Question

我将样本存储在以下列表中

 sample = [AAAA,CGCG,TTTT,AT-T,CATC]

..为了说明问题，我将它们表示为＆＃34; Set＆＃34;以下

Set1 AAAA
Set2 CGCG
Set3 TTTT
Set4 AT-T
Set5 CATC

消除集合中每个元素与其自身相同的所有集合。

输出：

 Set2 CGCG
 Set4 AT-T
 Set5 CATC

执行集合之间的成对比较。（Set2 v Set4，Set 2v Set5，Set4 v Set5）
每个成对比较只能有两种类型的组合，如果没有，则消除那些成对比较。例如，
```
Set2    Set5
C       C
G       A
C       T 
G       C
```

这里，有两种以上的对（CC），（GA），（CT）和（GC）。所以这种成对比较不会发生。

每次比较只能有两种组合（AA，GG，CC，TT，AT，TA，AC，CA，AG，GA，GC，CG，GT，TG，CT，TC）...基本上所有订单重要的ACGT的可能组合。

在给定的示例中，找到了超过2个这样的组合。

因此，Set2和Set4;不能考虑Set4和Set5。因此，剩下的唯一对是：

Output
Set2 CGCG
Set4 AT-T

在这种成对比较中，删除任何元素＆＃34; - ＆＃34;及其在另一对中的对应元素
```
Output    
Set2 CGG
Set4 ATT
```
计算Set2和Set4中元素的频率。计算集合（CA和GT对）中对的类型出现的频率
```
Output
Set2 (C = 1/3, G = 2/3)
Set4 (A = 1/3, T = 2/3)
Pairs (CA = 1/3, GT = 2/3)
```
计算相应元素的float（a）=（Pairs） - （Set2）*（Set4）（任何一对就足够了）
```
eg. For CA pairs, float (a) = (freq of CA pairs) - (freq of C) * (freq of A)
```

注意：如果该对是AAAC和CCCA，则C的频率为1/4，即它是其中一对的基频率

计算

float (b) = float(a)/ (freq of C in CGG) * (freq G in CGG) * (freq A in ATT) * (ATT==> freq of T in ATT)

对所有成对比较重复此操作

例如。

Set2 CGCG
Set4 AT-T
Set6 GCGC

Set2 v Set4，Set2 v Set6，Set4 v Set6

我的半生不熟的代码到现在为止： **如果建议的所有代码都采用标准的for-loop格式而不是理解**，我更希望**

#Step 1
for i in sample: 
    for j in range(i):
        if j = j+1    #This needs to be corrected to if all elements in i identical to each other i.e. if all "j's" are the same
                        del i 
    #insert line of code where sample1 = new sample with deletions as above

#Step 2
    for i,i+1 in enumerate(sample):
    #Step 3
    for j in range(i):
        for k in range (i+1):
        #insert line of code to say only two types of pairs can be included, if yes continue else skip
            #Step 4
            if j = "-" or k = "-":
                #Delete j/k and the corresponding element in the other pair
                #Step 5
                count_dict = {}
                    square_dict = {}
                for base in list(i):
                    if base in count_dict:
                            count_dict[base] += 1
                    else:
                            count_dict[base] = 1
                    for allele in count_dict:
                    freq = (count_dict[allele] / len(i)) #frequencies of individual alleles
                    #Calculate frequency of pairs 
                #Step 6
                No code yet

Answer 1

我认为这就是你想要的：

from collections import Counter

# Remove elements where all nucleobases are the same.
for index in range(len(sample) - 1, -1, -1):
    if sample[index][:1] * len(sample[index]) == sample[index]:
        del sample[index]

for indexA, setA in enumerate(sample):
    for indexB, setB in enumerate(sample):
        # Don't compare samples with themselves nor compare same pair twice.
        if indexA <= indexB:
            continue

        # Calculate number of unique pairs
        pair_count = Counter()
        for pair in zip(setA, setB):
            if '-' not in pair:
                pair_count[pair] += 1

        # Only analyse pairs of sets with 2 unique pairs.
        if len(pair_count) != 2:
            continue

        # Count individual bases.
        base_counter = Counter()
        for pair, count in pair_count.items():
            base_counter[pair[0]] += count
            base_counter[pair[1]] += count

        # Get the length of one of each item in the pair.
        sequence_length = sum(pair_count.values())

        # Convert counts to frequencies.
        base_freq = {}
        for base, count in base_counter.items():
            base_freq[base] = count / float(sequence_length)

        # Examine a pair from the two unique pairs to calculate float_a.
        pair = list(pair_count)[0]
        float_a = (pair_count[pair] / float(sequence_length)) - base_freq[pair[0]] * base_freq[pair[1]]

        # Step 7!
        float_b = float_a / float(base_freq.get('A', 0) * base_freq.get('T', 0) * base_freq.get('C', 0) * base_freq.get('G', 0))

或者，更多的Python（使用你不想要的list / dict理解）：

from collections import Counter

BASES = 'ATCG'

# Remove elements where all nucleobases are the same.
sample = [item for item in sample if item[:1] * len(item) != item]

for indexA, setA in enumerate(sample):
    for indexB, setB in enumerate(sample):
        # Don't compare samples with themselves nor compare same pair twice.
        if indexA <= indexB:
            continue

        # Calculate number of unique pairs
        relevant_pairs = [(elA, elB) for (elA, elB) in zip(setA, setB) if elA != '-' and elB != '-']
        pair_count = Counter(relevant_pairs)

        # Only analyse pairs of sets with 2 unique pairs.
        if len(pair_count) != 2:
            continue

        # setA and setB as tuples with pairs involving '-' removed.
        setA, setB = zip(*relevant_pairs)

        # Get the total for each base.
        seq_length = len(setA)

        # Convert counts to frequencies.
        base_freq = {base : count / float(seq_length) for (base, count) in (Counter(setA) + Counter(setB)).items()}

        # Examine a pair from the two unique pairs to calculate float_a.
        pair = list(pair_count)[0]
        float_a = (pair_count[pair] / float(seq_length)) - base_freq[pair[0]] * base_freq[pair[1]]

        # Step 7!
        denominator = 1
        for base in BASES:
            denominator *= base_freq.get(base, 0)

        float_b = float_a / denominator

如何计算Python中列表成对比较的元素频率？

1 个答案: