我正在尝试为每个用户名生成随机颜色。
当我编写此代码时,name1和name2具有相同的颜色。怎么了?
import random
from collections import defaultdict
def get_color():
print('call')
return ''.join([random.choice('0123456789ABCDEF') for j in range(6)])
colors = defaultdict(get_color)
msg1 = {'name' : 'name1'}
msg2 = {'name' : 'name2'}
for msg in [msg1, msg2]:
msg = '{colors[name]} {name}'.format(colors=colors, **msg)
print(msg)
输出:
致电
72C44D名称1
72C44D名称2
谢谢
答案 0 :(得分:3)
在循环完成后打印colors
会显示
defaultdict(<function get_color at 0x7f6e7faed1e0>, {'name': '67C80A'})
即color
只有一个密钥。您正在通过硬编码密钥colors['name']
访问'name'
,而不是通过动态名称访问colors[name]
。
您还需要一个格式化步骤。一个用于构建模板,另一个用于将colors[name]
插入模板。
import random
from collections import defaultdict
def get_color():
print('call')
return ''.join([random.choice('0123456789ABCDEF') for j in range(6)])
colors = defaultdict(get_color)
msg1 = {'name' : 'name1'}
msg2 = {'name' : 'name2'}
for msg in [msg1, msg2]:
msg_template = '{{colors[{name}]}} {{name}}'.format(**msg)
print(msg_template) # for demo purposes
msg = msg_template.format(colors=colors, **msg)
print(msg)
输出
{colors[name1]} {name}
call
A70B47 name1
{colors[name2]} {name}
call
55709A name2
答案 1 :(得分:0)
您仅呼叫get_colors()
一次,并且将存储的颜色(在colors
中分配给这两个消息。像下面这样为每次迭代生成随机颜色:
for msg in [msg1, msg2]:
msg = '{colors[name]} {name}'.format(colors=defaultdict(get_color), **msg)
print(msg)
答案 2 :(得分:0)
我将生成物与印刷品分开,以使其更易于阅读。您可以简化代码,这将变得更加容易。
import random
from collections import defaultdict
def generate_color():
color = ''.join([random.choice('0123456789ABCDEF') for j in range(6)])
print ('generate_color returns', color)
return color
colors = defaultdict(generate_color)
# simplify this. It's just a list of names
names = ['name1', 'name2']
# to call generate_color once for each name,
# all you need to do is access that name in the default dict
print ('building colors')
for name in names:
colors[name]
# now you have populated colors.
print ('printing colors')
for key in colors:
# I went with simplifying the format string
print ('{} {}'.format(key, colors[key]))