如何在python中壓縮列表的列表？

Question

我有一個列表的列表

sample = [['A','T','N','N'],['T', 'C', 'C', 'C']],[['A','T','T','N'],['T', 'T', 'C', 'C']]. 

我試圖壓縮文件，使得只有A/T/G/C在列表和輸出必須是一個列表

[['AT','TCCC'],['ATT','TTCC']] 

當我使用此代碼：

tt = ["".join(y for y in x if y in {'A','G','T','C'}) for x in sample] 

不過，我只得到輸出：

['ATT','TTCC'] 

任何建議，我哪裏錯了？

在我實際的代碼，我第一次調換名單：

seq_list = [['TCCGGGGGTATC', 'TCCGTGGGTATC', ...]] # one nested list 

numofpops = len(seq_list) 

### Tranposing. Moving along the columns only 

#column_list = [] 
for k in range(len(seq_list)): 
    column_list = [[] for i in range(len(seq_list[k][0]))] 
    for seq in seq_list[k]: 
     for i, nuc in enumerate(seq): 
      column_list[i].append(nuc) 
      ddd = column_list 
    print ddd 

tt = ["".join(y for y in x if y in {'A','G','T','C'}) for x in ddd] 
print tt 

Answer 1

您的實際代碼丟棄列表。您只處理最後一項。

否則，您的代碼正常工作。只要做的是，在環，然後將結果追加到一些最後的名單：

results = [] 

for k in range(len(seq_list)): 
    column_list = [[] for i in range(len(seq_list[k][0]))] 
    for seq in seq_list[k]: 
     for i, nuc in enumerate(seq): 
      column_list[i].append(nuc) 
    # process `column_list` here, in the loop (no need to assign to ddd) 
    tt = ["".join(y for y in x if y in {'A','G','T','C'}) for x in column_list] 

    results.append(tt) 

請注意，您可以使用，而不是你換位列表中zip()功能：

results = [] 
for sequence in seq_list: 
    for column_list in zip(*sequence): 
     tt = [''.join([y for y in x if y in 'AGTC']) for x in column_list] 
     results.append(tt) 

Answer 2

我想你已經給定了輸入是在一個列表中的兩個項目。然後，你將使用一個列表理解與2層嵌套。在最深層次，你過濾掉不A, T, G或C然後join其他項目：

sample = [[['A','T','N','N'],['T', 'C', 'C', 'C']], [['A','T','T','N'],['T', 'T', 'C', 'C']]] 

result = [[''.join(i for i in lst if i in 'ATGC') for lst in sub] for sub in sample] 
#         ^<- filter items 
#   ^<- join the matching items 
print(result) 
# [['AT', 'TCCC'], ['ATT', 'TTCC']] 

Answer 3

你要做到以下幾點：

sample = [[['A','T','N','N'],['T', 'C', 'C', 'C']], [['A','T','T','N'],['T', 'T', 'C', 'C']]] 

然後：

tt = [[''.join([c for c in sublist if c in 'AGTC']) for sublist in doublet] for doublet in sample] 

也許這樣更具可讀性：

tt = [ 
    [''.join([c for c in sublist if c in 'AGTC']) 
    for sublist in doublet] 
    for doublet in sample 
] 

它提供所需的結果：

[['AT', 'TCCC'], ['ATT', 'TTCC']] 

Answer 4

您可以先創建一個輔助函數：

def filterJoin(s): 
    return ''.join(x for x in s if x in 'ATGC') 

然後：

>>> sample = [['A','T','N','N'],['T', 'C', 'C', 'C']],[['A','T','T','N'],['T', 'T', 'C', 'C']] 
>>> [[filterJoin(s) for s in t] for t in sample] 
[['AT', 'TCCC'], ['ATT', 'TTCC']]