如何根据列名、另一列中的值和索引范围替换数据框中列中的值？

Question

我有一个具有这些特征的数据框（索引是浮点值）：

import pandas as pd

d = {'A': [1,2,3,4,5,6,7,8,9,10],
     'B': [1,2,3,4,5,6,7,8,9,10],
     'C': [1,2,3,4,5,6,7,8,9,10],
     'D': ['one','one','one','one','one','two','two','two','two','two']}

df = pd.DataFrame(data=d)
df
        A   B   C    D
50.0    1   1   1  one
50.2    2   2   2  one
50.4    3   3   3  one
50.6    4   4   4  one
50.8    5   5   5  one
51.0    6   6   6  two
51.2    7   7   7  two
51.4    8   8   8  two
51.6    9   9   9  two
51.8   10  10  10  two

以及具有这些值的偏移量列表（它们也是浮点数）：

offsets = [[0.4, 0.6, 0.8], [0.2, 0.4, 0.6]]

我需要在 A、B 和 C 列上遍历我的数据框，从 D 列中选择分类值，根据它们与我的列表，结果是这样的数据框：

        A   B   C    D
50.0    1   1   1  one
50.2    2   2  nan one
50.4    3  nan nan one
50.6   nan nan nan one
50.8   nan nan nan one
51.0    6   6   6  two
51.2    7   7   7  two
51.4    8   8  nan two
51.6    9  nan nan two
51.8   nan nan nan two

偏移量的值是指从下往上必须设置nan的值。例如：offsets[0][0]=0.4，那么对于A列当D=='one'时，自下而上的两个值必须设置为nan（第4行和第3行，50.8-0.4 = 50.4 - 50.4 不变）。对于A，当D == 'two'时，offsets[1][0]=0.2，所以自下而上的一个值必须设置为nan（第9行，51.8-0.2 = 51.6 - 51.6不变). Offsets[1][0]=0.6，所以对于B列，当D=='one'时，自下而上的三个值必须设置为nan（第4、3、2行，50.8-0.6=50.2- 50.2 不变）。对于B，当D == 'two'时，offsets[1][1]=0.4，所以自下而上的两个值必须设置为nan（第9行和第8行，51.8-0.4 = 51.4 - 51.4不'改变）。 C 列相同。

知道怎么做吗？快速评论 - 我想在数据框本身中替换这些值，而不创建新值。

Answer 1

一种方法是使用 apply 将每列的最后一个值设置为 NaN：

import pandas as pd

# toy data
df = pd.DataFrame(data={'A': [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
                        'B': [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
                        'C': [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
                        'D': ['one', 'one', 'one', 'one', 'one', 'two', 'two', 'two', 'two', 'two']})
offsets = [2, 3, 4]
offset_lookup = dict(zip(df.columns[:3], offsets))


def funny_shift(x, ofs=None):
    """This function shift each column by the given offset in the ofs parameter"""
    for column, offset in ofs.items():
        x.loc[x.index[-1 * offset:], column] = None
    return x


df.loc[:, ["A", "B", "C"]] = df.groupby("D").apply(funny_shift, ofs=offset_lookup)
print(df)

输出

     A    B    C    D
0  1.0  1.0  1.0  one
1  2.0  2.0  NaN  one
2  3.0  NaN  NaN  one
3  NaN  NaN  NaN  one
4  NaN  NaN  NaN  one
5  6.0  6.0  6.0  two
6  7.0  7.0  NaN  two
7  8.0  NaN  NaN  two
8  NaN  NaN  NaN  two
9  NaN  NaN  NaN  two

更新

如果每个组有多个更新，您可以这样做：

offsets = [[2, 3, 4], [1, 2, 3]]
offset_lookup = (dict(zip(df.columns[:3], offset)) for offset in offsets)


def funny_shift(x, ofs=None):
    """This function shift each column by the given offset in the ofs parameter"""
    current = next(ofs)
    for column, offset in current.items():
        x.loc[x.index[-1 * offset:], column] = None
    return x


df.loc[:, ["A", "B", "C"]] = df.groupby("D").apply(funny_shift, ofs=offset_lookup)
print(df)

如何根据列名、另一列中的值和索引范围替换数据框中列中的值？

How to substitute values in a column in a dataframe based on its column name, values in another column and index range?

python

pandas

dataframe

indexing

substitution