[英]python merge list of dictionaries based on key
我正在寻找联接的python替代方案。
我正在尝试获取每一天的清单,并根据时间戳将数据加入其中。 到目前为止,我是这样的:
keys=('DRIP_ID','DESCR','OBJECT','TIMESTAMP','DRIP_R1','DRIP_R2','RT_DISP1','RT_DISP2','DAY','TIME')
键是列名
rawdata=[['242418',"242418 Rechts.BD242418: tot Oudkarspel - - ${pijlop} N242 10 min - N508 14 min${pijlr}",'BD242418','20150701063825','N242','N508','10','14','20150701','063825'],
['242418',"242418 Rechts.BD242418: tot Oudkarspel - - ${pijlop} N242 10 min - N508 14 min${pijlr}",'BD242418','20150701064327','N242','N508','10','14','20150701','064327'],
['242418',"242418 Rechts.BD242418: tot Oudkarspel - - ${pijlop} N242 10 min - N508 14 min${pijlr}",'BD242418','20150701085717','N242','N508','10','14','20150701','085717'],
['242418',"242418 Rechts.BD242418: tot Oudkarspel - - ${pijlop} N242 10 min - N508 14 min${pijlr}",'BD242418','20150701100116','N242','N508','10','14','20150701','100116'],
['242418',"242418 Rechts.BD242418: tot Oudkarspel - - ${pijlop} N242 10 min - N508 14 min${pijlr}",'BD242418','20150701191611','N242','N508','10','14','20150701','191611'],
['242418',"242418 Rechts.BD242418: tot Oudkarspel - - ${pijlop} N242 10 min - N508 14 min${pijlr}",'BD242418','20150701213616','N242','N508','10','14','20150701','213616']]
原始数据是软件产生的
sec = ['00','01','02','03','04','05','06','07','08','09','10','11','12','13','14','15','16','17','18','19','20','21','22','23','24','25','26','27','28','29','30','31','32','33','34','35','36','37','38','39','40','41','42','43','44','45','46','47','48','49','50','51','52','53','54','55','56','57','58','59']
mm = sec
hh = ['00','01','02','03','04','05','06','07','08','09','10','11','12','13','14','15','16','17','18','19','20','21','22','23']
timestamp=()
time = []
dictData = []
# Dictionary with all seconds (HHMMSS) in 1 day
for ih, uur in enumerate(hh):
if ih < 24:
for im, minutes in enumerate(mm):
if im < 60:
for isec, secs in enumerate(sec):
if isec < 60:
timestamp = str(uur)+str(minutes)+str(secs)
timeDict = dict()
timeDict['DRIP_ID']=""
timeDict['DESCR']=""
timeDict['OBJECT']=""
timeDict['TIMESTAMP']=""
timeDict['DRIP_R1']=""
timeDict['DRIP_R2']=""
timeDict['RT_DISP1']=""
timeDict['RT_DISP2']=""
timeDict['DAY']=""
timeDict['TIME']=timestamp
time.append(timeDict)
在这里,我一天中的所有时间都是秒,并为它们提供了相同的键,以便于进行匹配
# Turn raw data into dictionary
for row in rawdata:
dictionary = dict(zip(keys, row))
dictData.append(dictionary)
然后我将原始数据也转换成字典
#Join, sort off
compleet=()
for t in time:
t.update(dictData)
compleet.append(t)
print len(compleet)
print compleet[1]
但是,当我运行此命令时,出现错误:
ValueError: dictionary update sequence element #0 has length 10; 2 is required
这使我相信我一次只能更新key:value对,但是我不确定这是正确的。
此外:这是1:1的联接。 1个时间戳只能进行1个测量。 并非一天中的每一秒钟都有测量结果。 “加入”时间为“ TIME”
该文件说:
dict.update =更新(...)
D.update([E,] ** F)->无。 从dict /可迭代E和F更新D。
如果E存在并且具有.keys()方法,则执行以下操作:对于E中的k:D [k] = E [k]
如果E存在并且缺少.keys()方法,则执行:对于k,E中的v:D [k] = v
无论哪种情况,都紧随其后:对于F中的k:D [k] = F [k]
因为dictData
是一个列表并且没有keys()
方法,所以for k, v in dictData: t[k] = v
在update
方法中运行,并导致异常。
实际上,我不太了解您的代码,因此无法对此提供具体帮助。
如果您能解释代码(例如,执行后正确的t
变量),我想为您提供帮助。
#Same result as a join, by iterating.
for iTime, t in enumerate(time):
for iData, d in enumerate(dictData):
if t['TIME'] == d['TIME']:
t.update(d)
在意识到出了什么问题之后,并没有看到加入,这就是最好的下一件事。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.