I play with a dataset in pandas.
At some point I use it as matrix (df.as_matrix()) , then I do some transformations (with sklearn) and I want to go back to DataFrame.
How can I go back from df.as_matrix() back to df this the most straightworward way and with preserving indexes and col names?
Consider the data frame df
df = pd.DataFrame(1, list('xyz'), list('abc'))
df
a b c
x 1 1 1
y 1 1 1
z 1 1 1
as_matrix gives you:
df.as_matrix()
array([[1, 1, 1],
[1, 1, 1],
[1, 1, 1]])
It is completely reasonable to go back to a data frame with
pd.DataFrame(df.as_matrix())
0 1 2
0 1 1 1
1 1 1 1
2 1 1 1
But you lose the index and column information.
If you still have that info lying around
pd.DataFrame(df.as_matrix(), df.index, df.columns)
a b c
x 1 1 1
y 1 1 1
z 1 1 1
And you are back where you started.
Related
My dataframe is
a b
1 2 1 2
0 0.281045 0.975469 -0.538213 -0.180008
1 0.128696 1.875480 0.247637 -0.047927
I want to insert the matrix to (a,3), (b, 3)
[[1, 1],
[1, 1]]
a b
1 2 3 1 2 3
0 0.281045 0.975469 1. -0.538213 -0.180008 1.
1 0.128696 1.875480 1. 0.247637 -0.047922 1.
It seems like there is no decent way to add value to the multiindex dataframe, Here is the code that I tried:
df[:,:,3] = [[1, 1],
[1, 1]]```
But it didn't work...
You can create new DataFrame with MultiIndex and then append to data by DataFrame.join with sorting MultiIndex:
arr = np.array([[1, 1],[1, 1]])
df1 = pd.DataFrame(arr,
index=df.index,
columns= pd.MultiIndex.from_product([df.columns.levels[0], [3]]))
df = df.join(df1).sort_index(axis=1)
print (df)
a b
1 2 3 1 2 3
0 0.281045 0.975469 1 -0.538213 -0.180008 1
1 0.128696 1.875480 1 0.247637 -0.047927 1
I need to break a column in a DataFrame that at present collects multiple values (someone else's excel sheet unfortunately) for a categorical data field that can have multiple values.
As you can see below the column has 15 category codes seen in the column header.
Original DataFrame
I want to split the column based on the category codes seen in the column header ['Pamphlet'] and then transform the values collected for each record in the original column to be mapped to there respective new columns as a (1) for checked and (0) for unchecked instead of the raw value [1,2,4,5].
This is the code to split based on , between values but I need to put these into the new columns I need to set up by splitting the column ['Pamphlet'] up by the values in the header [15: 1) OSA\n2) Nutrition\n3) Activity\n4) etc.].
'''df_old['Pamphlets'].str.split(pat = ',', n = -1, expand = True)'''
Shape of desired DatFrame
If I could just get an outline of whats the best approach, if it is even possible to do this within Pandas, Thanks.
You need to go through your columns one by one and divide the headers, then create a new dataframe for each column made up of split columns, then join all that back to the original dataframe. It's a bit messy but doable.
You need to use a function and some loops to go through the columns.
First lets define the dataframe. (It would be much appreciated if in future questions you supply a replicatable dataframe and any other data.
data = {
"1) Mail\n2) Email \n3) At PAC/TPAC": [2, 1, 3, 2, 3, 1, 3, 2, 3, 1],
"1) ACC\n2) IM \n3) PT\n4) Smoking, \n5) Cessation": [5, 1, 4, 4, 2, 5, 1, 4, 3, 2],
}
df_full = pd.DataFrame(data)
print(df_full)
1) Mail\n2) Email \n3) At PAC/TPAC 1) ACC\n2) IM \n3) PT\n4) Smoking, \n5) Cessation
0 2 5
1 1 1
2 3 4
3 2 4
4 3 2
5 1 5
6 3 1
7 2 4
8 3 3
9 1 2
We will go through the dataframe column by column using a function. For now let's build the column manually for the first column. After we'll turn this next part into a function.
First, let's grab the first column.
s_col = df_full.iloc[:, 0]
print(s_col)
0 2
1 1
2 3
3 2
4 3
5 1
6 3
7 2
8 3
9 1
Name: 1) Mail\n2) Email \n3) At PAC/TPAC, dtype: int64
Split the header into individual pieces.
col = s_col.name.split("\n")
print(col)
['1) Mail', '2) Email ', '3) At PAC/TPAC']
Clean up any leading or trailing white space.
col = [x.strip() for x in col]
print(col)
['1) Mail', '2) Email', '3) At PAC/TPAC']
Create a new dataframe from series and column heads.
data = {col[x]: s_col.to_list() for x in range(len(col))}
df = pd.DataFrame(data)
print(df)
1) Mail 2) Email 3) At PAC/TPAC
0 2 2 2
1 1 1 1
2 3 3 3
3 2 2 2
4 3 3 3
5 1 1 1
6 3 3 3
7 2 2 2
8 3 3 3
9 1 1 1
Create a copy to make changes to the values.
df_res = df.copy()
Go through the column headers, get the first number, then filter and apply bool.
for col in df.columns:
value = pd.to_numeric(col[0])
df_res.loc[df[col] == value, col] = 1
df_res.loc[df[col] != value, col] = 0
print(df_res)
1) Mail 2) Email 3) At PAC/TPAC
0 0 1 0
1 1 0 0
2 0 0 1
3 0 1 0
4 0 0 1
5 1 0 0
6 0 0 1
7 0 1 0
8 0 0 1
9 1 0 0
Now we have split a column into its components and assigned a bool value.
Let's step back and make the above a function so we can use it for each column in the original dataframe.
def split_column(s_col):
# Split the header into individual pieces.
col = s_col.name.split("\n")
# Clean up any leading or trailing white space.
col = [x.strip() for x in col]
# Create a new dataframe from series and column heads.
data = {col[x]: s_col.to_list() for x in range(len(col))}
df = pd.DataFrame(data)
# Create a copy to make changes to the values.
df_res = df.copy()
# Go through the column headers, get the first number, then filter and apply bool.
for col in df.columns:
value = pd.to_numeric(col[0])
df_res.loc[df[col] == value, col] = 1
df_res.loc[df[col] != value, col] = 0
return df_res
Now for the last step. Let's create a loop to go through the columns in the original dataframe, call the function to split each column, and then concat it to the original dataframe less the columns that were split.
for c in df_full.columns:
# Call the function to get the split columns in a new dataframe.
df_split = split_column(df_full[c])
# Join it with the origianl full dataframe but drop the current column.
df_full = pd.concat([df_full.loc[:, ~df_full.columns.isin([c])], df_split], axis=1)
print(df_full)
1) Mail 2) Email 3) At PAC/TPAC 1) ACC 2) IM 3) PT 4) Smoking, 5) Cessation
0 0 1 0 0 0 0 0 1
1 1 0 0 1 0 0 0 0
2 0 0 1 0 0 0 1 0
3 0 1 0 0 0 0 1 0
4 0 0 1 0 1 0 0 0
5 1 0 0 0 0 0 0 1
6 0 0 1 1 0 0 0 0
7 0 1 0 0 0 0 1 0
8 0 0 1 0 0 1 0 0
9 1 0 0 0 1 0 0 0
Here is the full code...
data = {
"1) Mail\n2) Email \n3) At PAC/TPAC": [2, 1, 3, 2, 3, 1, 3, 2, 3, 1],
"1) ACC\n2) IM \n3) PT\n4) Smoking, \n5) Cessation": [5, 1, 4, 4, 2, 5, 1, 4, 3, 2],
}
df_full = pd.DataFrame(data)
def split_column(s_col):
# Split the header into individual pieces.
col = s_col.name.split("\n")
# Clean up any leading or trailing white space.
col = [x.strip() for x in col]
# Create a new dataframe from series and column heads.
data = {col[x]: s_col.to_list() for x in range(len(col))}
df = pd.DataFrame(data)
# Create a copy to make changes to the values.
df_res = df.copy()
# Go through the column headers, get the first number, then filter and apply bool.
for col in df.columns:
value = pd.to_numeric(col[0])
df_res.loc[df[col] == value, col] = 1
df_res.loc[df[col] != value, col] = 0
return df_res
for c in df_full.columns:
# Call the function to get the split columns in a new dataframe.
df_split = split_column(df_full[c])
# Join it with the origianl full dataframe but drop the current column.
df_full = pd.concat([df_full.loc[:, ~df_full.columns.isin([c])], df_split], axis=1)
print(df_full)
Can I insert a column at a specific column index in pandas?
import pandas as pd
df = pd.DataFrame({'l':['a','b','c','d'], 'v':[1,2,1,2]})
df['n'] = 0
This will put column n as the last column of df, but isn't there a way to tell df to put n at the beginning?
see docs: http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.insert.html
using loc = 0 will insert at the beginning
df.insert(loc, column, value)
df = pd.DataFrame({'B': [1, 2, 3], 'C': [4, 5, 6]})
df
Out:
B C
0 1 4
1 2 5
2 3 6
idx = 0
new_col = [7, 8, 9] # can be a list, a Series, an array or a scalar
df.insert(loc=idx, column='A', value=new_col)
df
Out:
A B C
0 7 1 4
1 8 2 5
2 9 3 6
If you want a single value for all rows:
df.insert(0,'name_of_column','')
df['name_of_column'] = value
Edit:
You can also:
df.insert(0,'name_of_column',value)
df.insert(loc, column_name, value)
This will work if there is no other column with the same name. If a column, with your provided name already exists in the dataframe, it will raise a ValueError.
You can pass an optional parameter allow_duplicates with True value to create a new column with already existing column name.
Here is an example:
>>> df = pd.DataFrame({'b': [1, 2], 'c': [3,4]})
>>> df
b c
0 1 3
1 2 4
>>> df.insert(0, 'a', -1)
>>> df
a b c
0 -1 1 3
1 -1 2 4
>>> df.insert(0, 'a', -2)
Traceback (most recent call last):
File "", line 1, in
File "C:\Python39\lib\site-packages\pandas\core\frame.py", line 3760, in insert
self._mgr.insert(loc, column, value, allow_duplicates=allow_duplicates)
File "C:\Python39\lib\site-packages\pandas\core\internals\managers.py", line 1191, in insert
raise ValueError(f"cannot insert {item}, already exists")
ValueError: cannot insert a, already exists
>>> df.insert(0, 'a', -2, allow_duplicates = True)
>>> df
a a b c
0 -2 -1 1 3
1 -2 -1 2 4
You could try to extract columns as list, massage this as you want, and reindex your dataframe:
>>> cols = df.columns.tolist()
>>> cols = [cols[-1]]+cols[:-1] # or whatever change you need
>>> df.reindex(columns=cols)
n l v
0 0 a 1
1 0 b 2
2 0 c 1
3 0 d 2
EDIT: this can be done in one line ; however, this looks a bit ugly. Maybe some cleaner proposal may come...
>>> df.reindex(columns=['n']+df.columns[:-1].tolist())
n l v
0 0 a 1
1 0 b 2
2 0 c 1
3 0 d 2
Here is a very simple answer to this(only one line).
You can do that after you added the 'n' column into your df as follows.
import pandas as pd
df = pd.DataFrame({'l':['a','b','c','d'], 'v':[1,2,1,2]})
df['n'] = 0
df
l v n
0 a 1 0
1 b 2 0
2 c 1 0
3 d 2 0
# here you can add the below code and it should work.
df = df[list('nlv')]
df
n l v
0 0 a 1
1 0 b 2
2 0 c 1
3 0 d 2
However, if you have words in your columns names instead of letters. It should include two brackets around your column names.
import pandas as pd
df = pd.DataFrame({'Upper':['a','b','c','d'], 'Lower':[1,2,1,2]})
df['Net'] = 0
df['Mid'] = 2
df['Zsore'] = 2
df
Upper Lower Net Mid Zsore
0 a 1 0 2 2
1 b 2 0 2 2
2 c 1 0 2 2
3 d 2 0 2 2
# here you can add below line and it should work
df = df[list(('Mid','Upper', 'Lower', 'Net','Zsore'))]
df
Mid Upper Lower Net Zsore
0 2 a 1 0 2
1 2 b 2 0 2
2 2 c 1 0 2
3 2 d 2 0 2
A general 4-line routine
You can have the following 4-line routine whenever you want to create a new column and insert into a specific location loc.
df['new_column'] = ... #new column's definition
col = df.columns.tolist()
col.insert(loc, col.pop()) #loc is the column's index you want to insert into
df = df[col]
In your example, it is simple:
df['n'] = 0
col = df.columns.tolist()
col.insert(0, col.pop())
df = df[col]
I have to create an id in a pandas df where a counter resets itself.
My data looks like
counter
0
0
1
1
1
2
0
1
1
My desired output looks like
counter id
0 0
0 0
1 1
1 1
1 1
2 2
0 3
1 4
1 4
I have tried the following, which does not help. Any help will be appreciated.
df['id'] = df.groupby(df.counter.tolist(), sort=False).ngroup()
Check diff and cumsum
df['id'] = df.diff().ne(0).cumsum()-1
Another way of using itertools.groupby
from itertools import groupby
sum([ [y]*len(list(g)) for y,(_,g) in enumerate(groupby(df.counter))],[])
Out[46]: [0, 0, 1, 1, 1, 2, 3, 4, 4]
I've got a dataframe (result from groupby par "nr")
id lap nr time
1 1 2 10
4 2 2 100
I need to rearrange this dataframe to folowing format
nr lap1 time1 lap2 time2
2 1 10 2 100
Any Idea how can I do it?
You can think of this as a pivot. If your DataFrame had an extra column called, say, colnum:
lap nr time colnum
0 1 2 10 1
1 2 2 100 2
then
df.pivot(index='nr', columns='colnum')
moves the nr column values into the row index, and the colnum column values into the column index:
lap time
colnum 1 2 1 2
nr
2 1 2 10 100
This is basically the desired result. All we need to do is fix up the column labels:
df.columns = ['{}{}'.format(col, num) for col,num in df.columns]
Thus,
import pandas as pd
df = pd.DataFrame({'id': [1, 4], 'lap': [1, 2], 'nr': [2, 2], 'time': [10, 100]})
df['colnum'] = df.groupby('nr').cumcount()+1
df = df[['lap','nr','time','colnum']]
df = df.pivot(index='nr', columns='colnum')
df.columns = ['{}{}'.format(col, num) for col,num in df.columns]
df = df.reset_index()
yields
nr lap1 lap2 time1 time2
0 2 1 2 10 100