直接在字典上调用len()是有效的，并且比构建迭代器d.s ekeys()并在其上调用len()快，但与程序所做的其他事情相比，这两者的速度都可以忽略不计。

d = {x: x**2 for x in range(1000)}

len(d)
# 1000

len(d.keys())
# 1000

%timeit len(d)
# 41.9 ns ± 0.244 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

%timeit len(d.keys())
# 83.3 ns ± 0.41 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

len(yourdict.keys())

或者只是

len(yourdict)

如果你想统计文件中唯一的单词，你可以使用set和do like

len(set(open(yourdictfile).read().split()))

直接在字典上调用len()是有效的，并且比构建迭代器d.s ekeys()并在其上调用len()快，但与程序所做的其他事情相比，这两者的速度都可以忽略不计。

d = {x: x**2 for x in range(1000)}

len(d)
# 1000

len(d.keys())
# 1000

%timeit len(d)
# 41.9 ns ± 0.244 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

%timeit len(d.keys())
# 83.3 ns ± 0.41 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

为了计算字典中关键字的数量:

def dict_finder(dict_finders):
    x=input("Enter the thing you want to find: ")
    if x in dict_finders:
        print("Element found")
    else:
        print("Nothing found:")

如果问题是关于计算关键字的数量，然后会推荐一些像

def countoccurrences(store, value):
    try:
        store[value] = store[value] + 1
    except KeyError as e:
        store[value] = 1
    return

在main函数中有一些东西可以循环遍历数据并将值传递给countooccurrence函数

if __name__ == "__main__":
    store = {}
    list = ('a', 'a', 'b', 'c', 'c')
    for data in list:
        countoccurrences(store, data)
    for k, v in store.iteritems():
        print "Key " + k + " has occurred "  + str(v) + " times"

代码输出

Key a has occurred 2 times
Key c has occurred 2 times
Key b has occurred 1 times

不同单词的数量(即字典中条目的数量)可以使用len()函数找到。

> a = {'foo':42, 'bar':69}
> len(a)
2

要获取所有不同的单词(即键)，请使用.keys()方法。

> list(a.keys())
['foo', 'bar']