Python 列表理解，值（如果不在第二个列表中）

如果每周存在于

sales

else 0中，是否可以使用列表理解来获取每周的销售额

weeks = [{'week': 31}, {'week': 32}, {'week': 33}, 
         {'week': 34}, {'week': 35}, {'week': 36}]
weeks = [x['week'] for x in weeks]

sales = [
    {'week': 32, 'sales': 1167.26}, 
    {'week': 33, 'sales': 1373.61}, 
    {'week': 36, 'sales': 1491.43}, 
]

expected = [0, 1167.26, 1373.61, 0, 0, 1491.43]

---

您可以从

sales

创建一个字典，用作

weeks

中每周的查找：

weeks = [{'week': 31}, {'week': 32}, {'week': 33}, 
     {'week': 34}, {'week': 35}, {'week': 36}]

sales = [
{'week': 32, 'sales': 1167.26}, 
{'week': 33, 'sales': 1373.61}, 
{'week': 36, 'sales': 1491.43}, 
]
_sales = {i['week']:i['sales'] for i in sales}
results = [_sales.get(i['week'], 0) for i in weeks]

输出：

[0, 1167.26, 1373.61, 0, 0, 1491.43]

---

如果您保证任何给定的

周

值只有一个

销售

元素，则您可以获得一系列销售值，并将其相加以获得任何给定周的值：

out=[
总额（如果销售['周]==周，则销售中销售的销售['销售]）
周而复始
]

如果对于任何给定的

周数

在

销售

中有多个条目，则它们的结果将相加以生成销售输出的合计值

这为您提供了一个具有所需输出的单一列表理解：

输出：

[0, 1167.26, 1373.61, 0, 0, 1491.43]

[01167.26、1373.61、0、0、1491.43]

比如说，创建一个周数到销售额的中间映射，比如

week_sales = {s['week']: s['sales'] for s in sales}

expected = [week_sales.get(w['week'], 0) for w in weeks]

---

如果您可以使用像pandas这样的第三方库

import pandas as pd
weeks = [{'week': 31}, {'week': 32}, {'week': 33}, 
         {'week': 34}, {'week': 35}, {'week': 36}]
df1=pd.DataFrame(weeks)    

sales = [
    {'week': 32, 'sales': 1167.26}, 
    {'week': 33, 'sales': 1373.61}, 
    {'week': 36, 'sales': 1491.43}, 
]
df2=pd.DataFrame(sales)

df1.merge(df2,how='left').fillna(0)

输出

week    sales
31      0.00
32      1167.26
33      1373.61
34      0.00
35      0.00
36      1491.43

[0, 1167.26, 1373.61, 0, 0, 1496.43]

如果您在同一周的销售列表中有多个条目

import pandas as pd
weeks = [{'week': 31}, {'week': 32}, {'week': 33}, 
         {'week': 34}, {'week': 35}, {'week': 36}]
df1=pd.DataFrame(weeks) 

sales = [
    {'week': 32, 'sales': 1167.26}, 
    {'week': 33, 'sales': 1373.61}, 
    {'week': 36, 'sales': 1491.43}, 
    {'week': 36, 'sales': 5}, 
]
df2=pd.DataFrame(sales)

pd.DataFrame(df1.merge(df2,how='left').fillna(0).groupby('week')['sales'].sum(name='sales'))

week    sales
31      0.00
32      1167.26
33      1373.61
34      0.00
35      0.00
36      1496.43

受@Ajax1234解决方案的启发，如果您在一周内有多个销售价值，您可以创建一个包含分组价值的dict

weeks = [{'week': 31}, {'week': 32}, {'week': 33}, 
         {'week': 34}, {'week': 35}, {'week': 36}]
sales = [
    {'week': 32, 'sales': 1167.26}, 
    {'week': 33, 'sales': 1373.61}, 
    {'week': 36, 'sales': 1491.43}, 
    {'week': 36, 'sales': 5}, 
]

sales_dict={}
for i in sales:
    if i['week'] in sales_dict:
        sales_dict[i['week']]=sales_dict[i['week']]+i['sales']
    else:
        sales_dict[i['week']]=i['sales']
[sales_dict.get(i['week'],0) for i in weeks] # sales_dict= {32: 1167.26, 33: 1373.61, 36: 1496.43}

输出

week    sales
31      0.00
32      1167.26
33      1373.61
34      0.00
35      0.00
36      1491.43

[0, 1167.26, 1373.61, 0, 0, 1496.43]

---

销售

中的所有项目是否保证在

周

中有匹配的项目？@PatrickHaugh在示例中没有。您尝试了哪些不起作用的内容？@PatrickHaugh是的所有

销售

周必须存在于

周

中它们是否始终都会被排序？

如果销售['week'，则销售中销售的金额（'sale['sales']==周）

更具可读性。我喜欢这个解决方案，但如果销售列表在同一周有多个条目，会发生什么情况。我知道OP没有澄清这一点。