I'd like to convert a Pandas DataFrame that is derived from a pivot table into a row representation as shown below.
This is where I'm at:
import pandas as pd
import numpy as np
df = pd.DataFrame({
'goods': ['a', 'a', 'b', 'b', 'b'],
'stock': [5, 10, 30, 40, 10],
'category': ['c1', 'c2', 'c1', 'c2', 'c1'],
'date': pd.to_datetime(['2014-01-01', '2014-02-01', '2014-01-06', '2014-02-09', '2014-03-09'])
})
# we don't care about year in this example
df['month'] = df['date'].map(lambda x: x.month)
piv = df.pivot_table(["stock"], "month", ["goods", "category"], aggfunc="sum")
piv = piv.reindex(np.arange(piv.index[0], piv.index[-1] + 1))
piv = piv.ffill(axis=0)
piv = piv.fillna(0)
print piv
which results in
stock
goods a b
category c1 c2 c1 c2
month
1 5 0 30 0
2 5 10 30 40
3 5 10 10 40
And this is where I want to get to.
goods category month stock
a c1 1 5
a c1 2 0
a c1 3 0
a c2 1 0
a c2 2 10
a c2 3 0
b c1 1 30
b c1 2 0
b c1 3 10
b c2 1 0
b c2 2 40
b c2 3 0
Previously, I used
piv = piv.stack()
piv = piv.reset_index()
print piv
to get rid of the multi-indexes, but this results in this because I pivot now on two columns (["goods", "category"]):
month category stock
goods a b
0 1 c1 5 30
1 1 c2 0 0
2 2 c1 5 30
3 2 c2 10 40
4 3 c1 5 10
5 3 c2 10 40
Does anyone know how I can get rid of the multi-index in the column and get the result into a DataFrame of the exemplified format?
>>> piv.unstack().reset_index().drop('level_0', axis=1)
goods category month 0
0 a c1 1 5
1 a c1 2 5
2 a c1 3 5
3 a c2 1 0
4 a c2 2 10
5 a c2 3 10
6 b c1 1 30
7 b c1 2 30
8 b c1 3 10
9 b c2 1 0
10 b c2 2 40
11 b c2 3 40
then all you need is to change last column name from 0 to stock.
It seems to me that melt (aka unpivot) is very close to what you want to do:
In [11]: pd.melt(piv)
Out[11]:
NaN goods category value
0 stock a c1 5
1 stock a c1 5
2 stock a c1 5
3 stock a c2 0
4 stock a c2 10
5 stock a c2 10
6 stock b c1 30
7 stock b c1 30
8 stock b c1 10
9 stock b c2 0
10 stock b c2 40
11 stock b c2 40
There's a rogue column (stock), that appears here that column header is constant in piv. If we drop it first the melt works OOTB:
In [12]: piv.columns = piv.columns.droplevel(0)
In [13]: pd.melt(piv)
Out[13]:
goods category value
0 a c1 5
1 a c1 5
2 a c1 5
3 a c2 0
4 a c2 10
5 a c2 10
6 b c1 30
7 b c1 30
8 b c1 10
9 b c2 0
10 b c2 40
11 b c2 40
Edit: The above actually drops the index, you need to make it a column with reset_index:
In [21]: pd.melt(piv.reset_index(), id_vars=['month'], value_name='stock')
Out[21]:
month goods category stock
0 1 a c1 5
1 2 a c1 5
2 3 a c1 5
3 1 a c2 0
4 2 a c2 10
5 3 a c2 10
6 1 b c1 30
7 2 b c1 30
8 3 b c1 10
9 1 b c2 0
10 2 b c2 40
11 3 b c2 40
I know that the question has already been answered, but for my dataset multiindex column problem, the provided solution was unefficient. So here I am posting another solution for unpivoting multiindex columns using pandas.
Here is the problem I had:
As one can see, the dataframe is composed of 3 multiindex, and two levels of multiindex columns.
The desired dataframe format was:
When I tried the options given above, the pd.melt function didn't allow to have more than one column in the var_name attribute. Therefore, every time that I tried a melt, I would end up losing some attribute from my table.
The solution I found was to apply a double stacking function over my dataframe.
Before the coding, it is worth notice that the desired var_name for my unpivoted table column was "Populacao residente em domicilios particulares ocupados" (see in the code below). Therefore, for all my value entries, they should be stacked in this newly created var_name new column.
Here is a snippet code:
import pandas as pd
# reading my table
df = pd.read_excel(r'my_table.xls', sep=',', header=[2,3], encoding='latin3',
index_col=[0,1,2], na_values=['-', ' ', '*'], squeeze=True).fillna(0)
df.index.names = ['COD_MUNIC_7', 'NOME_MUN', 'TIPO']
df.columns.names = ['sexo', 'faixa_etaria']
df.head()
# making the stacking:
df = pd.DataFrame(pd.Series(df.stack(level=0).stack(), name='Populacao residente em domicilios particulares ocupados')).reset_index()
df.head()
Another solution I found was to first apply a stacking function over the dataframe and then apply the melt.
Here is an alternative code:
df = df.stack('faixa_etaria').reset_index().melt(id_vars=['COD_MUNIC_7', 'NOME_MUN','TIPO', 'faixa_etaria'],
value_vars=['Homens', 'Mulheres'],
value_name='Populacao residente em domicilios particulares ocupados',
var_name='sexo')
df.head()
Sincerely yours,
Philipe Riskalla Leal
Related
I have a column that is positioned in the middle of a dataframe. I need to split it into multiple columns, and replace it with the new columns. I'm able to do it with the following code:
df = df.join(df[col_to_split].str.split(', ', expand=True).add_prefix(col_to_split + '_'))
However, the new columns are placed at the end of the dataframe, rather than replacing the original column. I need a way to place the new columns at the same position of original columns.
Note that I don't want to manually order ALL columns (i.e. df = df[[c1, c2, c3 ... cn]]) because of many reasons, i.e.it's not known how many new columns are going to be generated, and dataframe contains hundreds of columns.
Sample data:
c1 c2 c3 col_to_split c4 c5 ... cn
1 a b 1,5,3 1 1 ... 1
2 a c 5,10 3 3 ... 4
3 z c 3 2 3 ... 4
Desired output:
c1 c2 c3 col_to_split_0 col_to_split_1 col_to_split_2 c4 c5 ... cn
1 a b 1 5 3 1 1 ... 1
2 a c 5 10 3 3 ... 4
3 z c 3 2 3 ... 4
Idea is use your solution with dynamic insert df1.columns to original columns with cols[pos:pos] trick, position of original column is count by Index.get_loc:
col_to_split = 'col_to_split'
cols = df.columns.tolist()
pos = df.columns.get_loc(col_to_split)
df1 = df[col_to_split].str.split(',', expand=True).fillna("").add_prefix(col_to_split + '_')
cols[pos:pos] = df1.columns.tolist()
cols.remove(col_to_split)
print (cols)
['c1', 'c2', 'c3', 'col_to_split_0', 'col_to_split_1', 'col_to_split_2',
'c4', 'c5', 'cn']
df = df.join(df1).reindex(cols, axis=1)
print (df)
c1 c2 c3 col_to_split_0 col_to_split_1 col_to_split_2 c4 c5 cn
0 1 a b 1 5 3 1 1 1
1 2 a c 5 10 3 3 4
2 3 z c 3 2 3 4
Similar solution for join columsn names in lists:
col_to_split = 'col_to_split'
pos = df.columns.get_loc(col_to_split)
df1 = df[col_to_split].str.split(",", expand=True).fillna("").add_prefix(col_to_split + '_')
cols = df.columns.tolist()
cols = cols[:pos] + df1.columns.tolist() + cols[pos+1:]
print(cols)
['c1', 'c2', 'c3', 'col_to_split_0', 'col_to_split_1', 'col_to_split_2',
'c4', 'c5', 'cn']
df = df.join(df1).reindex(cols, axis=1)
print (df)
c1 c2 c3 col_to_split_0 col_to_split_1 col_to_split_2 c4 c5 cn
0 1 a b 1 5 3 1 1 1
1 2 a c 5 10 3 3 4
2 3 z c 3 2 3 4
We can wrap this operation to a function
import pandas as pd
import numpy as np
from io import StringIO
df = pd.read_csv(StringIO("""c1 c2 c3 col_to_split c4 c5 cn
1 a b 1,5,3 1 1 1
2 a c 5,10 3 3 4
3 z c 3 2 3 4"""), sep="\s+")
def split_by_col(df, colname):
pos = df.columns.tolist().index(colname)
df_tmp = df[colname].str.split(",", expand=True).fillna("")
df_tmp.columns=["col_to_split_" + str(i) for i in range(len(df_tmp.columns))]
return pd.concat([df.iloc[:,:pos], df_tmp, df.iloc[:,pos+1:]], axis=1)
With example:
>>> split_by_col(df, "col_to_split")
c1 c2 c3 col_to_split_0 col_to_split_1 col_to_split_2 c4 c5 cn
0 1 a b 1 5 3 1 1 1
1 2 a c 5 10 3 3 4
2 3 z c 3 2 3 4
Try this:
df = df.join(df[col_to_split].str.split(', ', expand=True).add_prefix(col_to_split + '_'))
df = df[["c1", "c2", "c3" "col_to_split_0" "col_to_split_1" "col_to_split_2" "c4" "c5" ... "cn"]]
I have a very large csv file with following structure:
a1 b1 c1 a2 b2 c2 a3 b3 c3 ..... a999 b999 c999
0 5 4 2 3 2 2 6 7 9 ....................
1 2 1 4 4 6 9 3 5 9 ....................
.
.
What I want to do is to group the columns in sets of N, for a, b and c, and check when the index of maximum value (argmax) of the set changes, in each row.
So in the above example, for N = 3, a1, b1, c1 is the first set in row 0, and argmax is 0, 2nd set is a2, b2, c2 and argmax is still 0, 3rd set is a3, b3, c3 but now the argmax is 2. I deally I am looking for a script that parses the whole csv file and returns [c3, c1]. c3 because thats where the argmax changes in row 0 and c1 becuase argmax doesn't change in row 1 but c1 is the largest value in that set.
I am doing this right now by using two for loops and its slow and looks very ugly, is there a better pandas pythonic way of doing this? I feel there must be.
I tried to keep to code as simple as possible. You can translate your dataframe and group by the sliced column name:
df = df.T.reset_index()
idx = df.groupby(df['index'].str.slice(1,2)).idxmax()
Output:
0 1
index
1 0 2
2 3 5
3 8 8
That means that for row 0 the max for group 1 is at index 0, the max group 2 is at index 3 (or 0 is you take the mod 3), the max for group 3 is at index 8, (or 2 if you take mod 3). Same reading for row 1 :)
If you need the actual column name:
df.columns[idx.values.flatten(order='F')]
Output:
['a1', 'a2', 'c3', 'c1', 'c2', 'c3']
You can groupby sets of columns and use .idxmax to find the column where the maximum occurs within each set. You can find where the first letter changes (if it ever does) to get your list.
n = 3
df2 = df.groupby([x//n for x in range(len(df.columns))], axis=1).idxmax(1)
mask = df2.applymap(lambda x: x[0]) # Case of 1-letter column prefix
## If possibility of words with different length ending in digits try
# import string
# mask = df2.applymap(lambda x: x.strip(string.digits))
df2.lookup(df2.index,
(mask.ne(mask.shift(-1, axis=1)).idxmax(1)+1) % (len(mask.columns))).tolist()
Sample Data
print(df)
a1 b1 c1 a2 b2 c2 a3 b3 c3
0 5 4 2 3 2 2 6 7 9
1 2 1 4 4 6 9 3 5 9
2 2 1 4 10 6 9 3 5 9
3 2 1 4 1 6 9 3 10 9
n = 3
df2 = df.groupby([x//n for x in range(len(df.columns))], axis=1).idxmax(1)
print(df2)
# 0 1 2
#0 a1 a2 c3
#1 c1 c2 c3
#2 c1 a2 c3
#3 c1 c2 b3
mask = df2.applymap(lambda x: x[0])
df2.lookup(df2.index, (mask.ne(mask.shift(-1, axis=1)).idxmax(1)+1) % (len(mask.columns))).tolist()
#['c3', 'c1', 'a2', 'b3']
I have the following example of dataframe.
c1 c2
0 1 a
1 2 b
2 3 c
3 4 d
4 5 e
Given a template c1 = [3, 2, 5, 4, 1], I want to change the order of the rows based on the new order of column c1, so it will look like:
c1 c2
0 3 c
1 2 b
2 5 e
3 4 d
4 1 a
I found the following thread, but the shuffle is random. Cmmiw.
Shuffle DataFrame rows
If values are unique in list and also in c1 column use reindex:
df = df.set_index('c1').reindex(c1).reset_index()
print (df)
c1 c2
0 3 c
1 2 b
2 5 e
3 4 d
4 1 a
General solution working with duplicates in list and also in column:
c1 = [3, 2, 5, 4, 1, 3, 2, 3]
#create df from list
list_df = pd.DataFrame({'c1':c1})
print (list_df)
c1
0 3
1 2
2 5
3 4
4 1
5 3
6 2
7 3
#helper column for count duplicates values
df['g'] = df.groupby('c1').cumcount()
list_df['g'] = list_df.groupby('c1').cumcount()
#merge together, create index from column and remove g column
df = list_df.merge(df).drop('g', axis=1)
print (df)
c1 c2
0 3 c
1 2 b
2 5 e
3 4 d
4 1 a
5 3 c
merge
You can create a dataframe with the column specified in the wanted order then merge.
One advantage of this approach is that it gracefully handles duplicates in either df.c1 or the list c1. If duplicates not wanted then care must be taken to handle them prior to reordering.
d1 = pd.DataFrame({'c1': c1})
d1.merge(df)
c1 c2
0 3 c
1 2 b
2 5 e
3 4 d
4 1 a
searchsorted
This is less robust but will work if df.c1 is:
already sorted
one-to-one mapping
df.iloc[df.c1.searchsorted(c1)]
c1 c2
2 3 c
1 2 b
4 5 e
3 4 d
0 1 a
Could you please help me in transforming the dataframe df
df=pd.DataFrame(data=[['a1',2,3],['b1',5,6],['c1',8,9]],columns=['A','B','C'])
df
Out[37]:
A B C
0 a1 2 3
1 b1 5 6
2 c1 8 9
in df2
df2=pd.DataFrame(data=[[2,5,8],[3,6,9]],columns=['a1','b1','c1'])
df2
Out[36]:
a1 b1 c1
0 2 5 8
1 3 6 9
The first column should become the column names
and then I should transpose the elements...is there a pythonic way?
A little trick with slicing, initialise a new DataFrame.
pd.DataFrame(df.values.T[1:], columns=df.A.tolist())
Or,
pd.DataFrame(df.values[:, 1:].T, columns=df.A.tolist())
a1 b1 c1
0 2 5 8
1 3 6 9
For general solution use set_index with transpose:
df1 = df.set_index('A').T.reset_index(drop=True).rename_axis(None)
Or remove column A, transpose and build new DataFrame by constructor:
df1 = pd.DataFrame(df.drop('A', 1).T.values, columns=df['A'].values)
print (df1)
a1 b1 c1
0 2 5 8
1 3 6 9
With these two data frames
df1 = pd.DataFrame({'c1':['a','b','c','d'],'c2':[10,20,10,22]})
df2 = pd.DataFrame({'c3':['e','f','a','g','b','c','r','j','d'],'c4':[1,2,3,4,5,6,7,8,9]})
I'm trying to add the values of c4 to df1 for only the elements in c3 that are also present in c1:
>>> df1
c1 c2 c4
a 10 3
b 20 5
c 10 6
d 22 9
Is there a simple way of doing this in pandas?
UPDATE:
If
df2 = pd.DataFrame({'c3':['e','f','a','g','b','c','r','j','d'],'c4':[1,2,3,4,5,6,7,8,9]},'c5':[10,20,30,40,50,60,70,80,90])
how can I achieve this result?
>>> df1
c1 c2 c4 c5
a 10 3 30
b 20 5 50
c 10 6 60
d 22 9 90
Doing:
>>> df1['c1'].map(df2.set_index('c3')['c4','c5'])
gives me a KeyError
You can call map on df2['c4'] after setting the index on df2['c3'], this will perform a lookup:
In [239]:
df1 = pd.DataFrame({'c1':['a','b','c','d'],'c2':[10,20,10,22]})
df2 = pd.DataFrame({'c3':['e','f','a','g','b','c','r','j','d'],'c4':[1,2,3,4,5,6,7,8,9]})
df1['c4'] = df1['c1'].map(df2.set_index('c3')['c4'])
df1
Out[239]:
c1 c2 c4
0 a 10 3
1 b 20 5
2 c 10 6
3 d 22 9