Forcing pandas .iloc to return a single-row dataframe?

Solution 1:

Use double brackets,

df.iloc[[0]]

Output:

   a  b
0  1  3

print(type(df.iloc[[0]])

<class 'pandas.core.frame.DataFrame'>

Short for df.iloc[[0],:]

Solution 2:

Accessing row(s) by label: loc

# Setup
df = pd.DataFrame({'X': [1, 2, 3], 'Y':[4, 5, 6]}, index=['a', 'b', 'c'])
df        

   X  Y
a  1  4
b  2  5
c  3  6

To get a DataFrame instead of a Series, pass a list of indices of length 1,

df.loc[['a']]
# Same as
df.loc[['a'], :] # selects all columns

   X  Y
a  1  4

To select multiple specific rows, use

df.loc[['a', 'c']] 

   X  Y
a  1  4
c  3  6

To select a contiguous range of rows, use

df.loc['b':'c'] 

   X  Y
b  2  5
c  3  6

Access row(s) by position: iloc

Specify a list of indices of length 1,

i = 1
df.iloc[[i]]

   X  Y
b  2  5

Or, specify a slice of length 1:

df.iloc[i:i+1] 

   X  Y
b  2  5

To select multiple rows or a contiguous slice you'd use a similar syntax as with loc.

Solution 3:

please use the below options:

df1 = df.iloc[[0],:]
#type(df1)
df1

or

df1 = df.iloc[0:1,:]
#type(df1)
df1

Solution 4:

The double-bracket approach doesn't always work for me (e.g. when I use a conditional to select a timestamped row with loc).

You can, however, just add to_frame() to your operation.

>>> df = pd.DataFrame({'a':[1,2], 'b':[3,4]})

>>> df2 = df.iloc[0, :].to_frame()

>>> type(df2)
<class 'pandas.core.frame.DataFrame'>