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可能重复：
如何将文本文件提取到字典中

我有一个文本文件，我想将其更改为 python 中的字典。文本文件如下。我想将键设置为“太阳”、“地球”和“月亮”，然后设置轨道半径、周期等值，以便我可以将动画太阳系实现为快速绘图。

RootObject: Sun

Object: Sun
Satellites: Mercury,Venus,Earth,Mars,Jupiter,Saturn,Uranus,Neptune,Ceres,Pluto,Haumea,Makemake,Eris
Radius: 20890260
Orbital Radius: 0

Object: Earth
Orbital Radius: 77098290
Period: 365.256363004
Radius: 6371000.0
Satellites: Moon

Object: Moon
Orbital Radius: 18128500
Radius: 1737000.10
Period: 27.321582

到目前为止我的代码是

def file():
    file = open('smallsolar.txt', 'r')
    answer = {}
    text = file.readlines() 
    print(text)



text = file() 
print (text)

我不确定现在做什么。有任何想法吗？

4

2 回答 2

3 
answer = {} # initialize an empty dict
with open('path/to/file') as infile: # open the file for reading. Opening returns a "file" object, which we will call "infile"

    # iterate over the lines of the file ("for each line in the file")
    for line in infile:

        # "line" is a python string. Look up the documentation for str.strip(). 
        # It trims away the leading and trailing whitespaces
        line = line.strip()

        # if the line starts with "Object"
        if line.startswith('Object'):

            # we want the thing after the ":" 
            # so that we can use it as a key in "answer" later on
            obj = line.partition(":")[-1].strip()

        # if the line does not start with "Object" 
        # but the line starts with "Orbital Radius"
        elif line.startswith('Orbital Radius'):

            # get the thing after the ":". 
            # This is the orbital radius of the planetary body. 
            # We want to store that as an integer. So let's call int() on it
            rad = int(line.partition(":")[-1].strip())

            # now, add the orbital radius as the value of the planetary body in "answer"
            answer[obj] = rad

希望这可以帮助

3.14有时，如果您的文件（等）中有一个十进制数（python-speak 中的“浮点数” ），则调用int它会失败。在这种情况下，使用float()代替int()

于 2012-11-21T17:18:46.793 回答
1

用一个字符串而不是 readlines() 读取文件，然后在 "\n\n" 上拆分，这样您将拥有一个项目列表，每个项目都描述您的对象。

然后您可能想要创建一个执行以下操作的类：

class SpaceObject:
  def __init__(self,string):
    #code here to parse the string

    self.name = name
    self.radius = radius
    #and so on...

然后你可以创建一个这样的对象列表

#items is the list containing the list of strings describing the various items
l = map(lambda x: SpaceObject(x),items).

然后只需执行以下操作

d= {}
for i in l:
  d[i.name] = i
于 2012-11-21T17:20:17.857 回答