Pandas：借助字典将变量子字符串从 A 列插入 B 列

Question

我有这个pandas 数据框：

df = pd.DataFrame(["LONG AAPL 2X CBZ","SHORT GOOG 10X VON"], columns=["Name"])

我想在Name 列中识别"AAPL"，将其传递给字典"AAPL":"Apple"，然后将其插入到新列Description 的字符串中。

期望的输出：

Name                   Description
"LONG AAPL 2X CBZ"     "Tracks Apple with 2X leverage."
"SHORT GOOG 10X VON"   "Tracks Google with -10X leverage."

我遇到问题的部分是将变量子字符串输入另一个字符串"Tracks X with Y leverage."

如果我不必这样做，只需从 name 提取到 description 是可能的：

df["Description"] = df["Name"].str.extract(r"\s(\S+)\s").map({"AAPL":"Apple", "GOOG":"Google"})

或提取杠杆：

df["Description"] = df["Name"].str.extract(r"(\d+X)")

如果可能的话，我想使用regex 来提取变量，因为实际上我会做一些更详细的正则表达式，例如用于检索不同格式的乘数，例如X2、2x 等等。

注意：我可能需要设置另一列来告知杠杆是正还是负，并用它来决定是否在乘数前面附加"-"作为-10X leverage.

df["direction"] = df["name"].map(lambda x: "Long" if "LONG" in x else "Short" if "SHORT " in x else "Long")

Name                   Direction      Description
"LONG AAPL 2X CBZ"     "Long"         "Tracks Apple with 2X leverage."
"SHORT GOOG 10X VON"   "Short"        "Tracks Google with -10X leverage."

Answer 1

因为我们只关心前两个和倒数第二个子字符串：

df = pd.DataFrame(["LONG AAPL 2X CBZ", "SHORT GOOG 10X VON", "BULL AXP UN X3 VON","LONG AXP X3 VON"], columns=["Name"])

maps = {"AAPL": "Apple", "GOOG": "Google"}
signs = {"SHORT": "-"}

def split(i):
    spl = i.split()
    a, b, c = spl[0], spl[1], spl[-2]
    val = maps.get(b, b) # if name is not to be replaced keep original
    return "Tracks  {} with {}{} leverage".format(val, signs.get(a, ""), c)

df["Description"]  = df["Name"].map(split)

输出：

                 Name                        Description
0    LONG AAPL 2X CBZ     Tracks  Apple with 2X leverage
1  SHORT GOOG 10X VON  Tracks  Google with -10X leverage
2  BULL AXP UN X3 VON       Tracks  AXP with X3 leverage
3     LONG AXP X3 VON       Tracks  AXP with X3 leverage

仅拆分比使用正则表达式更有效：

In [33]: df2 = pd.concat([df]*10000)
In [34]: timeit  df2["Name"].map(split)
10 loops, best of 3: 57.5 ms per loop

In [35]: timeit f2(df2['Name'])
10 loops, best of 3: 168 ms per loop

如果你想添加更多的词来替换，只需将它们添加到地图中，与符号相同。

Answer 2

您可以定义一个显式函数以应用于整个Name 系列。

df = pd.DataFrame(["LONG AAPL 2X CBZ","SHORT GOOG 10X VON"], columns=["Name"])

dmap = {"AAPL":"Apple", "GOOG":"Google"}
signmap = {"LONG": "", "SHORT": "-"}

def f(strseries):
    company = strseries.str.extract(r"\s(\S+)\s").map(dmap)
    leverage = strseries.str.extract(r"(\d+X)")
    sign = strseries.str.extract(r"(\S+)\s").map(signmap)
    return "Tracks " + company + " with " + sign + leverage + " leverage."

df['Description'] = f(df['Name'])

编辑：一次执行正则表达式提取可以将速度提高大约 2 倍，但会牺牲一些可读性。

def f2(strseries):
    sub_df = strseries.str.extract('(?P<sign>\S+)\s(?P<company>\S+)\s(?P<leverage>\d+X)')
    return "Tracks " + sub_df.company.map(dmap) + " with " + sub_df.sign.map(signmap) + sub_df.leverage + " leverage."

df['Description'] = f2(df['Name'])