如何在python中从dict内部列表中提取值

Question

我有下面的dict内部列表，dictlist有g1到g4的键，在许多键中有一对。我想打印带有组标题的a和23以及每个组的值之和，希望输出如下：

dictlist = { "g1" : [], "g2" : [] , "g3" : [] }

a = ["a", 23]
a1 = ["asd", 3]
a2 = ["asdf", 10]
a3 = ["adg", 5]

b1 = ["df", 5]
b2 = ["dfg", 1]

c = ["dfg", 50]

dictlist.["g1"].append[a]
dictlist.["g1"].append[a1]
dictlist.["g1"].append[a2]
dictlist.["g1"].append[a3]

dictlist.["g2"].append[b]
dictlist.["g2"].append[b1]

dictlist.["g3"].append[c]

我尝试了以下方法：

for i in dictlist.items():
  for k,v in i.item():
     print(k,v)

期望输出：

g1
a, 23
asd,3
asdf,10
adg,5
total, 41

g2
df,5
dfg,1
total, 6

 ...

有人能帮助提取每组数据的总和吗

Answer 1

如果要将dictlist中的每个项目视为字典，为什么它都是一个列表？我建议采用dict中的dict方法：

dictlist = { "g1" : {}, "g2" : {}, "g3" : {} }
a = ["a", 23] # does this really need to be a list?

# assign `a` value (23) into `a` key ("a")
dictlist['g1'][a[0]] = a[1]

# print all items from dictlist
for key, val in dictlist.items():
    print(key)
    for subkey, subval in val.items():
        print(subkey, ",", subval)
    

如果您想将列表保留在dict中，它将如下所示

dictlist = { "g1" : {}, "g2" : {}, "g3" : {} }
a = ["a", 23] # does this really need to be a list?

# assign `a` value (23) into `a` key ("a")
dictlist['g1'].append(a)

# print all items from dictlist
for key, val in dictlist.items():
    print(key)
    for subkey, subval in val: # only difference: dont use `.items()`
        print(subkey, ",", subval)
    

Answer 2

注意：要添加total，请阅读至结尾

首先，要获取dict中某个值的值，索引如下

dict["key"]而不是dict.["key"]

其次，要在列表中添加内容，请使用括号而不是方括号。所以，
.append(x)不是{}

修复我们目前拥有的两件事：

dictlist = { "g1" : [], "g2" : [] , "g3" : [] }

a = ["a", 23]
a1 = ["asd", 3]
a2 = ["asdf", 10]
a3 = ["adg", 5]

b1 = ["df", 5]
b2 = ["dfg", 1]

c = ["dfg", 50]

dictlist["g1"].append(a)
dictlist["g1"].append(a1)
dictlist["g1"].append(a2)
dictlist["g1"].append(a3)

dictlist["g2"].append(b1)
dictlist["g2"].append(b2)

dictlist["g3"].append(c)

#Now, you want to print the key of the dict as the heading, so our first loop will be:
for heading in dict.keys(): # dictlist.keys() returns a list of all the keys
    print(heading)

#The next thing you want is to print the values of the respective key seperated by ",":
    for value in dict[heading]:
        print(*value, sep=",")  # we can use the sep keyword to print it as we require

# The at the end, an empty print() to seperate the headings
    print()

输出：

g1
a, 23
asd, 3
asdf, 10
adg, 5

g2
df, 5
dfg, 1

g3
dfg, 50

现在，对于添加total：

for heading in dict.keys():
    print(heading)

    total = 0  # We make a variable, becomes zero for every heading
    for value in dict[heading]: # value is a list like ["a", 23], has str and int.. So;
        total += sum(item for item in value if type(item) == int)
        print(*value, sep=",")

    #print the total before starting the next heading
    print("total," total)
    print()

输出：

g1
a, 23
asd, 3
asdf, 10
adg, 5
total,  41

g2
df, 5
dfg, 1
total,  6

g3
dfg, 50
total,  50

Answer 3

列表中没有dict。您在dict中有列表。一旦您修复了拼写错误（append是一个函数，应该使用括号，而不是括号，并且您没有b和b1，您有b1和b2），您的代码将生成您描述的结构。它不是这样打印的，因为Python不是这样打印列表的，但数据都在那里：

dictlist = { "g1" : [], "g2" : [] , "g3" : [] }

a = ["a", 23]
a1 = ["asd", 3]
a2 = ["asdf", 10]
a3 = ["adg", 5]

b1 = ["df", 5]
b2 = ["dfg", 1]

c = ["dfg", 50]

dictlist["g1"].append(a)
dictlist["g1"].append(a1)
dictlist["g1"].append(a2)
dictlist["g1"].append(a3)

dictlist["g2"].append(b1)
dictlist["g2"].append(b2)

dictlist["g3"].append(c)
print(dictlist)

for k, v in dictlist.items():
   print(k)
   for i in v:
       print(i)

输出：

timr@tims-gram:~/src$ python x.py
{'g1': [['a', 23], ['asd', 3], ['asdf', 10], ['adg', 5]], 'g2': [['df', 5], ['dfg', 1]], 'g3': [['dfg', 50]]}
g1
['a', 23]
['asd', 3]
['asdf', 10]
['adg', 5]
g2
['df', 5]
['dfg', 1]
g3
['dfg', 50]
timr@tims-gram:~/src$