Related
This question already has answers here:
How can I replicate rows of a Pandas DataFrame?
(10 answers)
Closed 11 months ago.
I want to replicate rows in a Pandas Dataframe. Each row should be repeated n times, where n is a field of each row.
import pandas as pd
what_i_have = pd.DataFrame(data={
'id': ['A', 'B', 'C'],
'n' : [ 1, 2, 3],
'v' : [ 10, 13, 8]
})
what_i_want = pd.DataFrame(data={
'id': ['A', 'B', 'B', 'C', 'C', 'C'],
'v' : [ 10, 13, 13, 8, 8, 8]
})
Is this possible?
You can use Index.repeat to get repeated index values based on the column then select from the DataFrame:
df2 = df.loc[df.index.repeat(df.n)]
id n v
0 A 1 10
1 B 2 13
1 B 2 13
2 C 3 8
2 C 3 8
2 C 3 8
Or you could use np.repeat to get the repeated indices and then use that to index into the frame:
df2 = df.loc[np.repeat(df.index.values, df.n)]
id n v
0 A 1 10
1 B 2 13
1 B 2 13
2 C 3 8
2 C 3 8
2 C 3 8
After which there's only a bit of cleaning up to do:
df2 = df2.drop("n", axis=1).reset_index(drop=True)
id v
0 A 10
1 B 13
2 B 13
3 C 8
4 C 8
5 C 8
Note that if you might have duplicate indices to worry about, you could use .iloc instead:
df.iloc[np.repeat(np.arange(len(df)), df["n"])].drop("n", axis=1).reset_index(drop=True)
id v
0 A 10
1 B 13
2 B 13
3 C 8
4 C 8
5 C 8
which uses the positions, and not the index labels.
You could use set_index and repeat
In [1057]: df.set_index(['id'])['v'].repeat(df['n']).reset_index()
Out[1057]:
id v
0 A 10
1 B 13
2 B 13
3 C 8
4 C 8
5 C 8
Details
In [1058]: df
Out[1058]:
id n v
0 A 1 10
1 B 2 13
2 C 3 8
It's something like the uncount in tidyr:
https://tidyr.tidyverse.org/reference/uncount.html
I wrote a package (https://github.com/pwwang/datar) that implements this API:
from datar import f
from datar.tibble import tribble
from datar.tidyr import uncount
what_i_have = tribble(
f.id, f.n, f.v,
'A', 1, 10,
'B', 2, 13,
'C', 3, 8
)
what_i_have >> uncount(f.n)
Output:
id v
0 A 10
1 B 13
1 B 13
2 C 8
2 C 8
2 C 8
Not the best solution, but I want to share this: you could also use pandas.reindex() and .repeat():
df.reindex(df.index.repeat(df.n)).drop('n', axis=1)
Output:
id v
0 A 10
1 B 13
1 B 13
2 C 8
2 C 8
2 C 8
You can further append .reset_index(drop=True) to reset the .index.
import pandas as pd
df1 = pd.DataFrame({"A":[14, 4, 5, 4],"B":[1,2,3,4]})
df2 = pd.DataFrame({"A":[14, 4, 5, 4],"C":[5,6,7,8]})
df = pd.concat([df1,df2],axis=1)
Let's see the concated df,the first column and third column shares the same column name A.
df
A B A C
0 14 1 14 5
1 4 2 4 6
2 5 3 5 7
3 4 4 4 8
I want to get the following format.
df
A B C
0 14 1 5
1 4 2 6
2 5 3 7
3 4 4 8
Drop column by id.
result = df.drop(df.columns[2],axis=1)
result
B C
0 1 5
1 2 6
2 3 7
3 4 8
I can get what i expect this way:
import pandas as pd
df1 = pd.DataFrame({"A":[14, 4, 5, 4],"B":[1,2,3,4]})
df2 = pd.DataFrame({"A":[14, 4, 5, 4],"C":[5,6,7,8]})
df2 = df2.drop(df2.columns[0],axis=1)
df = pd.concat([df1,df2],axis=1)
It is so strange that both the first and third column removed when to drop specified column by id.
1.Please tell me the reason of dataframe's this action.
2.How can i remove the third column at the same time keep the first column undeleted?
Here's a way using indexes:
index_to_drop = 2
# get indexes to keep
col_idxs = [en for en, _ in enumerate(df.columns) if en != index_to_drop]
# subset the df
df = df.iloc[:,col_idxs]
A B C
0 14 1 5
1 4 2 6
2 5 3 7
3 4 4 8
I have a dataframe:
import pandas as pd
d = {'A-101': ['1','2','4'], 'B-102':
['5','7','8'],'A-102': ['1','2','9']}
df = pd.DataFrame(data=d)
which is
A-101 B-102 A-102
0 1 5 1
1 2 7 2
2 4 8 9
how to change above into the follow:
company number '0' '1' '2'
A 101 1 2 4
B 102 5 7 8
A 102 1 2 9
Here I wan to split the column name A-101 into two columns and transpose the column into a row with name column name '0', '1', '2'
Solution
output = pd.DataFrame(columns=['company', 'number', '0', '1', '2'])
output['company'] = [col.split('-')[0] for col in df.columns]
output['number'] = [col.split('-')[1] for col in df.columns]
output['0'] = df.iloc[0].values
output['1'] = df.iloc[1].values
output['2'] = df.iloc[2].values
I have code which works but gives me data without header is there a way I can write this code so header is not removed? I know one way will be to add back header, but is there a better way?
My code:
df = pd.read_csv(“_data.csv",skiprows=[0], header=None)
df = df.groupby([2])[10].sum().astype(float)
Data:
A B
1 2
1 1
2 3
2 4
I have data like above trying to get this result:
A B
1 3
2 7
Try to use the function reset_index after the sum:
data = [{'a': 1, 'b': 2},{'a': 1, 'b': 1},{'a': 2, 'b': 3},{'a': 2, 'b': 4}]
df = pd.DataFrame(data)
df
a b
0 1 2
1 1 1
2 2 3
3 2 4
df.groupby('a').sum().reset_index()
a b
0 1 3
1 2 7
You should specify the separator (several spaces in your case) and that the header is the first row (=0, with python indexing), than groupby the column you want.
df = pd.read_csv("_data.csv", sep='\s*', header=0)
A B
0 1 2
1 1 1
2 2 3
3 2 4
df = df.groupby(['A']).sum()
B
A
1 3
2 7
I say to dataframes.
df_A has columns A__a, B__b, C. (shape 5,3)
df_B has columns A_a, B_b, D. (shape 4,3)
How can I unify them (without having to iterate over all columns) to get one df with columns A,B ? (shape 9,2) - meaning A__a and A_a should be unified to the same column.
I need to use merge with applying the function lambda x: x.replace("_",""). Is it possible?
import pandas as pd
df = pd.DataFrame(np.random.randint(0,5,size=(5, 3)), columns=['A__a', 'B__b', 'C'])
df:
A__a B__b C
0 3 0 2
1 0 3 4
2 0 4 4
3 4 2 1
4 3 4 3
df2:
df2 = pd.DataFrame(np.random.randint(0,4,size=(4, 3)), columns=['A__a', 'B__b', 'D'])
A__a B__b D
0 3 2 0
1 3 1 1
2 0 2 0
3 3 2 0
df3 = pd.concat([df, df2], join='inner', ignore_index=True)
df_final = df3.rename(lambda x: str(x).split("__")[0],axis='columns')
df_final
df_final:
A B
0 3 0
1 0 3
2 0 4
3 4 2
4 3 4
5 3 2
6 3 1
7 0 2
8 3 2
A simple concatenation will do
pd.concat([df_A, df_B], join='outer')[['A', 'B']].copy().
or
'pd.concat([df_A, df_B], join='inner')
You have to merge Dataframe using 'outer'
import pandas as pd
import numpy as np
df_A = pd.DataFrame(np.random.randint(10,size=(5,3)), columns=['A','B','C'])
df_B = pd.DataFrame(np.random.randint(10,size=(4,3)), columns=['A','B','D'])
print(df_A.shape,df_B.shape)
#(5, 3) (4, 3)
new_df = df_A.merge(df_B , how= 'outer', on = ['A','B'])[['A','B']]
print(new_df.shape)
#(9,2)
If you cant change the name of the columns in advance and you want to use lambda x: x.replace("_",""), this is a way:
df = pd.concat([df1.rename_axis(lambda x: str(x).replace("_",""),axis='columns'), df2.rename_axis(lambda x: str(x).replace("_",""),axis='columns')], join='inner', ignore_index=True)
Example:
d1 = {'A__a' : ('A', 'B', 'C', 'D', 'E') , 'B__b' : ('a', 'b', 'c', 'd', 'e') ,'C': (1,2,3,4,5)}
df1 = pd.DataFrame(d1)
A__a B__b C
0 A a 1
1 B b 2
2 C c 3
3 D d 4
4 E e 5
d2 = {'A_a' : ('B', 'C', 'D','G') , 'B_b' : ('l','m','n','o') ,'D': (6,7,8,9)}
df2=pd.DataFrame(d2)
A_a B_b D
0 B l 6
1 C m 7
2 D n 8
3 G o 9
Output:
Aa Bb
0 A a
1 B b
2 C c
3 D d
4 E e
5 B l
6 C m
7 D n
8 G o
Alternative with:
df = pd.concat([df1.rename(columns={'A__a':'A', 'B__b':'B'}), df2.rename(columns={'A_a':'A', 'B_b':'B'})], join='inner', ignore_index=True)