熊猫:根据栏内容有条件地生成描述

Question

我试图解决一个函数的一些问题，该函数使用pandas regex通过str.extract获取列"name"中的每一行以生成列"description"。我使用的是regex，而不是split，因为代码必须能够管理各种格式。

必须修改该函数以承认各种条件。

DataFrame：

import pandas as pd
import re

df = pd.DataFrame(["LONG AXP UN X3 VON", "SHORT BIDU UN 5x VON", "SHORT GOOG VON", "LONG GOOG VON"], columns=["name"])

输入：

name
"LONG AXP UN X3 VON"
"SHORT BIDU UN 5x VON"
"SHORT GOOG VON"
"LONG GOOG VON"

当前代码：

description_map = {"AXP":"American Express", "BIDU":"Baidu"}
sign_map = {"LONG": "", "SHORT": "-"}
def f(strseries):
    stock = strseries.str.extract(r"\s(\S+)\s").map(description_map)
    leverage = strseries.str.extract(r"(X\d+|\d+X)\s", flags=re.IGNORECASE)
    sign = strseries.str.extract(r"(\S+)\s").map(sign_map)
    return "Tracks " + stock + " with " + sign + leverage + " leverage"

df["description"] = f(df["name"])

当前产出：

name                        description
"LONG AXP UN X3 VON"        "Tracks American Express with X3 leverage"
"SHORT BIDU UN 5x VON"      "Tracks Baidu with -5x leverage"
"SHORT GOOG VON"            ""
"LONG GOOG VON"             ""

期望产出：

name                        description
"LONG AXP UN X3 VON"        "Tracks American Express with 3x leverage"
"SHORT BIDU UN 5x VON"      "Tracks Baidu inversely with -5x leverage"
"SHORT GOOG VON"            "Tracks inversely"
"LONG GOOG VON"             "Tracks"

所涉问题：

• 如果sign是"-"，我如何使它将direction = "inversely"添加到字符串中？
• 如果stock在name中与字典description_map没有匹配，则设置stock = ""并返回字符串。
• 如果在leverage中没有找到name：忽略部分"with" + sign + leverage + " leverage"。
• 对sign + leverage进行拆分和重新排序，使其始终以-5x"顺序显示，而不管它是否输入为"SHORT X5"。

Answer 1

我花了一些时间编写这个函数：

description_map = {"AXP":"American Express", "BIDU":"Baidu"}
sign_map = {"LONG": "", "SHORT": "-"}

stock_match = re.compile(r"\s(\S+)\s")
leverage_match = re.compile("[0-9]x|x[0-9]|X[0-9]|[0-9]X")

def f(value):

    f1 = lambda x: description_map[stock_match.findall(x)[0]] if stock_match.findall(x)[0] in description_map else ''
    f2 = lambda x: leverage_match.findall(x)[0] if len(leverage_match.findall(x)) > 0 else ''
    f3 = lambda x: '-' if 'SHORT' in x else ''

    stock = f1(value)
    leverage = f2(value)
    sign = f3(value)

    statement = "Tracks " + stock

    if stock == "":
        if sign == '-':
            return statement + "{}".format('inversely')
        else:
            return "Tracks"

    if leverage[0].replace('X','x') == 'x':
        leverage = leverage[1]+leverage[0].replace('X','x')

    if leverage != '' and sign == '-':
        statement += " {} with {}{} leverage".format('inversely', sign, leverage)
    elif leverage != '' and sign == '':
        statement += " with {} leverage".format(leverage)
    else:
        if sign == '-':
            statement += " {} ".format('Inversely')

    return statement

df["description"] = df["name"].map(lambda x:f(x))

输出：

In [97]: %paste
import pandas as pd
import re

df = pd.DataFrame(["LONG AXP UN X3 VON", "SHORT BIDU UN 5x VON", "SHORT GOOG VON", "LONG GOOG VON"], columns=["name"])

## -- End pasted text --

In [98]: df
Out[98]: 
                   name
0    LONG AXP UN X3 VON
1  SHORT BIDU UN 5x VON
2        SHORT GOOG VON
3         LONG GOOG VON

In [99]: df["description"] = df["name"].map(lambda x:f(x))

In [100]: df
Out[100]: 
                   name                               description
0    LONG AXP UN X3 VON  Tracks American Express with 3x leverage
1  SHORT BIDU UN 5x VON  Tracks Baidu inversely with -5x leverage
2        SHORT GOOG VON                          Tracks inversely
3         LONG GOOG VON                                    Tracks