如何计算两个字符型年月列的月份差值?
计算字符型年月列的月份差值
方法1:手动提取年月计算(无依赖,通用)
直接从字符串中拆分年份和月份,通过公式 (年差)*12 + 月差 计算差值:
ym1 = "202301" ym2 = "202201" # 拆分年份和月份 year1, month1 = int(ym1[:4]), int(ym1[4:]) year2, month2 = int(ym2[:4]), int(ym2[4:]) # 计算月份差 month_diff = (year1 - year2) * 12 + (month1 - month2) print(month_diff) # 输出:12
方法2:Python datetime + relativedelta(适合复杂日期场景)
如果需要处理更复杂的日期逻辑,可借助dateutil库的relativedelta直接计算月份差:
from datetime import datetime from dateutil.relativedelta import relativedelta ym1 = "202301" ym2 = "202201" # 转成datetime对象 date1 = datetime.strptime(ym1, "%Y%m") date2 = datetime.strptime(ym2, "%Y%m") # 计算月份差 diff = relativedelta(date1, date2) month_diff = diff.years * 12 + diff.months print(month_diff) # 输出:12
方法3:SQL实现(以MySQL/PostgreSQL为例)
MySQL
使用STR_TO_DATE转日期类型,再用TIMESTAMPDIFF直接计算月份差:
SELECT TIMESTAMPDIFF(MONTH, STR_TO_DATE(ym2, '%Y%m'), STR_TO_DATE(ym1, '%Y%m')) AS month_diff FROM your_table;
PostgreSQL
通过TO_DATE转日期,结合age函数提取年月差:
SELECT (DATE_PART('year', age(TO_DATE(ym1, 'YYYYMM'), TO_DATE(ym2, 'YYYYMM'))) * 12 + DATE_PART('month', age(TO_DATE(ym1, 'YYYYMM'), TO_DATE(ym2, 'YYYYMM'))))::int AS month_diff FROM your_table;
内容的提问来源于stack exchange,提问作者Amandeep Singh
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