通过分隔符将 Pandas 系列拆分为 DataFrame

Split Pandas Series into DataFrame by delimiter

提问by O.rka

I'm trying to split a pandasseriesobject by a particular delimiter "; "in this case. I want to turn it into a dataframethere will always be the same amount of "columns" or to be more exact, same amount of "; "that will indicate columns. I thought this would do the trick but it didnt python, how to convert a pandas series into a pandas DataFrame?I dont want to iterate through, I'm sure pandashas made a shortcut that's more effective.

在这种情况下，我试图pandasseries用特定的分隔符分割一个对象"; "。我想把它变成一个dataframe总是有相同数量的“列”，或者更准确地说，相同数量的“列”"; "将指示列。我认为这可以解决问题，但它没有使用python，如何将 Pandas 系列转换为 Pandas DataFrame？我不想重复，我肯定pandas已经做了一个更有效的捷径。

Does anyone know of the most efficient way to split this series into a dataframe by "; "?

有谁知道将这个系列拆分为数据帧的最有效方法是"; "？

#Example Data
SR_test = pd.Series(["a; b; c; d; e","aa; bb; cc; dd; ee","a1; b2; c3; d4; e5"])
# print(SR_test)
# 0         a; b; c; d; e
# 1    aa; bb; cc; dd; ee
# 2    a1; b2; c3; d4; e5

#Convert each row one at a time (not efficient)
tmp = []
for element in SR_test:
    tmp.append([e.strip() for e in element.split("; ")])
DF_split = pd.DataFrame(tmp)
# print(DF_split)
#     0   1   2   3   4
# 0   a   b   c   d   e
# 1  aa  bb  cc  dd  ee
# 2  a1  b2  c3  d4  e5

回答by jezrael

You can use str.split:

您可以使用str.split：

df = SR_test.str.split('; ', expand=True)
print df

    0   1   2   3   4
0   a   b   c   d   e
1  aa  bb  cc  dd  ee
2  a1  b2  c3  d4  e5

Another faster solution, if Serieshave no NaNvalues:

另一个更快的解决方案，如果Series没有NaN值：

print pd.DataFrame([ x.split('; ') for x in SR_test.tolist() ])
    0   1   2   3   4
0   a   b   c   d   e
1  aa  bb  cc  dd  ee
2  a1  b2  c3  d4  e5

Timings:

时间：

SR_test = pd.concat([SR_test]*1000).reset_index(drop=True)

In [21]: %timeit SR_test.str.split('; ', expand=True)
10 loops, best of 3: 34.5 ms per loop

In [22]: %timeit pd.DataFrame([ x.split('; ') for x in SR_test.tolist() ])
100 loops, best of 3: 9.59 ms per loop

回答by EdChum

Use the vectorised str.splitwith param expand=Trueand pass as the data arg to the DataFramector:

使用str.split带有 param 的向量化expand=True并作为数据 arg 传递给DataFramector：

In [4]:
df = pd.DataFrame(SR_test.str.split(';',expand=True))
df

Out[4]:
    0    1    2    3    4
0   a    b    c    d    e
1  aa   bb   cc   dd   ee
2  a1   b2   c3   d4   e5