根据 pandas 中的相等性计算数据的符咒长度 [英] computing spell lengths of data based on equality in pandas

问题描述

我想基于熊猫数据框中相邻列的相等性来计算spell长度.最好的方法是什么?

I would like to compute spell lengths based on equality of the adjacent column in a pandas dataframe. What is the best way to do this?

一个例子:

import pandas as pd
d1 = pd.DataFrame([['4', '4', '4', '5'], ['23', '23', '24', '24'], ['112', '112', '112', '112']], 
              index=['c1', 'c2', 'c3'], columns=[1962, 1963, 1964, 1965])

产生一个看起来像

我想返回一个如下所示的数据框.此输出记录每行发生的咒语数量.在这种情况下，c1有2个咒语，第一个咒语发生在1962年至1964年，第二个咒语始于1965年:

I would like to return a dataframe such as the following below. This output documents the number of spells that occur on each row. In this case c1 has 2 spells the first one occurs in 1962 to 1964 and the second starts and finishes in 1965:

和一个描述咒语长度的数据框，如下所示.例如，c1的一个咒语持续时间为3年，第二个咒语持续时间为1年.

And a dataframe that describes the spell length as shown below. For example c1 has one spell of 3 years and a second spell of 1 year long in duration.

此重新编码在生存分析中很有用.

This re-coding is useful in survival analysis.

推荐答案

以下内容适用于您的数据集，需要提出一个问题才能减少使用

The following works for your dataset, needed to ask a question in order to reduce my original answer to using list comprehensions and itertools:

In [153]:

def num_spells(x):
    t = list(x.unique())
    return [t.index(el)+1 for el in x]

d1.apply(num_spells, axis=1)

Out[153]:
    1962  1963  1964  1965
c1     1     1     1     2
c2     1     1     2     2
c3     1     1     1     1

In [144]:
from itertools import chain, repeat
def spell_len(x):
    t = list(x.value_counts())
    return list(chain.from_iterable(repeat(i,i) for i in t))

d1.apply(spell_len, axis=1)
Out[144]:
    1962  1963  1964  1965
c1     3     3     3     1
c2     2     2     2     2
c3     4     4     4     4

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