I have this dataframe df
where
>>> df = pd.DataFrame({'Date':['10/2/2011', '11/2/2011', '12/2/2011', '13/2/11'],
'Event':['Music', 'Poetry', 'Theatre', 'Comedy'],
'Cost':[10000, 5000, 15000, 2000],
'Name':['Roy', 'Abraham', 'Blythe', 'Sophia'],
'Age':['20', '10', '13', '17']})
I want to determine the column index with the corresponding name. I tried it with this:
>>> list(df.columns)
But the solution above only returns the column names without index numbers.
How can I code it so that it would return the column names and the corresponding index for that column? Like This:
0 Date
1 Event
2 Cost
3 Name
4 Age
CodePudding user response:
A nice short way to get a dictionary:
d = dict(enumerate(df))
output: {0: 'Date', 1: 'Event', 2: 'Cost', 3: 'Name', 4: 'Age'}
For a Series, pd.Series(list(df))
is sufficient as iteration occurs directly on the column names
CodePudding user response:
Simpliest is add pd.Series
constructor:
pd.Series(list(df.columns))
Or convert columns to Series and create default index:
df.columns.to_series().reset_index(drop=True)
Or:
df.columns.to_series(index=False)
CodePudding user response:
In addition to using enumerate
, this also can get a numbers in order using zip
, as follows:
import pandas as pd
df = pd.DataFrame({'Date':['10/2/2011', '11/2/2011', '12/2/2011', '13/2/11'],
'Event':['Music', 'Poetry', 'Theatre', 'Comedy'],
'Cost':[10000, 5000, 15000, 2000],
'Name':['Roy', 'Abraham', 'Blythe', 'Sophia'],
'Age':['20', '10', '13', '17']})
result = list(zip([i for i in range(len(df.columns))], df.columns.values,))
for r in result:
print(r)
#(0, 'Date')
#(1, 'Event')
#(2, 'Cost')
#(3, 'Name')
#(4, 'Age')
CodePudding user response:
You can use loop like this:
myList = list(df.columns)
index = 0
for value in myList:
print(index, value)
index = 1