有沒有辦法沿著熊貓系列聚合（例如，總和，平均值）間隔？我以漫長而不優雅的方式做到了，但我覺得必須有一種內置的方法來做到這一點，因為人們用熊貓做了很多時間序列分析（順便說一句，這不是時間序列）。也許使用pd.Interval物件？
import numpy as np
import pandas as pd
from collections import OrderedDict
data = pd.Series({0.0: 0.0027077328584676173,0.0041841004184100415: 0.002791971614505515,0.008368200836820083: 0.0029836729636686537,0.012552301255230124: 0.00310210056994597,0.016736401673640166: 0.003123592522767784,0.02092050209205021: 0.0031416073125521624,0.025104602510460247: 0.0031485613317755977,0.02928870292887029: 0.0031557637903831775,0.03347280334728033: 0.00326313482475896,0.03765690376569037: 0.003301728046497936,0.04184100418410042: 0.0033366539533600335,0.046025104602510455: 0.0033702672902052584,0.050209205020920494: 0.0033868374139273388,0.05439330543933054: 0.003396846064522763,0.05857740585774058: 0.0034311293771955432,0.06276150627615062: 0.0034554861912855788,0.06694560669456066: 0.0035232368934153594,0.0711297071129707: 0.0035777792118595695,0.07531380753138074: 0.003592954000065738,0.0794979079497908: 0.00361514147677572,0.08368200836820083: 0.003623149988144421,0.08786610878661087: 0.0037193787063349777,0.09205020920502091: 0.0037478924475459496,0.09623430962343095: 0.003916888466467329,0.10041841004184099: 0.00394360032517084,0.10460251046025104: 0.0039620222687681865,0.10878661087866108: 0.003999287617522048,0.11297071129707112: 0.004006491834823014,0.11715481171548116: 0.004013290425253716,0.1213389121338912: 0.004042553765475244,0.12552301255230125: 0.0040505029191929675,0.12970711297071127: 0.004055597770467728,0.13389121338912133: 0.004089750738907019,0.13807531380753138: 0.0041098023814089,0.1422594142259414: 0.004120096873249847,0.14644351464435146: 0.004131372273604568,0.15062761506276148: 0.004148097669469449,0.15481171548117154: 0.004157794773892295,0.1589958158995816: 0.004174682633106945,0.1631799163179916: 0.004192587889456918,0.16736401673640167: 0.004213104284729356,0.1715481171548117: 0.004221311516318688,0.17573221757322174: 0.004256186252391833,0.1799163179916318: 0.004261480055637909,0.18410041841004182: 0.004270623452687678,0.18828451882845187: 0.004272709261097341,0.1924686192468619: 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0.004360062449603033,0.7991631799163179: 0.004334990916415593,0.8033472803347279: 0.0043053152894044775,0.807531380753138: 0.004299540898698937,0.811715481171548: 0.004292780780235888,0.8158995815899581: 0.0042857760464103944,0.8200836820083681: 0.0042838512495085475,0.8242677824267781: 0.00427540230624067,0.8284518828451882: 0.004265220326618783,0.8326359832635982: 0.004258643325438144,0.8368200836820083: 0.004248216730453809,0.8410041841004183: 0.004238868110347041,0.8451882845188283: 0.004220947550812318,0.8493723849372384: 0.00416931971618869,0.8535564853556484: 0.004167417025061434,0.8577405857740585: 0.004165033746474712,0.8619246861924685: 0.00415209361043586,0.8661087866108785: 0.0041256577903094884,0.8702928870292886: 0.004103651664874053,0.8744769874476986: 0.00405500931549898,0.8786610878661087: 0.00404009152774367,0.8828451882845187: 0.0040284345896393825,0.8870292887029287: 0.004002117941651392,0.8912133891213389: 0.003998528797899053,0.8953974895397488: 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0.0035866395461282012,1.0: 0.0035637164646374914})
interval_points = np.linspace(0,1,11)
interval_aggregate = OrderedDict()
for i in range(len(interval_points) - 1):
lower = interval_points[i]
upper = interval_points[i 1]
if i == 0:
interval_aggregate[(lower, upper)] = data[data.index.map(lambda x: lower <= x < upper)].sum()
else:
interval_aggregate[(lower, upper)] = data[data.index.map(lambda x: lower < x <= upper)].sum()
aggregate_data = pd.Series(interval_aggregate)
aggregate_data.sum()
# 1.0
uj5u.com熱心網友回復：
您正在尋找 groupby pd.cut：
data.groupby(pd.cut(data.index, interval_points, include_lowest=True)).agg(['mean','count','sum'])
輸出：
mean count sum
(-0.001, 0.1] 0.003351 24 0.080414
(0.1, 0.2] 0.004136 24 0.099263
(0.2, 0.3] 0.004452 24 0.106853
(0.3, 0.4] 0.004528 24 0.108668
(0.4, 0.5] 0.004299 24 0.103173
(0.5, 0.6] 0.004595 24 0.110278
(0.6, 0.7] 0.004078 24 0.097881
(0.7, 0.8] 0.004233 24 0.101600
(0.8, 0.9] 0.004165 24 0.099967
(0.9, 1.0] 0.003829 24 0.091904
轉載請註明出處，本文鏈接：https://www.uj5u.com/qukuanlian/383038.html
標籤：Python 熊猫 pandas-groupby 总计的 间隔
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