In Python pandas, start row index from 1 instead of zero without creating additional column

Question

I know that I can reset the indices like so

df.reset_index(inplace=True)

but this will start the index from 0. I want to start

Answer 1

You can also specify the start value using index range like below. RangeIndex is supported in pandas.

#df.index

default value is printed, (start=0,stop=lastelement, step=1)

You can specify any start value range like this:

df.index = pd.RangeIndex(start=1, stop=600, step=1)

Refer: pandas.RangeIndex

Answer 2

For this, you can do the following(I created an example dataframe):

price_of_items = pd.DataFrame({
"Wired Keyboard":["$7","4.3","12000"],"Wireless Keyboard":["$13","4.6","14000"]
                             })
price_of_items.index += 1

Answer 3

Just assign directly a new index array:

df.index = np.arange(1, len(df) + 1)

Example:

In [151]:

df = pd.DataFrame({'a':np.random.randn(5)})
df
Out[151]:
          a
0  0.443638
1  0.037882
2 -0.210275
3 -0.344092
4  0.997045
In [152]:

df.index = np.arange(1,len(df)+1)
df
Out[152]:
          a
1  0.443638
2  0.037882
3 -0.210275
4 -0.344092
5  0.997045

Or just:

df.index = df.index + 1

If the index is already 0 based

TIMINGS

For some reason I can't take timings on reset_index but the following are timings on a 100,000 row df:

In [160]:

%timeit df.index = df.index + 1
The slowest run took 6.45 times longer than the fastest. This could mean that an intermediate result is being cached 
10000 loops, best of 3: 107 µs per loop


In [161]:

%timeit df.index = np.arange(1, len(df) + 1)
10000 loops, best of 3: 154 µs per loop

So without the timing for reset_index I can't say definitively, however it looks like just adding 1 to each index value will be faster if the index is already 0 based