blocks|key|4274314|text|如果您不想导入Counter，请使用字典来跟踪每个部门的员工数量：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4274315|employee_list+=+[{
++++'name':+'John',
++++'empID':+'102',
++++'dpt':+'tech',
++++'title':+'programmer',
++++'salary':+'75'
},+{
++++'name':+'Jane',
++++'empID':+'202',
++++'dpt':+'tech',
++++'title':+'programmer',
++++'salary':+'80'
},+{
++++'name':+'Joe',
++++'empID':+'303',
++++'dpt':+'accounting',
++++'title':+'accountant',
++++'salary':+'85'
}]

department_to_count+=+dict()
for+employee+in+employee_list:
++++department+=+employee["dpt"]

++++if+department+not+in+department_to_count:
++++++++department_to_count[department]+=+0

++++department_to_count[department]+%2B=+1

for+department,+employee_count+in+department_to_count.items():
++++print("Department+{}+has+{}+employees".format(department,
++++++++++++++++++++++++++++++++++++++++++++++++++employee_count))|code-block|syntax|javascript|4274316|entityMap^0|7|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

If you don't want to import <code>Counter</code>, use a dictionary to keep track of the employee count per department:

<pre><code>employee_list = [{
 'name': 'John',
 'empID': '102',
 'dpt': 'tech',
 'title': 'programmer',
 'salary': '75'
}, {
 'name': 'Jane',
 'empID': '202',
 'dpt': 'tech',
 'title': 'programmer',
 'salary': '80'
}, {
 'name': 'Joe',
 'empID': '303',
 'dpt': 'accounting',
 'title': 'accountant',
 'salary': '85'
}]

department_to_count = dict()
for employee in employee_list:
 department = employee["dpt"]

 if department not in department_to_count:
 department_to_count[department] = 0

 department_to_count[department] += 1

for department, employee_count in department_to_count.items():
 print("Department {} has {} employees".format(department,
 employee_count))
</code></pre>

blocks|key|4273899|text|使用collections.Counter|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4273900|from+collections+import+Counter

employees+=+[{'name':+'John',+'empID':+'102',+'dpt':+'tech',+'title':+'programmer',+'salary':+'75'},
+++++++++++++{'name':+'Jane',+'empID':+'202',+'dpt':+'tech',+'title':+'programmer',+'salary':+'80'},
+++++++++++++{'name':+'Joe',+'empID':+'303',+'dpt':+'accounting',+'title':+'accountant',+'salary':+'85'}]

dpts+=+[x['dpt']+for+x+in+employees]
print(Counter(dpts))

#+Counter({'tech':+2,+'accounting':+1})|code-block|syntax|javascript|4273901|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/collections.html#collections.Counter^0|2|J|2|J|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Use <a href="https://docs.python.org/3/library/collections.html#collections.Counter" rel="nofollow noreferrer"><code>collections.Counter</code></a>:

<pre><code>from collections import Counter

employees = [{'name': 'John', 'empID': '102', 'dpt': 'tech', 'title': 'programmer', 'salary': '75'},
 {'name': 'Jane', 'empID': '202', 'dpt': 'tech', 'title': 'programmer', 'salary': '80'},
 {'name': 'Joe', 'empID': '303', 'dpt': 'accounting', 'title': 'accountant', 'salary': '85'}]

dpts = [x['dpt'] for x in employees]
print(Counter(dpts))

# Counter({'tech': 2, 'accounting': 1})
</code></pre>

blocks|key|4274378|text|如果您正在寻找一个特定关键字的合计，您可以使用.count()+list+method作为直接查询。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4274379|给出一个字典列表：|4274380|>>>+employees+=+[{'name':+'John',+'empID':+'102',+'dpt':+'tech',+'title':+'programmer',+'salary':+'75'},
+++++++++{'name':+'Jane',+'empID':+'202',+'dpt':+'tech',+'title':+'programmer',+'salary':+'80'},
+++++++++{'name':+'Joe',+'empID':+'303',+'dpt':+'accounting',+'title':+'accountant',+'salary':+'85'}]|code-block|syntax|javascript|4274381|您可以使用列表理解创建每种类型的列表：|4274382|>>>+[d['dpt']+for+d+in+employees]
['tech',+'tech',+'accounting']|4274383|然后这就可以被记录下来：|4274384|>>>+[d['dpt']+for+d+in+employees].count('tech')
2
>>>+[d['dpt']+for+d+in+employees].count('accounting')
1
>>>+[d['dpt']+for+d+in+employees].count('nothing')
0|4274385|或者，您可以构造一个生成器来计算特定值：|4274386|>>>+sum(1+for+d+in+employees+if+d['dpt']=='tech')
2|4274387|如果您想要获得所有元素的计数(而不是使用计数器)，您可以这样做：|4274388|>>>+lot=[d['dpt']+for+d+in+employees]
>>>+{c:lot.count(c)+for+c+in+set(lot)}
{'accounting':+1,+'tech':+2}|4274389|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.6/tutorial/datastructures.html^0|N|8|W|B|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|F|3|G|5|6|7|1G|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1H|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1I|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1J|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1K|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1L|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1M|8|@]|D|@]|E|$]]|$1|W|3|X|5|J|7|1N|8|@]|D|@]|E|$K|L]]|$1|Y|3|Z|5|6|7|1O|8|@]|D|@]|E|$]]|$1|10|3|11|5|J|7|1P|8|@]|D|@]|E|$K|L]]|$1|12|3|-4|5|6|7|1Q|8|@]|D|@]|E|$]]]|13|$14|$5|15|16|17|E|$18|19]]]]

If you are looking for one specific key's total, you can use the <code>.count()</code> <a href="https://docs.python.org/3.6/tutorial/datastructures.html" rel="nofollow noreferrer">list method</a> as a direct query.

Given a list of dicts:

<pre><code>&gt;&gt;&gt; employees = [{'name': 'John', 'empID': '102', 'dpt': 'tech', 'title': 'programmer', 'salary': '75'},
 {'name': 'Jane', 'empID': '202', 'dpt': 'tech', 'title': 'programmer', 'salary': '80'},
 {'name': 'Joe', 'empID': '303', 'dpt': 'accounting', 'title': 'accountant', 'salary': '85'}]
</code></pre>

You can create a list of each with a list comprehension:

<pre><code>&gt;&gt;&gt; [d['dpt'] for d in employees]
['tech', 'tech', 'accounting']
</code></pre>

And that can then be tallied:

<pre><code>&gt;&gt;&gt; [d['dpt'] for d in employees].count('tech')
2
&gt;&gt;&gt; [d['dpt'] for d in employees].count('accounting')
1
&gt;&gt;&gt; [d['dpt'] for d in employees].count('nothing')
0
</code></pre>

Alternatively, you can construct a generator that will tally specific values:

<pre><code>&gt;&gt;&gt; sum(1 for d in employees if d['dpt']=='tech')
2
</code></pre>

If you want to have the count of all elements (other than using a Counter) you can do:

<pre><code>&gt;&gt;&gt; lot=[d['dpt'] for d in employees]
&gt;&gt;&gt; {c:lot.count(c) for c in set(lot)}
{'accounting': 1, 'tech': 2}
</code></pre>

python is new language for me so this question might sound simple, but if someone can point me in the right direction, I would appreciate it! I created a dictionary call employees and it holds some value about them:

I'm trying to read how many people are in each department, for example: tech-2, accounting-1. 

I have something like this, but it prints out blank.

<pre><code> def main():
 employees= {'name': 'John', 'empID': '102', 'dpt': 'tech', 'title': 
 'programmer', 'salary': '75'}
 {'name': 'Jane', 'empID': '202', 'dpt': 'tech', 'title': 'programmer', 
 'salary': '80'}
 {'name': 'Joe', 'empID': '303', 'dpt': 'accounting', 'title': 
 'accountant', 'salary': '85'}
 for item in employees:
 dic = employees[item]
 if dic['dpt'[0]]==dic['dpt'[1]]:
 duplicate += 1
 print("there are "+ duplicate)
 else:
 print("There are no duplicate")
</code></pre>

Counting values in an employee dictionary database in python

python对我来说是一门新语言，所以这个问题听起来可能很简单，但如果有人能给我指明正确的方向，我将不胜感激！我创建了一个名为employees的字典，它包含了关于它们的一些值：我试着读出每个部门有多少人，例如:技术-2，会计-1。我有类似的东西，但打印出来是空白的。  def main():   employees=...

问在python中对雇员字典数据库中的值进行计数
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