Question

我有一个列表：

name_list = ["David Joyner", "David Zuber", "Brenton Joyner",
             "Brenton Zuber", "Nicol Barthel", "Shelba Barthel",
             "Shelba Crowley", "Shelba Fernald", "Shelba Odle",
             "Shelba Fry", "Maren Fry"]

我想编写一个函数来返回以名字为键的字典，所有与名字相关的全名作为值：

 {'Shelba': ['Shelba Barthel', 'Shelba Crowley', 'Shelba Fernald', 
 'Shelba Odle', 'Shelba Fry'], 'David': ['David Joyner', 'David Zuber'], 
 'Brenton': ['Brenton Joyner', 'Brenton Zuber'], 'Maren': ['Maren Fry'], 
 'Nicol': ['Nicol Barthel']}

谢谢。

Answer 1

智能方式（1条通过您的列表）：

from collections import defaultdict

name_list = ["David Joyner", "David Zuber", "Brenton Joyner",
             "Brenton Zuber", "Nicol Barthel", "Shelba Barthel",
             "Shelba Crowley", "Shelba Fernald", "Shelba Odle",
             "Shelba Fry", "Maren Fry"]

rv = defaultdict(list)
for elem in name_list:
    rv[elem.split()[0]].append(elem) 

print(dict(rv))

输出：

{'David':  ['David Joyner', 'David Zuber'], 
 'Brenton':['Brenton Joyner', 'Brenton Zuber'], 
 'Nicol':  ['Nicol Barthel'], 
 'Shelba': ['Shelba Barthel','Shelba Crowley','Shelba Fernald','Shelba Odle','Shelba Fry'], 
 'Maren':  ['Maren Fry']}

慢速（可能看起来不错，但效率低下）：

d = {key:[n for n in name_list if n.startswith(key)] 
          for key in set( p.split()[0] for p in name_list)}

在计算上，它比defaultdict版本差很多，因为它多次遍历列表：

一次创建set( p.split()[0] for p in name_list)
对于我们刚刚收集的每个名称，它再次遍历完整列表以创建字典值[n for n in name_list if n.startswith(key)]

如果以后需要排序，只需在list.sort()的值上使用dict：

# sorting the data afterwards
for k in rv:
    rv[k].sort(key = lambda x:x.split()[1]) # to sort by last names

输出：

# only difference when sorted: see key `Shelba`
{'David':  ['David Joyner', 'David Zuber'], 
 'Brenton':['Brenton Joyner', 'Brenton Zuber'], 
 'Nicol':  ['Nicol Barthel'], 
 'Shelba': ['Shelba Barthel','Shelba Crowley','Shelba Fernald','Shelba Fry','Shelba Odle'], 
 'Maren':  ['Maren Fry']}

Answer 2

如Patrick Artner在评论中所建议，请使用import numpy as np import pandas as pd from PyQt5 import QtCore, QtGui, QtWidgets def pandas_to_str(): df = pd.DataFrame({ 'A' : 1., 'B' : pd.Timestamp('20130102'), 'C' : pd.Series(1,index=list(range(4)),dtype='float32'), 'D' : np.array([3] * 4,dtype='int32'), 'E' : pd.Categorical(["test","train","test","train"]), 'F' : 'foo' }) return df.to_string(col_space =14,justify = "justify") if __name__ == '__main__': import sys app = QtWidgets.QApplication(sys.argv) w = QtWidgets.QTextBrowser() w.setStyleSheet('color: blue') w.setFont(QtGui.QFont("Monospace")) w.setWordWrapMode(QtGui.QTextOption.NoWrap) w.setText(pandas_to_str()) w.showMaximized() sys.exit(app.exec_())。怎么样：

defaultdict

Answer 3

尝试一下：

def build_dict(name_list)
result_dict = {}
for name in name_list:
    names = name.split(" ")
    if names[0] in result_dict:
        result_dict[names[0]].append(names[1])
    else:
        result_dict[names[0]] = [names[1]]

return result_dict

这个想法是迭代名称列表，将每个名称分为名字和姓氏，看看名字是否在字典中。如果是这样，请查找它，并将姓氏添加到姓氏列表中。如果不是，请为其分配一个列表，该列表的唯一成员是姓氏

Answer 4

def name_lists(name_list):
    d = {}
    
    for name in name_list:
                
        first_name = name.split()[0]
        if first_name in d:
            
            d[first_name].append(name)
            d[first_name].sort()
            
        else:
            d[first_name]=[name]

    return d

Answer 5

def name_counts(name_list):
    result_dict = {}
    for name in name_list:
        names = name.split(" ")
        if names[0] in result_dict:
            result_dict[names[0]] += 1
        else:
            result_dict[names[0]] = 1
    return result_dict

Answer 6

def name_lists(n_list):
    result_dict = {}
    n_list.sort()
    for name in n_list:
        names = name.split(" ")
        if names[0] in result_dict:
            result_dict[names[0]].append(name)
        else:
            result_dict[names[0]] = [name]
    return result_dict

创建一个以名字为键的字典，将与该键关联的全名作为值

6 个答案: