在 Python pandas 中，从 1 而不是 0 开始行索引而不创建额外的列
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In Python pandas, start row index from 1 instead of zero without creating additional column
提问by Bram Vanroy
I know that I can reset the indices like so
我知道我可以像这样重置索引
df.reset_index(inplace=True)
but this will start the index from 0. I want to start it from 1. How do I do that without creating any extra columns and by keeping the index/reset_index functionality and options? I do notwant to create a new dataframe, so inplace=Trueshould still apply.
但这将从0. 我想从1. 如何在不创建任何额外列并保留 index/reset_index 功能和选项的情况下做到这一点？我不希望创建一个新的数据帧，所以inplace=True应该仍然适用。
采纳答案by EdChum
Just assign directly a new index array:
只需直接分配一个新的索引数组：
df.index = np.arange(1, len(df) + 1)
Example:
例子：
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:
要不就：
df.index = df.index + 1
If the index is already 0 based
如果索引已经基于 0
TIMINGS
时机
For some reason I can't take timings on reset_indexbut the following are timings on a 100,000 row df:
出于某种原因，我无法计时，reset_index但以下是 100,000 行 df 的计时：
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_indexI can't say definitively, however it looks like just adding 1 to each index value will be faster if the index is already 0based
所以没有时间reset_index我不能明确地说，但是如果索引已经0基于，看起来只是向每个索引值添加 1 会更快
回答by hakuna_code
You can also specify the start value using index range like below. RangeIndex is supported in pandas.
您还可以使用如下所示的索引范围指定起始值。熊猫支持 RangeIndex。
#df.index
default value is printed, (start=0,stop=lastelement, step=1)
打印默认值，(start=0,stop=lastelement, step=1)
You can specify any start value range like this:
您可以指定任何起始值范围，如下所示：
df.index = pd.RangeIndex(start=1, stop=600, step=1)
Refer: pandas.RangeIndex
