继here之后
import pandas as pd
data = {'date': ['1998-03-01 00:00:01', '2001-04-01 00:00:01','1998-06-01 00:00:01','2001-08-01 00:00:01','2001-05-03 00:00:01','1994-03-01 00:00:01'],
'node1': [1, 1, 2,2,3,2],
'node2': [8,316,26,35,44,56],
'weight': [1,1,1,1,1,1], }
df = pd.DataFrame(data, columns = ['date', 'node1','node2','weight'])
df['date'] = pd.to_datetime(df['date'])
mask = df.groupby('node1').apply(lambda x : (x['date'].dt.year.isin([1998,1999,2000])).any())
mask2 = df.groupby('node1').apply(lambda x : (x['date'].dt.year.isin([2001,2002,2003])).any())
print df[df['node1'].isin(mask[mask & mask2].index)]
我需要的输出是年份范围 (98-00) 和 (01-03) 中的节点,但它也应该只显示这两个范围中的行。
预期输出-
node1 node2 date 1 8 1998-03-01 1 316 2001-04-01 2 26 1998-06-01 2 35 2001-08-01
现在这段代码也打印了这一行:2 56 1994-03-01。
最佳答案
一个简单的解决方案是首先删除不在两个日期范围内的日期,然后应用掩码,即
l1 = [1998,1999,2000]
l2 = [2001,2002,2003]
ndf = df[df['date'].dt.year.isin(l1+l2)]
获取ndf后:
选项 1:您可以采用基于双 groupby 掩码的方法,即
mask = ndf.groupby('node1').apply(lambda x : (x['date'].dt.year.isin(l1)).any())
mask2 = ndf.groupby('node1').apply(lambda x : (x['date'].dt.year.isin(l2)).any())
new = ndf[ndf['node1'].isin(mask[mask & mask2].index)]
谢谢@零
选项 2:您可以进行 groupby 转换
new = ndf[ndf.groupby('node1')['date'].transform(lambda x: x.dt.year.isin(l1).any() & x.dt.year.isin(l2).any())]
选项 3: groupby 过滤器
new = ndf.groupby('node1').filter(lambda x: x['date'].dt.year.isin(l1).any() & x['date'].dt.year.isin(l2).any())
输出:
date node1 node2 weight 0 1998-03-01 00:00:01 1 8 1 1 2001-04-01 00:00:01 1 316 1 2 1998-06-01 00:00:01 2 26 1 3 2001-08-01 00:00:01 2 35 1
关于python - 仅显示 pandas 中日期范围行的交集,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46500619/