使用python创建字典时的值错误3

我是一个对Python完全陌生的学生，目前正在学习它。所以我被要求从列表中创建一个包含文件中的一行数据的字典。使用python创建字典时的值错误3

record = ["Name", "ID", "datetime", "Type", "Lat", "Long", "Central Pressure", "Mean radius gf wind", "Max wind speed", "Comment"]

文件已经导入，它由对应的元素record列表数据的多行的。数据

例子：

r1 = ["unnamed", "AU190607_01U", "1907-01-17 23:00", "T", "-13", "146.5", "994", "100", "10.3"]   
r2 = ["unnamed", "AU190607_01U", "1907-01-17 23:00", "T", "-13", "146.5", "994", "", "10.3"]

所以我的代码是这样

def parse_record(record): 
key = ["id", "name", "year", "month", "day", "hour", "central pressure", 
     "radius", "speed", "lat", "long"] 
value = [str(record[1]), str(record[0]), int((record[2][:4]), 
     int((record[2])[5:7]),int((record[2])[8:10]), 
     int((record[2][10:13]),float(record[6]), 
     float(record[7]),float(record[8]),float(record[4]), float(record[5])] 

record_dictionary = dict(zip(key,value)) 

return record_dictionary

我期待着得到

a1 = {"id": "AU190607_01U", "name": "unnamed", "year": 1907, "month": 1, 
     "day": 17, "hour": 23,"central pressure": 994.0, "radius": 100.0, 
     "speed": 10.3, "lat": -13.0, "long": 146.5}

，但我得到一个错误信息说：

int((record[2])[8:10]),int((record[2])[10:13]), float(record[6]), float(record[7]), 
ValueError: could not convert string to float:

并且如果数据中存在空白值（例如在上面的r2中），它应该被排除在字典之外，我应该怎么做才能做到这一点？

感谢您的帮助！

来源

2017-08-23 jiajia hong

你有一个'return'语句，但我没有看到你声明一个函数。 –

'float（record [...]）'你可以用0到9之间的任何整数替换'...'会抛出这个异常，因为你试图从一个无关的字符串中创建一个float浮动。 'float（record [6]）'实际上是float（“Central Pressure”）'，因此是错误。 – zezollo

@zezollo，记录涉及r1而非密钥 –

尝试实施这一

def parse_record(record): 
    key = ["id", "name", "year", "month", "day", "hour", "central pressure", 
      "radius", "speed", "lat", "long"] 
    record = [i if i else 0 for i in record] 
    value = [str(record[1]), str(record[0]), int((record[2])[:4]), 
      int((record[2])[5:7]),int((record[2])[8:10]), 
      int((record[2])[10:13]),float(record[6]), 
      float(record[7]),float(record[8]),float(record[4]), float(record[5])] 

    record_dictionary = dict(zip(key,value)) 

    return record_dictionary 

record = ["unnamed", "AU190607_01U", "1907-01-17 23:00", "T", "-13", "146.5", "994", "", "10.3"]   

r1 = ["unnamed", "AU190607_01U", "1907-01-17 23:00", "T", "-13", "146.5", "994", "100", "10.3"]   
r2 = ["unnamed", "AU190607_01U", "1907-01-17 23:00", "T", "-13", "146.5", "994", "", "10.3"] 

parse_record(r1) 
parse_record(r2)

注意，在输出第二个记录，您将有零作为radius值。这是因为此记录没有值。

编辑

我所做的更改。

def parse_record(record): 
    key = ["id", "name", "year", "month", "day", "hour", "central pressure", 
      "radius", "speed", "lat", "long"] 
    record = [i if i else 0 for i in record] 

    value = [str(record[1]), str(record[0]), int((record[2])[:4]), 
      int((record[2])[5:7]),int((record[2])[8:10]), 
      int((record[2])[10:13]),float(record[6]), 
      float(record[7]),float(record[8]),float(record[4]), float(record[5])] 
    _ = [(a,b) for a,b in zip(key,value) if b] 
    record_dictionary = dict(_) 

    return record_dictionary 

record = ["unnamed", "AU190607_01U", "1907-01-17 23:00", "T", "-13", "146.5", "994", "", "10.3"]   

r1 = ["unnamed", "AU190607_01U", "1907-01-17 23:00", "T", "-13", "146.5", "994", "100", "10.3"]   
r2 = ["unnamed", "AU190607_01U", "1907-01-17 23:00", "T", "-13", "146.5", "994", "", "10.3"] 

parse_record(r1) 
parse_record(r2)

来源

2017-08-23 18:29:29

嗨，感谢您的帮助，请问如果我想从字典中删除这个值，我该怎么办？ –

哪个值？你的意思是删除？你想用其他值替换它，或者你想从记录中删除它？ –

它在r2中为半径它是“”，这是空的，我应该把它从字典中删除 –

错误在于

value = […, float(record[7])…]

因为在你的第二个值列表，record[7]是""这不是一个有效的float文字。

来源

2017-08-23 14:34:52

谢谢！如果我想从字典中删除一对值和密钥，如果密钥的值是“”，这是空的，该怎么办？ –

以下是帮助您使用空白值的方式（在制作字典时不改变类型）。

record_dictionary = {a:b for a,b in zip(key,value) if b != '' }

上面有一个很好的建议，由@tdube首先压缩键和值，然后才改变那里的类型。

来源

2017-08-23 14:50:08 Nibuton

new_list = zip（key，value） record_dictionary = {a：b for a，b in new_list if b！=“”}我试图将这两个列表压缩在一起，但我认为这仍然无法删除空键值被检测到..我可以问这条线应该做什么？谢谢。 –

@jiajiahong嗨！我刚刚尝试过，它对我的例子很有帮助，也许我误解了这个任务。 r1和r2是两个不同的行，它们应该是最终结果，但不会有r2中空白值的键/值对 - 是吗？这个方法做了什么：它使一个字典从一个条件压缩，b（值）不是空白。 – Nibuton

使用python创建字典时的值错误3

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