补全：锁过期了吗

👁️ 2 人浏览 💬 0 人评论 ❤️ 添加收藏

（每道题开头都有同一段：parse_ver 把「1.4.0」变成能比大小的元组，satisfies(版本, 约束) 判一个版本满不满足「>=1.4,<2」这种约束，split_req 把「httpkit>=1.4」切成名字和约束；REPO 是上面那个小仓库。）

（candidates 列出仓库里满足约束的版本（从低到高）；resolve(清单) 逐个选最高可用版本、把它的依赖加进队列、约束越收越紧，交回 {包: 版本}，无解交回 None。它不回溯——pip 会。）

（这一节多了：parse_requirements 读清单文本（跳过空行和 # 注释），freeze 把选好的版本写成锁文件文本，parse_lock 读回来，diff_lock 比两份锁。）

补全 lock_ok(reqs, lock)：清单里每一条，锁里都得有这个包、而且锁住的版本满足清单的约束；交回不满足的包名列表（按清单顺序）。

提交你的答案

请登录后提交答案。

去登录

← 补全：读清单为什么要隔离 →

更多题目

代码编辑器

语言： Ctrl + Enter 运行

def parse_ver(s):
    return tuple(int(x) for x in s.split("."))

def satisfies(ver, spec):
    v = parse_ver(ver)
    for part in spec.split(","):
        part = part.strip()
        if not part:
            continue
        for op in (">=", "<=", "==", "!=", ">", "<"):
            if part.startswith(op):
                w = parse_ver(part[len(op):])
                break
        ok = {">=": v >= w, "<=": v <= w, "==": v == w, "!=": v != w, ">": v > w, "<": v < w}[op]
        if not ok:
            return False
    return True

def split_req(line):
    i = 0
    while i < len(line) and (line[i].isalnum() or line[i] in "_-"):
        i += 1
    return line[:i], line[i:].replace(" ", "")

REPO = {
    "urlkit":  {"1.9.0": [], "2.0.0": [], "2.1.0": []},
    "charkit": {"4.0.0": [], "5.1.0": []},
    "httpkit": {"1.3.0": ["urlkit<2", "charkit<5"], "1.4.0": ["urlkit<2", "charkit>=4"], "1.5.0": ["urlkit>=2", "charkit>=5"]},
    "templ":   {"3.0.0": [], "3.1.0": []},
    "webapp":  {"2.3.0": ["httpkit>=1.3,<1.5", "templ>=3"], "3.0.0": ["httpkit>=1.5", "templ>=3.1"]},
}

def candidates(name, spec):
    return sorted((v for v in REPO.get(name, {}) if satisfies(v, spec)), key=parse_ver)

def resolve(reqs):
    spec = {}
    for r in reqs:
        n, s = split_req(r)
        spec[n] = s if n not in spec else spec[n] + "," + s
    chosen = {}
    todo = list(spec)
    while todo:
        n = todo.pop(0)
        cands = candidates(n, spec.get(n, ""))
        if not cands:
            return None
        v = cands[-1]
        if chosen.get(n) == v:
            continue
        chosen[n] = v
        for d in REPO[n][v]:
            dn, ds = split_req(d)
            spec[dn] = ds if dn not in spec else spec[dn] + "," + ds
            todo.append(dn)
    for n, v in chosen.items():
        if not satisfies(v, spec[n]):
            return None
    return chosen

def parse_requirements(text):
    reqs = []
    for line in text.split("\n"):
        line = line.split("#")[0].strip()
        if line:
            reqs.append(line)
    return reqs

def freeze(chosen):
    return "\n".join(n + "==" + chosen[n] for n in sorted(chosen))

def parse_lock(text):
    lock = {}
    for line in parse_requirements(text):
        n, s = split_req(line)
        lock[n] = s[2:]
    return lock

def diff_lock(a, b):
    changed = sorted(n for n in a if n in b and a[n] != b[n])
    added = sorted(n for n in b if n not in a)
    removed = sorted(n for n in a if n not in b)
    return changed, added, removed

def lock_ok(reqs, lock):
    bad = []
    # TODO
    return bad

lock = parse_lock("charkit==5.1.0\nhttpkit==1.4.0\ntempl==3.1.0\nurlkit==1.9.0\nwebapp==2.3.0\n")
print(",".join(lock_ok(["webapp<3", "httpkit>=1.5", "templ"], lock)) + "/" + ",".join(lock_ok(["urlkit<2", "newkit"], lock)) + "/" + str(lock_ok(["webapp"], lock)))

本次输入：

输出：

👩‍🏫

AI

💬 题目评论

全部评论