blocks|key|801184|text|正如您所期望的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801185|sum(d.values())|code-block|syntax|javascript|801186|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

As you'd expect:

<pre><code>sum(d.values())
</code></pre>

blocks|key|801227|text|当然有。下面是一种对字典的值进行求和的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801228|>>>+d+=+{'key1':1,'key2':14,'key3':47}
>>>+sum(d.values())
62|code-block|syntax|javascript|801229|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Sure there is. Here is a way to sum the values of a dictionary.

<pre><code>&gt;&gt;&gt; d = {'key1':1,'key2':14,'key3':47}
&gt;&gt;&gt; sum(d.values())
62
</code></pre>

blocks|key|801287|text|在Python2中，您可以使用itervalues()字典方法来避免创建所有值的临时副本，该方法返回字典键的迭代器：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|801288|sum(d.itervalues())|code-block|syntax|javascript|801289|在Python3中，您可以只使用d.values()，因为该方法已更改为这样做(并且删除了itervalues()，因为不再需要它)。|801290|为了使编写与版本无关的代码更容易，这些代码总是迭代字典的键的值，实用函数可能会很有帮助：|801291|import+sys

def+itervalues(d):
++++return+iter(getattr(d,+('itervalues',+'values')[sys.version_info[0]>2])())

sum(itervalues(d))|801292|这基本上就是本杰明·彼得森的six模块所做的事情。|801293|entityMap|0|LINK|mutability|MUTABLE|url|https://pypi.python.org/pypi/six/^0|F|C|0|0|G|A|19|C|0|0|0|E|3|E|3|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|19|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|S|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

In Python 2 you can avoid making a temporary copy of all the values by using the <code>itervalues()</code> dictionary method, which returns an iterator of the dictionary's keys:

<pre><code>sum(d.itervalues())
</code></pre>

In Python 3 you can just use <code>d.values()</code> because that method was changed to do that (and <code>itervalues()</code> was removed since it was no longer needed).

To make it easier to write version independent code which always iterates over the values of the dictionary's keys, a utility function can be helpful:

<pre><code>import sys

def itervalues(d):
 return iter(getattr(d, ('itervalues', 'values')[sys.version_info[0]&gt;2])())

sum(itervalues(d))
</code></pre>

This is essentially what Benjamin Peterson's <a href="https://pypi.python.org/pypi/six/" rel="noreferrer"><code>six</code></a> module does.

blocks|key|1015814|text|d+=+{'key1':+1,'key2':+14,'key3':+47}
sum1+=+sum(d[item]+for+item+in+d)
print(sum1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1015815|您可以使用for循环来完成此操作|unstyled|1015816|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>d = {'key1': 1,'key2': 14,'key3': 47}
sum1 = sum(d[item] for item in d)
print(sum1)
</code></pre>

you can do it using the for loop

blocks|key|801382|text|我觉得sum(d.values())是最有效的求和方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|801383|您还可以尝试使用reduce函数与lambda表达式一起计算总和：|801384|reduce(lambda+x,y:x%2By,d.values())|code-block|syntax|javascript|801385|entityMap^0|3|F|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|S|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

I feel <code>sum(d.values())</code> is the most efficient way to get the sum.

You can also try the reduce function to calculate the sum along with a lambda expression:

<pre><code>reduce(lambda x,y:x+y,d.values())
</code></pre>

blocks|key|1015927|text|->+(d.values())-+"d“sum你的字典变量|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1015928|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

sum(d.values()) - "d" -> Your dictionary Variable

blocks|key|801521|text|您可以获得字典中所有值的生成器，然后将其转换为一个列表，并使用sum()函数来获得所有值的总和。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801522|示例：|801523|c={"a":123,"b":4,"d":4,"c":-1001,"x":2002,"y":1001}

sum(list(c.values()))|code-block|syntax|javascript|801524|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can get a generator of all the values in the dictionary, then cast it to a list and use the sum() function to get the sum of all the values.

Example:

<pre class="lang-py prettyprint-override"><code>c={"a":123,"b":4,"d":4,"c":-1001,"x":2002,"y":1001}

sum(list(c.values()))
</code></pre>

blocks|key|801607|text|你可以考虑使用'for+loop‘：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801608|++d+=+{'data':+100,+'data2':+200,+'data3':+500}
++total+=+0
++for+i+in+d.values():
++++++++total+%2B=+i|code-block|syntax|javascript|801609|总数=+800|801610|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You could consider 'for loop' for this:

<pre><code> d = {'data': 100, 'data2': 200, 'data3': 500}
 total = 0
 for i in d.values():
 total += i
</code></pre>

total = 800

blocks|key|1016101|text|使用sum()对字典中的值求和。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1016102|调用dict.values()返回字典字典的值。使用sum(+values+)返回上一步的值的总和。|1016103|d+=+{'key1':1,'key2':14,'key3':47}
values+=+d.values()
#Return+values+of+a+dictionary++++
total+=+sum(values)
print(total)|code-block|syntax|javascript|1016104|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

USE sum() TO SUM THE VALUES IN A DICTIONARY.
Call dict.values() to return the values of a dictionary dict.
Use sum(values) to return the sum of the values from the previous step.
<pre><code>d = {'key1':1,'key2':14,'key3':47}
values = d.values()
#Return values of a dictionary 
total = sum(values)
print(total)
</code></pre>

blocks|key|1016132|text|最简单/最愚蠢的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1016133|https://trinket.io/python/a8a1f25353|offset|length|1016134|d+=+{'key1':+1,'key2':+14,'key3':+47}
s+=+0
for+k+in+d:
++++s+%2B=+d[k]

print(s)|code-block|syntax|javascript|1016135|或者，如果你想让它更花哨：|1016136|https://trinket.io/python/5fcd379536|1016137|import+functools

d+=+{'key1':+1,'key2':+14,'key3':+47}
s+=+functools.reduce(lambda+acc,k:+acc%2Bd[k],+d,+0)

print(s)|1016138|entityMap|0|LINK|mutability|MUTABLE|url|1^0|0|0|10|0|0|0|0|0|10|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@$D|10|E|11|1|12]]|A|$]]|$1|F|3|G|5|H|7|13|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@$D|16|E|17|1|18]]|A|$]]|$1|O|3|P|5|H|7|19|8|@]|9|@]|A|$I|J]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|C]]|X|$5|T|U|V|A|$W|N]]]]

simplest/silliest solution:
<a href="https://trinket.io/python/a8a1f25353" rel="nofollow noreferrer">https://trinket.io/python/a8a1f25353</a>
<pre><code>d = {'key1': 1,'key2': 14,'key3': 47}
s = 0
for k in d:
 s += d[k]

print(s)
</code></pre>
or if you want it fancier:
<a href="https://trinket.io/python/5fcd379536" rel="nofollow noreferrer">https://trinket.io/python/5fcd379536</a>
<pre><code>import functools

d = {'key1': 1,'key2': 14,'key3': 47}
s = functools.reduce(lambda acc,k: acc+d[k], d, 0)

print(s)
</code></pre>

Let's say I have a dictionary in which the keys map to integers like:

<pre><code>d = {'key1': 1,'key2': 14,'key3': 47}
</code></pre>

Is there a syntactically minimalistic way to return the sum of the values in <code>d</code>—i.e. <code>62</code> in this case?

How to sum all the values in a dictionary?

假设我有一个字典，其中的键映射到整数，如下所示：d = {'key1': 1,'key2': 14,'key3': 47}在这种情况下，有没有一种语法上最小的方法来返回d-i.e. 62中所有值的总和？

问如何对字典中的所有值求和？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何对字典中的所有值求和？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何对字典中的所有值求和？
EN