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JL Peyret's questions

Martin Hope
JL Peyret
Asked: 2025-04-28 05:26:15 +0800 CST
  • 6

背景（但不是仅限 Django 的问题）是 Django 测试服务器不会在其响应和请求 URL 中返回方案或 netloc。

例如，我得到了/foo/bar，并且我希望以 结束http://localhost:8000/foo/bar。

urllib.parse.urlparse （但并没有那么复杂urllib.parse.urlsplit）使得从测试 URL 和我已知的服务器地址收集相关信息变得容易。看起来比必要更复杂的是，通过urllib.parse.urlcompose添加 scheme 和 netloc 来重新组合一个新的 URL ，这个 URL 需要位置参数，但没有文档说明它们是什么，也不支持命名参数。同时，解析函数返回的是不可变的元组……

def urlunparse(components):
    """Put a parsed URL back together again.  This may result in a ..."""

我确实让它工作了，请参阅下面的代码，但它看起来真的很笨拙，围绕我需要首先将解析元组转换为列表，然后在所需的索引位置修改列表的部分。

有没有更 Pythonic 的方式？

示例代码：


from urllib.parse import urlsplit, parse_qs, urlunparse, urlparse, urlencode, ParseResult, SplitResult

server_at_ = "http://localhost:8000"
url_in = "/foo/bar"  # this comes from Django test framework I want to change this to "http://localhost:8000/foo/bar"

from_server = urlparse(server_at_)
print("  scheme and netloc from server:",from_server)


print(f"{url_in=}")
from_urlparse = urlparse(url_in)

print("  missing scheme and netloc:",from_urlparse)

#this works
print("I can rebuild it unchanged :",urlunparse(from_urlparse))

#however, using the modern urlsplit doesnt work (I didn't know about urlunsplit when asking)
try:
    print("using urlsplit", urlunparse(urlsplit(url_in)))
#pragma: no cover pylint: disable=unused-variable
except (Exception,) as e: 
    print("no luck with urlsplit though:", e)


#let's modify the urlparse results to add the scheme and netloc
try:
    from_urlparse.scheme = from_server.scheme
    from_urlparse.netloc = from_server.netloc
    new_url = urlunparse(from_urlparse)
except (Exception,) as e: 
    print("can't modify tuples:", e)


# UGGGH, this works, but is there a better way?
parts = [v for v in from_urlparse]
parts[0] = from_server.scheme
parts[1] = from_server.netloc

print("finally:",urlunparse(parts))

示例输出：

  scheme and netloc from server: ParseResult(scheme='http', netloc='localhost:8000', path='', params='', query='', fragment='')
url_in='/foo/bar'
  missing scheme and netloc: ParseResult(scheme='', netloc='', path='/foo/bar', params='', query='', fragment='')
I can rebuild it unchanged : /foo/bar
no luck with urlsplit though: not enough values to unpack (expected 7, got 6)
can't modify tuples: can't set attribute
finally: http://localhost:8000/foo/bar
python
Martin Hope
JL Peyret
Asked: 2024-09-24 00:06:34 +0800 CST
  • 7

我正在尝试通过比较两个数据框dfcompare = (df0 == df1)，并且空值永远不会被视为相同（不像join没有允许空值匹配的选项）。

我对其他字段的方法是使用适合其数据类型的“空值”填充它们。我应该对结构体使用什么？

import polars as pl

df = pl.DataFrame(
    {
        "int": [1, 2, None],
        "data" : [dict(a=1,b="b"),dict(a=11,b="bb"),None]
    }
)

df.describe()
print(df)

df2 = df.with_columns(pl.col("int").fill_null(0))

df2.describe()
print(df2)

# these error out:...
try:
    df3 = df2.with_columns(pl.col("data").fill_null(dict(a=0,b="")))
except (Exception,) as e: 
    print("try#1", e)


try:
    df3 = df2.with_columns(pl.col("data").fill_null(pl.struct(dict(a=0,b=""))))
except (Exception,) as e: 
    print("try#2", e)

输出：


shape: (3, 2)
┌──────┬─────────────┐
│ int  ┆ data        │
│ ---  ┆ ---         │
│ i64  ┆ struct[2]   │
╞══════╪═════════════╡
│ 1    ┆ {1,"b"}     │
│ 2    ┆ {11,"bb"}   │
│ null ┆ {null,null} │
└──────┴─────────────┘
shape: (3, 2)
┌─────┬─────────────┐
│ int ┆ data        │
│ --- ┆ ---         │
│ i64 ┆ struct[2]   │
╞═════╪═════════════╡
│ 1   ┆ {1,"b"}     │
│ 2   ┆ {11,"bb"}   │
│ 0   ┆ {null,null} │
└─────┴─────────────┘
try#1 invalid literal value: "{'a': 0, 'b': ''}"
try#2 a

Error originated just after this operation:
DF ["int", "data"]; PROJECT */2 COLUMNS; SELECTION: "None"

我令人满意的解决方法是改为使用unnest列。这很好用（甚至更好，因为它允许逐个子字段填充）。不过，我仍然好奇如何实现可以传递给这些类型的函数的合适的“结构文字”。

你也可以想象想要添加一个硬编码的列，如下所示df4 = df.with_columns(pl.lit("0").alias("zerocol"))

python