Pandas 数据框的 Concat 列表，但忽略列名
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Concat list of pandas data frame, but ignoring column name
提问by Darren Cook
Sub-title: Dumb it down pandas, stop trying to be clever.
副标题：让Pandas们闭嘴，别再想聪明了。
I've a list (res) of single-column pandas data frames, each containing the same kind of numeric data, but each with a different column name. The row indices have no meaning. I want to put them into a single, very long, single-column data frame.
我有一个res单列 Pandas 数据框的列表 ( )，每个都包含相同类型的数字数据，但每个都有不同的列名。行索引没有意义。我想将它们放入一个非常长的单列数据框中。
When I do pd.concat(res)I get one column per input file (and loads and loads of NaN cells). I've tried various values for the parameters (*), but none that do what I'm after.
当我这样做时，pd.concat(res)我会为每个输入文件获取一列（以及加载和加载 NaN 单元格）。我已经尝试了参数 (*) 的各种值，但没有一个能满足我的要求。
Edit: Sample data:
编辑：示例数据：
res = [
pd.DataFrame({'A':[1,2,3]}),
pd.DataFrame({'B':[9,8,7,6,5,4]}),
pd.DataFrame({'C':[100,200,300,400]}),
]
I have an ugly-hack solution: copy every data frame and giving it a new column name:
我有一个丑陋的解决方案：复制每个数据框并给它一个新的列名：
newList = []
for r in res:
r.columns = ["same"]
newList.append(r)
pd.concat( newList, ignore_index=True )
Surely that is not the best way to do it??
当然，这不是最好的方法吗？？
BTW, pandas: concat data frame with different column nameis similar, but my question is even simpler, as I don't want the index maintained. (I also start with a list of N single-column data frames, not a single N-column data frame.)
顺便说一句，pandas：具有不同列名的 concat 数据框是相似的，但我的问题更简单，因为我不希望维护索引。（我也从 N 个单列数据框的列表开始，而不是单个 N 列数据框。）
*: E.g. axis=0is default behaviour. axis=1gives an error. join="inner"is just silly (I only get the index). ignore_index=Truerenumbers the index, but I stil gets lots of columns, lots of NaNs.
*：例如axis=0是默认行为。axis=1给出一个错误。join="inner"只是愚蠢（我只得到索引）。ignore_index=True重新编号索引，但我仍然得到很多列，很多 NaN。
UPDATE for empty lists
更新空列表
I was having problems (with all the given solutions) when the data had an empty list, something like:
当数据有一个空列表时，我遇到了问题（所有给定的解决方案），例如：
res = [
pd.DataFrame({'A':[1,2,3]}),
pd.DataFrame({'B':[9,8,7,6,5,4]}),
pd.DataFrame({'C':[]}),
pd.DataFrame({'D':[100,200,300,400]}),
]
The trick was to force the type, by adding .astype('float64'). E.g.
诀窍是通过添加.astype('float64'). 例如
pd.Series(np.concatenate([df.values.ravel().astype('float64') for df in res]))
or:
或者：
pd.concat(res,axis=0).astype('float64').stack().reset_index(drop=True)
采纳答案by Steven G
I would use list comphrension such has:
我会使用列表理解，例如：
import pandas as pd
res = [
pd.DataFrame({'A':[1,2,3]}),
pd.DataFrame({'B':[9,8,7,6,5,4]}),
pd.DataFrame({'C':[100,200,300,400]}),
]
x = []
[x.extend(df.values.tolist()) for df in res]
pd.DataFrame(x)
Out[49]:
0
0 1
1 2
2 3
3 9
4 8
5 7
6 6
7 5
8 4
9 100
10 200
11 300
12 400
I tested speed for you.
我为你测试了速度。
%timeit x = []; [x.extend(df.values.tolist()) for df in res]; pd.DataFrame(x)
10000 loops, best of 3: 196 μs per loop
%timeit pd.Series(pd.concat(res, axis=1).values.ravel()).dropna()
1000 loops, best of 3: 920 μs per loop
%timeit pd.concat(res, axis=1).stack().reset_index(drop=True)
1000 loops, best of 3: 902 μs per loop
%timeit pd.DataFrame(pd.concat(res, axis=1).values.ravel(), columns=['col']).dropna()
1000 loops, best of 3: 1.07 ms per loop
%timeit pd.Series(np.concatenate([df.values.ravel() for df in res]))
10000 loops, best of 3: 70.2 μs per loop
looks like
好像
pd.Series(np.concatenate([df.values.ravel() for df in res]))
is the fastest.
是最快的。
回答by jezrael
I think you need concatwith stack:
print (pd.concat(res, axis=1))
A B C
0 1.0 9 100.0
1 2.0 8 200.0
2 3.0 7 300.0
3 NaN 6 400.0
4 NaN 5 NaN
5 NaN 4 NaN
print (pd.concat(res, axis=1).stack().reset_index(drop=True))
0 1.0
1 9.0
2 100.0
3 2.0
4 8.0
5 200.0
6 3.0
7 7.0
8 300.0
9 6.0
10 400.0
11 5.0
12 4.0
dtype: float64
Another solution with numpy.ravelfor flattening:
numpy.ravel用于展平的另一种解决方案：
print (pd.Series(pd.concat(res, axis=1).values.ravel()).dropna())
0 1.0
1 9.0
2 100.0
3 2.0
4 8.0
5 200.0
6 3.0
7 7.0
8 300.0
10 6.0
11 400.0
13 5.0
16 4.0
dtype: float64
print (pd.DataFrame(pd.concat(res, axis=1).values.ravel(), columns=['col']).dropna())
col
0 1.0
1 9.0
2 100.0
3 2.0
4 8.0
5 200.0
6 3.0
7 7.0
8 300.0
10 6.0
11 400.0
13 5.0
16 4.0
Solution with list comprehension:
解决方案list comprehension：
print (pd.Series(np.concatenate([df.values.ravel() for df in res])))
0 1
1 2
2 3
3 9
4 8
5 7
6 6
7 5
8 4
9 100
10 200
11 300
12 400
dtype: int64
