Python Pandas 在连接后重新计算索引
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Pandas recalculate index after a concatenation
提问by Christopher
I have a problem where I produce a pandas dataframe by concatenating along the row axis (stacking vertically).
我有一个问题，我通过沿行轴连接（垂直堆叠）来生成熊猫数据框。
Each of the constituent dataframes has an autogenerated index (ascending numbers).
每个组成数据帧都有一个自动生成的索引（升序数字）。
After concatenation, my index is screwed up: it counts up to n (where n is the shape[0] of the corresponding dataframe), and restarts at zero at the next dataframe.
连接后，我的索引被搞砸了：它计数到 n（其中 n 是相应数据帧的形状 [0]），并在下一个数据帧处从零重新开始。
I am trying to "re-calculate the index, given the current order", or "re-index" (or so I thought). Turns out that isn't exactly what DataFrame.reindexseems to be doing.
我正在尝试“根据当前订单重新计算索引”或“重新索引”（或者我认为）。事实证明，这并不完全是DataFrame.reindex看起来在做的事情。
Here is what I tried to do:
这是我尝试做的：
train_df = pd.concat(train_class_df_list)
train_df = train_df.reindex(index=[i for i in range(train_df.shape[0])])
It failed with "cannot reindex from a duplicate axis." I don't want to change the order of my data... just need to delete the old index and set up a new one, with the order of rows preserved.
它因“无法从重复轴重新索引”而失败。我不想改变我的数据的顺序......只需要删除旧索引并设置一个新索引，并保留行的顺序。
采纳答案by Ami Tavory
After vertical concatenation, if you get an index of [0, n)followed by [0, m), all you need to do is call reset_index:
垂直串联后，如果您得到[0, n)后跟[0, m)的索引，您需要做的就是调用reset_index：
train_df.reset_index(drop=True)
(you can do this in place using inplace=True).
（您可以使用 就地执行此操作inplace=True）。
import pandas as pd
>>> pd.concat([
pd.DataFrame({'a': [1, 2]}),
pd.DataFrame({'a': [1, 2]})]).reset_index(drop=True)
a
0 1
1 2
2 1
3 2
回答by Mike Müller
回答by ilmarinen
If your index is autogenerated and you don't want to keep it, you can use the ignore_indexoption.
`
如果您的索引是自动生成的并且您不想保留它，则可以使用该ignore_index选项。`
train_df = pd.concat(train_class_df_list, ignore_index=True)
This will autogenerate a new index for you, and my guess is that this is exactly what you are after.
这将为您自动生成一个新索引，我猜这正是您所追求的。
