[英]Combine two columns in pandas dataframe but in specific order
For example, I have a dataframe where two of the columns are "Zeroes" and "Ones" that contain only zeroes and ones, respectively.例如,我有一个数据框,其中两列是“零”和“一”,分别只包含零和一。 If I combine them into one column I get first all the zeroes, then all the ones.如果我将它们组合成一列,我首先会得到所有的零,然后是所有的。
I want to combine them in a way that I get each element from both columns, not all elements from the first column and all elements from the second column.我想以一种从两列中获取每个元素的方式组合它们,而不是第一列中的所有元素和第二列中的所有元素。 So I don't want the result to be [0, 0, 0, 1, 1, 1], I need it to be [0, 1, 0, 1, 0, 1].所以我不希望结果是 [0, 0, 0, 1, 1, 1],我需要它是 [0, 1, 0, 1, 0, 1]。
I process 100K+ rows of data.我处理了 100K+ 行数据。 What is the fastest or optimal way to achieve this?实现这一目标的最快或最佳方法是什么? Thanks in advance!提前致谢!
Try:尝试:
import pandas as pd
df = pd.DataFrame({ "zeroes" : [0, 0, 0], "ones": [1, 1, 1], "some_other" : list("abc")})
res = df[["zeroes", "ones"]].to_numpy().ravel(order="C")
print(res)
Output输出
[0 1 0 1 0 1]
Micro-Benchmarks微基准
import pandas as pd
from itertools import chain
df = pd.DataFrame({ "zeroes" : [0] * 10_000, "ones": [1] * 10_000})
%timeit df[["zeroes", "ones"]].to_numpy().ravel(order="C").tolist()
672 µs ± 8.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit [v for vs in zip(df["zeroes"], df["ones"]) for v in vs]
2.57 ms ± 54 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit list(chain.from_iterable(zip(df["zeroes"], df["ones"])))
2.11 ms ± 73 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
You can use numpy.flatten()
like below as alternative:您可以使用numpy.flatten()
如下所示作为替代:
import numpy as np
import pandas as pd
df[["zeroes", "ones"]].to_numpy().flatten()
Benchmark (runnig on colab ) :基准(在colab上运行) :
df = pd.DataFrame({ "zeroes" : [0] * 10_000_000, "ones": [1] * 10_000_000})
%timeit df[["zeroes", "ones"]].to_numpy().flatten().tolist()
1 loop, best of 5: 320 ms per loop
%timeit df[["zeroes", "ones"]].to_numpy().ravel(order="C").tolist()
1 loop, best of 5: 322 ms per loop
I don't know if this is the most optimal solution but it should solve your case.我不知道这是否是最佳解决方案,但它应该可以解决您的问题。
df = pd.DataFrame([[0 for x in range(10)], [1 for x in range(10)]]).T
l = [[x, y] for x, y in zip(df[0], df[1])]
l = [x for y in l for x in y]
l
This may help you: Alternate elements of different columns using Pandas这可能对您有所帮助: 使用 Pandas 替换不同列的元素
pd.concat(
[df1, df2], axis=1
).stack().reset_index(1, drop=True).to_frame('C').rename(index='CC{}'.format)
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