试试这个：

import pandas as pd

pd.set_option('display.max_colwidth', None)

jeopardy = pd.DataFrame(
    [
        [4680, "2004-12-31", "Jeopardy!", "HISTORY", "$200", "For the last 8 years of his life, Galileo was under house arrest for espousing this man's theory", "Copernicus"],
        [4680, "2004-12-31", "Jeopardy!", "ESPN's TOP ALL-TIME ATHLETES", "$200", "No. 2: 1912 Olympian; football star at Carlisle Indian School; 6 MLB seasons with the Reds, Giants & Braves", "Jim Thorpe"],
        [4680, "2004-12-31", "Jeopardy!", "EVERYBODY TALKS ABOUT IT...", "$200", "The city of Yuma in this state has a record average of 4,055 hours of sunshine each year", "Arizona"],
        [4680, "2004-12-31", "Jeopardy!", "THE COMPANY LINE", "$1,300.50", 'In 1963, live on "The Art Linkletter Show", this company served its billionth burger', "McDonald's"],
        [4680, "2004-12-31", "Jeopardy!", "EPITAPHS & TRIBUTES", "$ 1,200 ", "Signer of the Dec. of Indep., framer of the Constitution of Mass., second President of the United States", "John Adams"],
    ],
    columns=["Show Number", "Air Date", "Round", "Category", "Value", "Question", "Answer"]
)
jeopardy.rename(columns=lambda x: x.strip(), inplace=True)

# Convert values to float by first removing the "$" sign and the "," (thousands separator).
# Then call `pandas.to_numeric()` function with `errors="coerce"` to convert values to numeric,
# and values that fail to convert to `None`.
jeopardy["Value_float"] = pd.to_numeric(jeopardy['Value'].str.replace("$", "", regex=False).str.replace(",", "", regex=False), errors="coerce")

# Find the values that failed to convert to float
errors = jeopardy[jeopardy["Value_float"].isna()]

jeopardy

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输出：

| 展会编号|播出日期|轮|类别|值|问题|回答|Value_float|
| --|--|--|--|--|--|--|--|
| 4680 |2004年12月31日|危险！|历史|两百块|在他生命的最后8年里，伽利略因为支持这个人的理论而被软禁。|哥白尼| 200 |
| 4680 |2004年12月31日|危险！|ESPN的顶级运动员|两百块|第2名：1912年奥运会选手;卡莱尔印第安学校的足球星星;在红军、巨人队和勇士队效力了6个MLB赛季|吉姆·索普| 200 |
| 4680 |2004年12月31日|危险！|每个人都在谈论它...|两百块|该州的尤马市每年平均日照时间达到创纪录的4,055小时|亚利桑那| 200 |
| 4680 |2004年12月31日|危险！|公司的路线|一千三百块五|1963年，在“艺术链接字母秀”的现场直播中，这家公司供应了第十亿个汉堡|麦当劳|一千三百点五|
| 4680 |2004年12月31日|危险！|墓志铭和颂词|一千二百块|印第安纳州12月的签署者，马萨诸塞州宪法的制定者，美国第二任总统|约翰·亚当斯| 1200 |

也许像这样的东西会起作用？它允许数字通过（如果它们还不是浮点数，就变成float），但它会删除不需要的字符：

import re

def to_float(x):
    if isinstance(x, str):
        x = re.sub('[$,%]', '', x)
    return float(x)

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范例：

v = ['$2,000.00', '12', 1, 1_500, '-$14']
>>> pd.Series(v).apply(to_float)
0    2000.0
1      12.0
2       1.0
3    1500.0
4     -14.0

型