I am struggling with the following: I have on dataset and have transposed it. After transposing, the first column was set as an index automatically, and from now one, this "index" column is not recognized as a variable. Here is an example of what I mean;
df = Date A B C
1/1/2021 1 2 3
1/2/2021 4 5 6
input: df_T = df.t
output: index 1/1/2021 1/2/2021
A 1 4
B 2 5
C 3 6
I would like to have a variable, and name if it is possible, instead of the generated "index".
To reproduce this dataset, I have used this chunk of code:
data = [['1/1/2021', 1, 2, 3], ['3/1/2021', 4, 5, 6]]
df = pd.DataFrame(data)
df.columns = ['Date', 'A', 'B', 'C']
df.set_index('Date', inplace=True)
To have meaningfull column inplace of index, the next line can be run:
df_T = df.T.reset_index()
To rename the column 'index', method rename can be used:
df_T.rename(columns={'index':'Variable'})
A Pandas Data Frame typically has names both for the colmns and for the rows. The list of names for the rows is called the "Index". When you do a transpose, rows and columns switch places. In your case, the dates column is for some reason the index, so it becomes the column names for the new data frame. You need to create a new index and turn the "Date" column into a regular column. As #sophocies wrote above, this is achived with df.reset_index(...). I hope the code example below will be helpful.
import pandas as pd
df = pd.DataFrame(columns=['Date', 'A', 'B', 'C'], data=[['1/1/2021', 1, 2,3], ['1/2/2021', 4,5,6]])
df.set_index('Date', inplace=True)
print(df.transpose(copy=True))
#Recreated the problem
df.reset_index(inplace=True)
print(df.transpose())
Output
0
1
Date
1/1/2021
1/2/2021
A
1
4
B
2
5
C
3
6
I hope this is what you wanted!
Related
I am using .size() on a groupby result in order to count how many items are in each group.
I would like the result to be saved to a new column name without manually editing the column names array, how can it be done?
This is what I have tried:
grpd = df.groupby(['A','B'])
grpd['size'] = grpd.size()
grpd
and the error I got:
TypeError: 'DataFrameGroupBy' object does not support item assignment
(on the second line)
The .size() built-in method of DataFrameGroupBy objects actually returns a Series object with the group sizes and not a DataFrame. If you want a DataFrame whose column is the group sizes, indexed by the groups, with a custom name, you can use the .to_frame() method and use the desired column name as its argument.
grpd = df.groupby(['A','B']).size().to_frame('size')
If you wanted the groups to be columns again you could add a .reset_index() at the end.
You need transform size - len of df is same as before:
Notice:
Here it is necessary to add one column after groupby, else you get an error. Because GroupBy.size count NaNs too, what column is used is not important. All columns working same.
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df['size'] = df.groupby(['A', 'B'])['A'].transform('size')
print (df)
A B size
0 x a 1
1 x c 2
2 x c 2
3 y b 2
4 y b 2
If need set column name in aggregating df - len of df is obviously NOT same as before:
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df = df.groupby(['A', 'B']).size().reset_index(name='Size')
print (df)
A B Size
0 x a 1
1 x c 2
2 y b 2
The result of df.groupby(...) is not a DataFrame. To get a DataFrame back, you have to apply a function to each group, transform each element of a group, or filter the groups.
It seems like you want a DataFrame that contains (1) all your original data in df and (2) the count of how much data is in each group. These things have different lengths, so if they need to go into the same DataFrame, you'll need to list the size redundantly, i.e., for each row in each group.
df['size'] = df.groupby(['A','B']).transform(np.size)
(Aside: It's helpful if you can show succinct sample input and expected results.)
You can set the as_index parameter in groupby to False to get a DataFrame instead of a Series:
df = pd.DataFrame({'A': ['a', 'a', 'b', 'b'], 'B': [1, 2, 2, 2]})
df.groupby(['A', 'B'], as_index=False).size()
Output:
A B size
0 a 1 1
1 a 2 1
2 b 2 2
lets say n is the name of dataframe and cst is the no of items being repeted.
Below code gives the count in next column
cstn=Counter(n.cst)
cstlist = pd.DataFrame.from_dict(cstn, orient='index').reset_index()
cstlist.columns=['name','cnt']
n['cnt']=n['cst'].map(cstlist.loc[:, ['name','cnt']].set_index('name').iloc[:,0].to_dict())
Hope this will work
I am using .size() on a groupby result in order to count how many items are in each group.
I would like the result to be saved to a new column name without manually editing the column names array, how can it be done?
This is what I have tried:
grpd = df.groupby(['A','B'])
grpd['size'] = grpd.size()
grpd
and the error I got:
TypeError: 'DataFrameGroupBy' object does not support item assignment
(on the second line)
The .size() built-in method of DataFrameGroupBy objects actually returns a Series object with the group sizes and not a DataFrame. If you want a DataFrame whose column is the group sizes, indexed by the groups, with a custom name, you can use the .to_frame() method and use the desired column name as its argument.
grpd = df.groupby(['A','B']).size().to_frame('size')
If you wanted the groups to be columns again you could add a .reset_index() at the end.
You need transform size - len of df is same as before:
Notice:
Here it is necessary to add one column after groupby, else you get an error. Because GroupBy.size count NaNs too, what column is used is not important. All columns working same.
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df['size'] = df.groupby(['A', 'B'])['A'].transform('size')
print (df)
A B size
0 x a 1
1 x c 2
2 x c 2
3 y b 2
4 y b 2
If need set column name in aggregating df - len of df is obviously NOT same as before:
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df = df.groupby(['A', 'B']).size().reset_index(name='Size')
print (df)
A B Size
0 x a 1
1 x c 2
2 y b 2
The result of df.groupby(...) is not a DataFrame. To get a DataFrame back, you have to apply a function to each group, transform each element of a group, or filter the groups.
It seems like you want a DataFrame that contains (1) all your original data in df and (2) the count of how much data is in each group. These things have different lengths, so if they need to go into the same DataFrame, you'll need to list the size redundantly, i.e., for each row in each group.
df['size'] = df.groupby(['A','B']).transform(np.size)
(Aside: It's helpful if you can show succinct sample input and expected results.)
You can set the as_index parameter in groupby to False to get a DataFrame instead of a Series:
df = pd.DataFrame({'A': ['a', 'a', 'b', 'b'], 'B': [1, 2, 2, 2]})
df.groupby(['A', 'B'], as_index=False).size()
Output:
A B size
0 a 1 1
1 a 2 1
2 b 2 2
lets say n is the name of dataframe and cst is the no of items being repeted.
Below code gives the count in next column
cstn=Counter(n.cst)
cstlist = pd.DataFrame.from_dict(cstn, orient='index').reset_index()
cstlist.columns=['name','cnt']
n['cnt']=n['cst'].map(cstlist.loc[:, ['name','cnt']].set_index('name').iloc[:,0].to_dict())
Hope this will work
I am using .size() on a groupby result in order to count how many items are in each group.
I would like the result to be saved to a new column name without manually editing the column names array, how can it be done?
This is what I have tried:
grpd = df.groupby(['A','B'])
grpd['size'] = grpd.size()
grpd
and the error I got:
TypeError: 'DataFrameGroupBy' object does not support item assignment
(on the second line)
The .size() built-in method of DataFrameGroupBy objects actually returns a Series object with the group sizes and not a DataFrame. If you want a DataFrame whose column is the group sizes, indexed by the groups, with a custom name, you can use the .to_frame() method and use the desired column name as its argument.
grpd = df.groupby(['A','B']).size().to_frame('size')
If you wanted the groups to be columns again you could add a .reset_index() at the end.
You need transform size - len of df is same as before:
Notice:
Here it is necessary to add one column after groupby, else you get an error. Because GroupBy.size count NaNs too, what column is used is not important. All columns working same.
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df['size'] = df.groupby(['A', 'B'])['A'].transform('size')
print (df)
A B size
0 x a 1
1 x c 2
2 x c 2
3 y b 2
4 y b 2
If need set column name in aggregating df - len of df is obviously NOT same as before:
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df = df.groupby(['A', 'B']).size().reset_index(name='Size')
print (df)
A B Size
0 x a 1
1 x c 2
2 y b 2
The result of df.groupby(...) is not a DataFrame. To get a DataFrame back, you have to apply a function to each group, transform each element of a group, or filter the groups.
It seems like you want a DataFrame that contains (1) all your original data in df and (2) the count of how much data is in each group. These things have different lengths, so if they need to go into the same DataFrame, you'll need to list the size redundantly, i.e., for each row in each group.
df['size'] = df.groupby(['A','B']).transform(np.size)
(Aside: It's helpful if you can show succinct sample input and expected results.)
You can set the as_index parameter in groupby to False to get a DataFrame instead of a Series:
df = pd.DataFrame({'A': ['a', 'a', 'b', 'b'], 'B': [1, 2, 2, 2]})
df.groupby(['A', 'B'], as_index=False).size()
Output:
A B size
0 a 1 1
1 a 2 1
2 b 2 2
lets say n is the name of dataframe and cst is the no of items being repeted.
Below code gives the count in next column
cstn=Counter(n.cst)
cstlist = pd.DataFrame.from_dict(cstn, orient='index').reset_index()
cstlist.columns=['name','cnt']
n['cnt']=n['cst'].map(cstlist.loc[:, ['name','cnt']].set_index('name').iloc[:,0].to_dict())
Hope this will work
I am using .size() on a groupby result in order to count how many items are in each group.
I would like the result to be saved to a new column name without manually editing the column names array, how can it be done?
This is what I have tried:
grpd = df.groupby(['A','B'])
grpd['size'] = grpd.size()
grpd
and the error I got:
TypeError: 'DataFrameGroupBy' object does not support item assignment
(on the second line)
The .size() built-in method of DataFrameGroupBy objects actually returns a Series object with the group sizes and not a DataFrame. If you want a DataFrame whose column is the group sizes, indexed by the groups, with a custom name, you can use the .to_frame() method and use the desired column name as its argument.
grpd = df.groupby(['A','B']).size().to_frame('size')
If you wanted the groups to be columns again you could add a .reset_index() at the end.
You need transform size - len of df is same as before:
Notice:
Here it is necessary to add one column after groupby, else you get an error. Because GroupBy.size count NaNs too, what column is used is not important. All columns working same.
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df['size'] = df.groupby(['A', 'B'])['A'].transform('size')
print (df)
A B size
0 x a 1
1 x c 2
2 x c 2
3 y b 2
4 y b 2
If need set column name in aggregating df - len of df is obviously NOT same as before:
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df = df.groupby(['A', 'B']).size().reset_index(name='Size')
print (df)
A B Size
0 x a 1
1 x c 2
2 y b 2
The result of df.groupby(...) is not a DataFrame. To get a DataFrame back, you have to apply a function to each group, transform each element of a group, or filter the groups.
It seems like you want a DataFrame that contains (1) all your original data in df and (2) the count of how much data is in each group. These things have different lengths, so if they need to go into the same DataFrame, you'll need to list the size redundantly, i.e., for each row in each group.
df['size'] = df.groupby(['A','B']).transform(np.size)
(Aside: It's helpful if you can show succinct sample input and expected results.)
You can set the as_index parameter in groupby to False to get a DataFrame instead of a Series:
df = pd.DataFrame({'A': ['a', 'a', 'b', 'b'], 'B': [1, 2, 2, 2]})
df.groupby(['A', 'B'], as_index=False).size()
Output:
A B size
0 a 1 1
1 a 2 1
2 b 2 2
lets say n is the name of dataframe and cst is the no of items being repeted.
Below code gives the count in next column
cstn=Counter(n.cst)
cstlist = pd.DataFrame.from_dict(cstn, orient='index').reset_index()
cstlist.columns=['name','cnt']
n['cnt']=n['cst'].map(cstlist.loc[:, ['name','cnt']].set_index('name').iloc[:,0].to_dict())
Hope this will work
I am using .size() on a groupby result in order to count how many items are in each group.
I would like the result to be saved to a new column name without manually editing the column names array, how can it be done?
This is what I have tried:
grpd = df.groupby(['A','B'])
grpd['size'] = grpd.size()
grpd
and the error I got:
TypeError: 'DataFrameGroupBy' object does not support item assignment
(on the second line)
The .size() built-in method of DataFrameGroupBy objects actually returns a Series object with the group sizes and not a DataFrame. If you want a DataFrame whose column is the group sizes, indexed by the groups, with a custom name, you can use the .to_frame() method and use the desired column name as its argument.
grpd = df.groupby(['A','B']).size().to_frame('size')
If you wanted the groups to be columns again you could add a .reset_index() at the end.
You need transform size - len of df is same as before:
Notice:
Here it is necessary to add one column after groupby, else you get an error. Because GroupBy.size count NaNs too, what column is used is not important. All columns working same.
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df['size'] = df.groupby(['A', 'B'])['A'].transform('size')
print (df)
A B size
0 x a 1
1 x c 2
2 x c 2
3 y b 2
4 y b 2
If need set column name in aggregating df - len of df is obviously NOT same as before:
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df = df.groupby(['A', 'B']).size().reset_index(name='Size')
print (df)
A B Size
0 x a 1
1 x c 2
2 y b 2
The result of df.groupby(...) is not a DataFrame. To get a DataFrame back, you have to apply a function to each group, transform each element of a group, or filter the groups.
It seems like you want a DataFrame that contains (1) all your original data in df and (2) the count of how much data is in each group. These things have different lengths, so if they need to go into the same DataFrame, you'll need to list the size redundantly, i.e., for each row in each group.
df['size'] = df.groupby(['A','B']).transform(np.size)
(Aside: It's helpful if you can show succinct sample input and expected results.)
You can set the as_index parameter in groupby to False to get a DataFrame instead of a Series:
df = pd.DataFrame({'A': ['a', 'a', 'b', 'b'], 'B': [1, 2, 2, 2]})
df.groupby(['A', 'B'], as_index=False).size()
Output:
A B size
0 a 1 1
1 a 2 1
2 b 2 2
lets say n is the name of dataframe and cst is the no of items being repeted.
Below code gives the count in next column
cstn=Counter(n.cst)
cstlist = pd.DataFrame.from_dict(cstn, orient='index').reset_index()
cstlist.columns=['name','cnt']
n['cnt']=n['cst'].map(cstlist.loc[:, ['name','cnt']].set_index('name').iloc[:,0].to_dict())
Hope this will work