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