我正在尝试从这个文件中创建一个字典:键是第一个单词,值是后面的所有单词。
andrew fred
fred
judy andrew fred
george judy andrew
john george
这是我的代码:
follows_file = open("C:\\Users\\Desktop\\Python\\follows.txt")
followers = {}
for line in follows_file: #==> [Judy Andrew Fred]
users = line.split(' ') #==> [Judy, andrew, Fred, ....]
follower = users[0] #==> [Judy]
followed_by = users[1:] #==> [Andrew, Fred]
for user in followed_by:
# Add the 'follower to the list of followers user
if user not in followers:
followers[user] = []
followers[user].append(follower)
print(followers.items())
当我打印关注者和后跟变量时,它们是正确的,但我无法正确地将它们添加到字典中;这是输出
dict_items([('fred\n', ['andrew', 'judy']), ('andrew', ['judy']), ('judy' ['george']), ('andrew\n', ['george']), ('george', ['john'])])
我想要的输出是
(Andrew[Fred])(Fred[])(judy[Andrew Fred])(George[Judy Fred])(john[george])
非常感谢任何帮助!
答案 0 :(得分:3)
您可以使用collections.defaultdict()
作为字典工厂,只需追踪一个人后面的用户,例如:
import collections
followers = collections.defaultdict(list) # use a dict factory to save some time on checks
with open("path/to/your_file", "r") as f: # open the file for reading
for line in f: # read the file line by line
users = line.split() # split on any white space
followers[users[0]] += users[1:] # append the followers for the current user
将为您的数据生成:
{'andrew': ['fred'],
'fred': [],
'judy': ['andrew', 'fred'],
'george': ['judy', 'andrew'],
'john': ['george']}
这也允许您在重复记录上为用户添加多个列表 - 否则您可以使用普通dict
作为followers
并将其设置为followers[users[0]] = users[1:]
。< / p>
您显示为所需输出的数据结构不是有效的Python,您真的希望以这种方式呈现吗?我的意思是,如果你坚持你可以这样做:
print("".join("({}[{}])".format(k, " ".join(v)) for k, v in followers.items()))
# (andrew[fred])(fred[])(judy[andrew fred])(george[judy andrew])(john[george])
答案 1 :(得分:1)
这是使用str.split
和try
/ except
子句捕获只存在密钥的实例的一种解决方案。
注意io.StringIO
让我们从字符串中读取,就像它是一个文件一样。
from io import StringIO
import csv
mystr = StringIO("""andrew fred
fred
judy andrew fred
george judy andrew
john george""")
# replace mystr with open("C:\\Users\\zacan\\Desktop\\Python\\follows.txt")
with mystr as follows_file:
d = {}
for users in csv.reader(follows_file):
try:
key, *value = users[0].split()
except ValueError:
key, value = users[0], []
d[key] = value
print(d)
{'andrew': ['fred'],
'fred': [],
'george': ['judy', 'andrew'],
'john': ['george'],
'judy': ['andrew', 'fred']}
答案 2 :(得分:0)
编辑回答,得益于@ PM2Ring和@IljaEverilä的评论。
这是我使用词典理解的原始解决方案
followers = {line.split()[0]: line.split()[1:] for line in follows_file}
@IljaEverilä提出的一种更有效的替代方法是避免两次调用split
:
followers = {follower: followees for follower, *followees in map(str.split, follows_file)}
结果:
{'andrew': ['fred'],
'fred': [],
'george': ['judy', 'andrew'],
'john': ['george'],
'judy': ['andrew', 'fred']}
请注意,上述两种解决方案均假设您的文件不包含重复的密钥。
之后不要忘记关闭文件:
follows_file.close()
或者更好的是,只需使用上下文管理器,它会为您处理文件关闭:
with open('C:\\Users\\zacan\\Desktop\\Python\\follows.txt', 'r') as follows_file:
followers = {follower: followees for follower, *followees in map(str.split, follows_file)}
答案 3 :(得分:0)
followers = dict()
with open('C:\\Users\\zacan\\Desktop\\Python\\follows.txt', 'r') as f:
for line in f:
users = line.split(' ')
followers[users[0]] = [_ for _ in users[1:]]
这应该有效,没有测试