=tuples_index)
df = df.pivot(index="tuples_index", columns=columns, values=values)
tuples_index = df.index # reduced
index = pd.MultiIndex.from_tuples(tuples_index, names=names)
df.index = index
df = df.reset_index() #me
df.columns.name = '' #me
return df
df = df.set_index(['id', 'num'])
df = multiindex_pivot(df, columns='q', values='v')