计算用一个以上的键找到一个字典值的次数 [英] Count how many times a dictionary value is found with more than one key

问题描述

我正在使用python.有没有一种方法可以计算使用一个以上的键在字典中找到值的次数，然后返回计数?

I'm working in python. Is there a way to count how many times values in a dictionary are found with more than one key, and then return a count?

因此，例如，如果我有50个值，并且运行了一个脚本来执行此操作，那么我将得到一个看起来像这样的计数:

So if for example I had 50 values and I ran a script to do this, I would get a count that would look something like this:

1: 23  
2: 15  
3: 7  
4: 5  

上面的内容告诉我，1个键中显示23个值，2个键中显示15个值，3个键中显示7个值，4个键中显示5个值.

The above would be telling me that 23 values appear in 1 key, 15 values appear in 2 keys, 7 values appear in 3 keys and 5 values appear in 4 keys.

此外，如果我的字典中每个键有多个值，这个问题是否会改变?

Also, would this question change if there were multiple values per key in my dictionary?

这是我字典的一个样本(它是细菌名称):

Here is a sample of my dictionary (it's bacteria names):

{'0': ['Pyrobaculum'], '1': ['Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium'], '3': ['Thermoanaerobacter', 'Thermoanaerobacter'], '2': ['Helicobacter', 'Mycobacterium'], '5': ['Thermoanaerobacter', 'Thermoanaerobacter'], '4': ['Helicobacter'], '7': ['Syntrophomonas'], '6': ['Gelria'], '9': ['Campylobacter', 'Campylobacter'], '8': ['Syntrophomonas'], '10': ['Desulfitobacterium', 'Mycobacterium']}

因此，从此示例中，有8个唯一值，我将获得理想的反馈:

So from this sample, there are 8 unique values, I the ideal feedback I would get be:

1:4
2:3
3:1

因此，一个键中只有4个细菌名称，两个键中有3个细菌，三个键中有1个细菌.

So 4 bacteria names are only in one key, 3 bacteria are found in two keys and 1 bacteria is found in three keys.

推荐答案

所以除非我读错了，否则你想知道:

So unless I'm reading this wrong you want to know:

• 对于原始字典中的每个值，每个不同的值计数会出现多少次?
• 本质上您想要的是频率 字典中的值

我采取了一种不太优雅的方法，但其他方法却无法解决，但已将问题分解为各个步骤:

I took a less elegant approach that the other answers, but have broken the problem down for you into individual steps:

d = {'0': ['Pyrobaculum'], '1': ['Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium', 'Mycobacterium'], '3': ['Thermoanaerobacter', 'Thermoanaerobacter'], '2': ['Helicobacter', 'Mycobacterium'], '5': ['Thermoanaerobacter', 'Thermoanaerobacter'], '4': ['Helicobacter'], '7': ['Syntrophomonas'], '6': ['Gelria'], '9': ['Campylobacter', 'Campylobacter'], '8': ['Syntrophomonas'], '10': ['Desulfitobacterium', 'Mycobacterium']}

# Iterate through and find out how many times each key occurs
vals = {}                       # A dictonary to store how often each value occurs.
for i in d.values():
  for j in set(i):              # Convert to a set to remove duplicates
    vals[j] = 1 + vals.get(j,0) # If we've seen this value iterate the count
                                # Otherwise we get the default of 0 and iterate it
print vals

# Iterate through each possible freqency and find how many values have that count.
counts = {}                     # A dictonary to store the final frequencies.
# We will iterate from 0 (which is a valid count) to the maximum count
for i in range(0,max(vals.values())+1):
    # Find all values that have the current frequency, count them
    #and add them to the frequency dictionary
    counts[i] = len([x for x in vals.values() if x == i])

for key in sorted(counts.keys()):
  if counts[key] > 0:
     print key,":",counts[key]

您还可以在键盘上测试此代码.

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