python 根据键合并字典列表

Question

我正在寻找 python 连接的替代方法。

我正在尝试获取一天中每一秒的列表，并根据时间戳将数据加入其中。到目前为止，我所拥有的是：

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']]

原始数据来自软件

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)

这里我做了一天所有的秒，并且给了他们相同的key，方便匹配

# Turn raw data into dictionary                        
for row in rawdata:
    dictionary = dict(zip(keys, row))
    dictData.append(dictionary)

然后我把原始数据也变成一个字典

#Join, sort off
compleet=()
for t in time:
    t.update(dictData)
    compleet.append(t)

print len(compleet)
print compleet[1]

然而，当我运行这个时，我得到错误：

ValueError: dictionary update sequence element #0 has length 10; 2 is required

这让我相信我一次只能更新 key:value 对，但我不确定这是正确的。

此外：这是一个 1:1 连接。 1 个时间戳只能有 1 个测量值。并非一天中的每一秒都有测量值。 'Join on' 会在 "TIME"

Answer 1

文档说：

dict.update = update(...)

D.update([E, ]**F) -> None. Update D from dict/iterable E and F.

If E is present and has a .keys() method, then does: for k in E: D[k] = E[k]

If E is present and lacks a .keys() method, then does: for k, v in E: D[k] = v

In either case, this is followed by: for k in F: D[k] = F[k]

因为dictData是一个列表，没有keys()方法，for k, v in dictData: t[k] = v在update方法内部运行，导致异常。

其实我不太理解你的代码，所以我不能给出具体的帮助。

如果你能解释一下代码（例如执行后正确的t变量），我想帮助你。

Answer 2

#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并且没有看到加入之后，这是最好的下一件事。

python 根据键合并字典列表

python merge list of dictionaries based on key

python

merge

dictionary

join

key