Python数据帧的分析,其中整数列用符号分隔

问题描述 投票:0回答:1

您能帮我处理python中具有定界整数值的列吗?

我们如何创建一个额外的列,例如“ PHR_INSTANTENEOUS_MIN”,该列将最小值存储在PHR_INSTANTENEOUS中。就像第一行:“-18”和第三行“ 14”

类似地:PHR_INSTANTENEOUS_MIN,PHR_INSTANTENEOUS_MEDIAN和PHR_INSTANTENEOUS_MODE派生值。

对于SINR_INSTANTENEOUS值重复类似的事情,我们需要形成派生值。

df1
START_TIME PRIMARY_KEY PHR_INSTANTANEOUS SINR_INSTANTANEOUS
2020-03-10 12:00:00 e7ca9da318f1 -18|-17 9|8
2020-03-10 12:01:00 68615e3db513 1 26
2020-03-10 12:05:00 7f250354808a 14|18|20|20 26|26|24|26
2020-03-10 12:07:00 9202ab7611d4 -8|-7|40 22|6|-2
2020-03-10 12:12:00 377bf955bdc0 4|9 26|20

完整的数据集图像如下:

enter image description here

python delimiter median number-with-delimiter
1个回答
1
投票

这是一种方法:

import pandas as pd
from statistics import median, mode
import numpy as np

df = pd.DataFrame(['-18|-17', '1', '14|18|20|20', '-8|-7|40', 5.2, np.nan], columns=['PHR_INSTANTANEOUS'])

# make sure the dtype is uniformly string
df['PHR_INSTANTANEOUS'] = df['PHR_INSTANTANEOUS'].astype(str)

# get the values
df['PHR_INSTANTANEOUS'].apply(lambda x: min(map(float, x.split('|'))))  # minimum
df['PHR_INSTANTANEOUS'].apply(lambda x: median(map(float, x.split('|'))))  # median
df['PHR_INSTANTANEOUS'].apply(lambda x: mode(map(float, x.split('|'))))  # mode
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