Python中如何获取排序后列表首尾行的首个元素

场景1：数据已在Python列表中 #

如果你的数据已经处理成Python列表（不管是拆分后的字段列表还是原始字符串行），直接用列表索引就能快速获取首尾元素的第一个字段：

情况A：数据是拆分后的字段列表

# 已按人口降序排好的列表，每个元素是拆分后的字段集合
countries_data = [
    ["Bahrain", "1701575", "0.0368", "60403", "760", "Asia"],
    ["Timor-Leste", "1318445", "0.0196", "25326", "14870", "Asia"],
    ["Cyprus", "1207359", "0.0073", "-8784", "9240", "Asia"],
    ["Bhutan", "771608", "0.0112", "8516", "38117", "Asia"],
    ["Macao", "649335", "0.0139", "8890", "30781", "Asia"],
    ["Maldives", "540544", "0.0181", "9591", "300", "Asia"]
]

# 取首行第一个元素（人口最多的国家）
most_populous = countries_data[0][0]
# 取末行第一个元素（人口最少的国家）
least_populous = countries_data[-1][0]

print(f"人口最多：{most_populous}")
print(f"人口最少：{least_populous}")

情况B：数据是原始字符串行的列表

# 已按人口降序排好的原始字符串行列表
country_lines = [
    "Bahrain,1701575,0.0368,60403,760,Asia",
    "Timor-Leste,1318445,0.0196,25326,14870,Asia",
    "Cyprus,1207359,0.0073,-8784,9240,Asia",
    "Bhutan,771608,0.0112,8516,38117,Asia",
    "Macao,649335,0.0139,8890,30781,Asia",
    "Maldives,540544,0.0181,9591,300,Asia"
]

# 拆分首行字符串，取第一个字段
most_populous = country_lines[0].split(",")[0]
# 拆分末行字符串，取第一个字段
least_populous = country_lines[-1].split(",")[0]

print(f"人口最多：{most_populous}")
print(f"人口最少：{least_populous}")

场景2：从文件读取已排序的数据 #

如果数据存在本地文本文件（比如countries.csv）且已经按人口降序排好，直接读取文件内容后处理：

with open("countries.csv", "r", encoding="utf-8") as f:
    # 读取所有非空行，自动去掉换行符
    lines = [line.strip() for line in f if line.strip()]

most_populous = lines[0].split(",")[0]
least_populous = lines[-1].split(",")[0]

print(f"人口最多：{most_populous}")
print(f"人口最少：{least_populous}")

关键要点 #

• Python列表用[0]取第一个元素，[-1]取最后一个元素，这是最直接的索引用法
• split(",")会把字符串按逗号拆分成列表，取索引[0]就能得到国家名称
• 因为数据已经按人口降序排好，不需要额外排序，直接取首尾行即可

内容的提问来源于stack exchange，提问作者Ruby W.