Question

如何在对象列表中搜索属性？假设我想要所有对象的fname以及年龄＆gt; 30或找到一些匹配

#!usr/bin/python
import sys
import pickle

class People:
    def __init__(self, fname=None, lname=None, age=None, gender=None):
        self.fname = fname
        self.lname = lname
        self.age = age
        self.gender = gender

    def display(self):

        #fp = open("abc","w")
        #fp.write("F.Name: "+ppl.fname+ "\nLName: "+ ppl.lname + "\nAge: "+     str(ppl.age)+ "\nGender: "+ ppl.gender) 




ppl = [People("Bean", "Sparrow", 22, "M"), People("Adam", "Sandler", 32, "M"),     People("Jack", "Marro", 28, "M")]

fp = open("abc","w")
for person in ppl:
    fp.write("F.Name: "+person.fname+ "\nLName: "+ person.lname+ "\nAge: "+     str(person.age)+ "\nGender: "+ person.gender+"\n\n")
fp.close()

Answer 1

在Python中有一个名为list comprehensions的东西。列表推导是你的朋友。

获取所有ppl.fname：

all_fnames = [person.fname for person in ppl]

让所有ppl.age超过30的人

all_greater_than_thirty = [person for person in ppl if person.age > 30]

等等。

编辑由于列表推导会返回一个列表，您只需使用它来代替原来的列表，例如写入文件：

with open("abc", "w") as fp:
    for person in [p for p in ppl if p.age > 30]:
        fp.write(...) # Fill it in with whatever you want to write

甚至更好，但更高级，你会为你的People类创建一个方法，它会返回一个格式化为写入文件的字符串，如People.to_string()，然后你可以这样做：

with open("abc", "w") as fp:
    fp.writelines("%s\n" % person.to_string() for person in ppl if person.age > 30)

优点是效率。而且它看起来更好。

Answer 2

您可以使用列表理解来获取30岁以上人员的姓名：

names_list = [person.name for person in ppl if person.age > 30]

并做任何你想做的事情：）

Answer 3

这就是你要找的东西：

for person in [p for p in ppl if (p.fname is "name" and p.age < 30)]:
    # Do whatever you want with it...

在Python中的对象列表属性中搜索

3 个答案: