Question

使用这个numpy矩阵：

>>> print matrix
[['L' 'G' 'T' 'G' 'A' 'P' 'V' 'I']
 ['A' 'A' 'S' 'G' 'P' 'S' 'S' 'G']
 ['A' 'A' 'S' 'G' 'P' 'S' 'S' 'G']
 ['G' 'L' 'T' 'G' 'A' 'P' 'V' 'I']]

我已经有了这段代码：

for i, j in itertools.combinations(range(len(matrix.T)), 2):
    c = matrix[:, [i,j]]
    counts = collections.Counter(map(tuple,c))
    print 'columns {} and {}'.format(i,j)
    for AB in counts:
      freq_AB = float(float(counts[AB])/len(c))
      print 'Frequency of {} = {}'.format(AB, freq_AB)
    print

产生

columns 0 and 1
Frequency of ('A', 'A') = 0.5
Frequency of ('G', 'L') = 0.25
Frequency of ('L', 'G') = 0.25

columns 0 and 2
Frequency of ('A', 'S') = 0.5
Frequency of ('G', 'T') = 0.25
Frequency of ('L', 'T') = 0.25

[...]

我想要添加到代码中的是检索列（i，j）中的频率，对于列i，j的字母对中的每个字母...我的意思是，输出类似于以下一个：

columns 0 and 1
Frequency of ('A', 'A') = 0.5
  Freq of 'A' in column 0 = 0.5
  Freq of 'A' in column 1 = 0.5
Frequency of ('G', 'L') = 0.25
  Freq of 'G' in column 0 = 0.25
  Freq of 'L' in column 1 = 0.25
Frequency of ('L', 'G') = 0.25
  Freq of 'L' in column 0 = 0.25
  Freq of 'G' in column 1 = 0.25

columns 0 and 2
Frequency of ('A', 'S') = 0.5
  Freq of 'A' in column 0 = 0.5
  Freq of 'S' in column 2 = 0.5
Frequency of ('G', 'T') = 0.25
  Freq of 'G' in column 0 = 0.25
  Freq of 'T' in column 2 = 0.5
Frequency of ('L', 'T') = 0.25
  Freq of 'L' in column 0 = 0.25
  Freq of 'T' in column 2 = 0.5

[...]

非常感谢任何帮助

Answer 1

如何扩展相同的方法并像这样做：

for i, j in itertools.combinations(range(len(matrix.T)), 2):
    c = matrix[:, [i,j]]
    combined_counts = collections.Counter(map(tuple,c))
    first_column_counts = collections.Counter(c[:,0])
    second_column_counts = collections.Counter(c[:,1])
    print 'columns {} and {}'.format(i,j)
    for AB in combined_counts:
      freq_AB = float(float(combined_counts[AB])/len(c))
      print 'Frequency of {} = {}'.format(AB, freq_AB)
      freq_A = float(first_column_counts[AB[0]])/len(c)
      print "  Freq of '{}' in column {} = {}".format(AB[0], i, freq_A)
      freq_B = float(second_column_counts[AB[1]])/len(c)
      print "  Freq of '{}' in column {} = {}".format(AB[1], i, freq_B)
    print

从python中的numpy矩阵计算这两个元素的两个元素的各个频率

1 个答案: