Question

我在Python工作。我有两个数据帧df1和df2：

items = []
items[0] = {id: "805", category: 'Apple'}
items[1] = {id: "804", category: 'Watermelon'}
items[2] = {id: "804", category: 'Plum'}


items.forEach(function(item) {
  console.log(item.category);
});

我想在df1中创建一个新列，其最小时间戳df2的值大于当前df1时间戳，其中df2 [＆＃39; col03＆＃39;]为零。这就是我这样做的方式：

d1 = {'timestamp1': [88148  , 5617900, 5622548, 5645748, 6603950, 6666502], 'col01': [1, 2, 3, 4, 5, 6]}
df1 = pd.DataFrame(d1)

d2 = {'timestamp2': [5629500, 5643050, 6578800, 6583150, 6611350], 'col02': [7, 8, 9, 10, 11], 'col03': [0, 1, 0, 0, 1]}
df2 = pd.DataFrame(d2)

它有效，但我不想使用for循环。有更好的方法吗？

Answer 1

尝试使用def func(x): values = df2['timestamp2'][(df2['timestamp2'] > x) & (df2['col03']==0)] if not values.empty: return values.iloc[0] else: np.NAN df1["timestamp1"].apply(func)方法。

0    5629500.0
1    5629500.0
2    5629500.0
3    6578800.0
4          NaN
5          NaN
Name: timestamp1, dtype: float64

您可以创建一个单独的函数来执行必须执行的操作。输出是您的新列

{{1}}

这不是一个单行解决方案，但它有助于保持组织有序。

Answer 2

使用pd.merge_asof( df1, df2.loc[df2.col03 == 0, ['timestamp2']], left_on='timestamp1', right_on='timestamp2', direction='forward' ).rename(columns=dict(timestamp2='colnew')) col01 timestamp1 colnew 0 1 88148 5629500.0 1 2 5617900 5629500.0 2 3 5622548 5629500.0 3 4 5645748 6578800.0 4 5 6603950 NaN 5 6 6666502 NaN正向

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    data.addRows([
        ['Team1',  {!Team1_l1}, {!Team1_l2}, {!Team1_l3}, {!Team1_l4}, {!Team1_cv}],
        ['Team2',  {!Team2_l1}, {!Team2_l2}, {!Team2_l3}, {!Team2_l4}, {!Team2_cv}],
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在Python Pandas中查找大于另一列值的列的最小值

2 个答案: