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i'm trying to shift the first row only by one cell to the right so the dates start under number 1 column,
also i'm trying to remove the tailing '\n' by doing this but its not working, any help please?
income_df2 = income_df2.replace('[\$,)]','', regex=True )\
.replace( '[(]','-', regex=True)\
.replace( '', 'NaN', regex=True)
Yes, you can do something like this shift he first row of a dataframe to the right one column. Use iloc to select this row all columns which returns a pd.Series, then use shift to shift the values of this series one position and assign this newly shifted series back to the first row of the dataframe.
df.iloc[0, :] = df.iloc[0, :].shift()
MCVE:
import pandas as pd
import numpy as np
df = pd.DataFrame([[*'ABCD']+[np.nan],[1,2,3,4,5],[5,6,7,9,10],[11,12,13,14,15]])
df
# Input DataFrame
# 0 1 2 3 4
# 0 A B C D NaN
# 1 1 2 3 4 5.0
# 2 5 6 7 9 10.0
# 3 11 12 13 14 15.0
df.iloc[0, :] = df.iloc[0, :].shift()
df
# Output DataFrame
# 0 1 2 3 4
# 0 NaN A B C D
# 1 1 2 3 4 5
# 2 5 6 7 9 10
# 3 11 12 13 14 15
Related
df1 = pd.DataFrame({'a':[1,2,3],'x':[4,5,6],'y':[7,8,9]})
df2 = pd.DataFrame({'b':[10,11,12],'x':[13,14,15],'y':[16,17,18]})
I'm trying to merge the two data frames using the keys from the df1. I think I should use pd.merge for this, but I how can I tell pandas to place the values in the b column of df2 in the a column of df1. This is the output I'm trying to achieve:
a x y
0 1 4 7
1 2 5 8
2 3 6 9
3 10 13 16
4 11 14 17
5 12 15 18
Just use concat and rename the column for df2 so it aligns:
In [92]:
pd.concat([df1,df2.rename(columns={'b':'a'})], ignore_index=True)
Out[92]:
a x y
0 1 4 7
1 2 5 8
2 3 6 9
3 10 13 16
4 11 14 17
5 12 15 18
similarly you can use merge but you'd need to rename the column as above:
In [103]:
df1.merge(df2.rename(columns={'b':'a'}),how='outer')
Out[103]:
a x y
0 1 4 7
1 2 5 8
2 3 6 9
3 10 13 16
4 11 14 17
5 12 15 18
Use numpy to concatenate the dataframes, so you don't have to rename all of the columns (or explicitly ignore indexes). np.concatenate also works on an arbitrary number of dataframes.
df = pd.DataFrame( np.concatenate( (df1.values, df2.values), axis=0 ) )
df.columns = [ 'a', 'x', 'y' ]
df
You can rename columns and then use functions append or concat:
df2.columns = df1.columns
df1.append(df2, ignore_index=True)
# pd.concat([df1, df2], ignore_index=True)
You can also concatenate both dataframes with vstack from numpy and convert the resulting ndarray to dataframe:
pd.DataFrame(np.vstack([df1, df2]), columns=df1.columns)
I have a pandas dataframe as below:
import pandas as pd
df = pd.DataFrame({'ORDER':["A", "A", "A", "B", "B","B"], 'GROUP': ["A_2018_1B1", "A_2018_1B1", "A_2018_1M1", "B_2018_I000_1C1", "B_2018_I000_1B1", "B_2018_I000_1C1H"], 'VAL':[1,3,8,5,8,10]})
df
ORDER GROUP VAL
0 A A_2018_1B1 1
1 A A_2018_1B1H 3
2 A A_2018_1M1 8
3 B B_2018_I000_1C1 5
4 B B_2018_I000_1B1 8
5 B B_2018_I000_1C1H 10
I want to create a column "CAL" as sum of 'VAL' where GROUP name is same for all the rows expect H character in the end. So, for example, 'VAL' column for 1st two rows will be added because the only difference between the 'GROUP' is 2nd row has H in the last. Row 3 will remain as it is, Row 4 and 6 will get added and Row 5 will remain same.
My expected output
ORDER GROUP VAL CAL
0 A A_2018_1B1 1 4
1 A A_2018_1B1H 3 4
2 A A_2018_1M1 8 8
3 B B_2018_I000_1C1 5 15
4 B B_2018_I000_1B1 8 8
5 B B_2018_I000_1C1H 10 15
Try with replace then transform
df.groupby(df.GROUP.str.replace('H','')).VAL.transform('sum')
0 4
1 4
2 8
3 15
4 8
5 15
Name: VAL, dtype: int64
df['CAL'] = df.groupby(df.GROUP.str.replace('H','')).VAL.transform('sum')
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 as follow:
import pandas as pd
d = {'location1': [1, 2,3,8,6], 'location2':
[2,1,4,6,8]}
df = pd.DataFrame(data=d)
The dataframe df means there is a road between two locations. look like:
location1 location2
0 1 2
1 2 1
2 3 4
3 8 6
4 6 8
The first row means there is a road between locationID1 and locationID2, however, the second row also encodes this information. The forth and fifth rows also have repeated information. I am trying the remove those repeated by keeping only one row. Any of row is okay.
For example, my expected output is
location1 location2
0 1 2
2 3 4
4 6 8
Any efficient way to do that because I have a large dataframe with lots of repeated rows.
Thanks a lot,
It looks like you want every other row in your dataframe. This should work.
import pandas as pd
d = {'location1': [1, 2,3,8,6], 'location2':
[2,1,4,6,8]}
df = pd.DataFrame(data=d)
print(df)
location1 location2
0 1 2
1 2 1
2 3 4
3 8 6
4 6 8
def Every_other_row(a):
return a[::2]
Every_other_row(df)
location1 location2
0 1 2
2 3 4
4 6 8
I am formatting some csv files, and I need to add columns that use other columns for arithmetic. Like in Excel, B3 = sum(A1:A3)/3, then B4 = sum(A2:A4)/3. I've looked up relative indexes and haven't found what I'm Trying to do.
def formula_columns(csv_list, dir_env):
for file in csv_list:
df = pd.read_csv(dir_env + file)
avg_12(df)
print(df[10:20])
# Create AVG(12) Column
def avg_12 ( df ):
df[ 'AVG(12)' ] = df[ 'Price' ]
# Right Here I want to set each value of 'AVG(12)' to equal
# the sum of the value of price from its own index plus the
# previous 11 indexes
df.loc[:10, 'AVG(12)'] = 0
I would imagine this to be a common task, I would assume I'm looking in the wrong places. If anyone has some advice I would appreciate it, Thank.
That can be done with the rolling method:
import numpy as np
import pandas as pd
np.random.seed(1)
df = pd.DataFrame(np.random.randint(1, 5, 10), columns = ['A'])
df
Out[151]:
A
0 2
1 4
2 1
3 1
4 4
5 2
6 4
7 2
8 4
9 1
Take the averages of A1:A3, A2:A4 etc:
df.rolling(3).mean()
Out[152]:
A
0 NaN
1 NaN
2 2.333333
3 2.000000
4 2.000000
5 2.333333
6 3.333333
7 2.666667
8 3.333333
9 2.333333
It requires pandas 18. For earlier versions, use pd.rolling_mean():
pd.rolling_mean(df['A'], 3)