Question

如何从2D字典将其转换为数据框或存在/不存在的矩阵，其中列表中的值是列，键是行名？累积列表中的值，我的目标是将它们组织成矩阵。

我一直在尝试，但是没有成功：

values = set()

for genome, info in dict_cluster.items():
    for v in info:
        #t = [genome, ([v for v in info])]
        t = [genome,v]
    print pd.DataFrame(t)

输入：

A ['arylpolyene', 'hserlactone', 'hserlactone', 'nrps', 'siderophore', 't1pks-nrps', 'terpene', 'thiopeptide', 'transatpks-nrps']
B ['hserlactone', 'hserlactone-arylpolyene', 'nrps', 'siderophore', 'thiopeptide']
C ['nrps', 'nrps', 'nrps', 'siderophore', 't1pks-nrps', 't1pks-nrps']
D ['nrps', 'siderophore', 't1pks-nrps']

输出：

    arylpolyene siderophore hserlactone-arylpolyene transatpks-nrps terpene thiopeptide hserlactone nrps    t1pks-nrps
A   1   2   0   1   1   1   2   1   1
B   0   1   1   0   0   1   1   1   0
C   0   1   0   0   0   0   0   3   2
D   0   1   0   0   0   0   0   1   1

我的输出是这样：

                 0
0  GCF_900068895.1
1  transatpks-nrps
                 0
0  GCA_002415165.1
1      thiopeptide
                 0
0  GCA_000367685.2
1       t1pks-nrps
                 0
0  GCA_002732135.1
1       t1pks-nrps

Answer 1

将Counter与dictionary comprehension一起使用并分配给DataFrame：

from collections import Counter

df = pd.DataFrame({k:Counter(v) for k, v in d.items()}).T.fillna(0).astype(int)
print (df)

   arylpolyene  hserlactone  hserlactone-arylpolyene  nrps  siderophore  \
A            1            2                        0     1            1   
B            0            1                        1     1            1   
C            0            0                        0     3            1   
D            0            0                        0     1            1   

   t1pks-nrps  terpene  thiopeptide  transatpks-nrps  
A           1        1            1                1  
B           0        0            1                0  
C           2        0            0                0  
D           1        0            0                0

编辑：

对于指标值，请使用MultiLabelBinarizer：

d = {'A': ['arylpolyene', 'hserlactone', 'hserlactone', 'nrps', 'siderophore', 't1pks-nrps', 'terpene', 'thiopeptide', 'transatpks-nrps'],
'B': ['hserlactone', 'hserlactone-arylpolyene', 'nrps', 'siderophore', 'thiopeptide'],
'C' :['nrps', 'nrps', 'nrps', 'siderophore', 't1pks-nrps', 't1pks-nrps'],
'D': ['nrps', 'siderophore', 't1pks-nrps']}

from sklearn.preprocessing import MultiLabelBinarizer

mlb = MultiLabelBinarizer()
df = pd.DataFrame(mlb.fit_transform(d.values()),columns=mlb.classes_, index=d.keys())
print (df)
   arylpolyene  hserlactone  hserlactone-arylpolyene  nrps  siderophore  \
A            1            1                        0     1            1   
B            0            1                        1     1            1   
C            0            0                        0     1            1   
D            0            0                        0     1            1   

   t1pks-nrps  terpene  thiopeptide  transatpks-nrps  
A           1        1            1                1  
B           0        0            1                0  
C           1        0            0                0  
D           1        0            0                0

Answer 2

也许您正在寻找这样的东西：

val = {'A': ['arylpolyene', 'hserlactone', 'hserlactone', 'nrps', 'siderophore', 't1pks-nrps', 'terpene', 'thiopeptide', 'transatpks-nrps'],
       'B': ['hserlactone', 'hserlactone-arylpolyene', 'nrps', 'siderophore', 'thiopeptide'],
       'C': ['nrps', 'nrps', 'nrps', 'siderophore', 't1pks-nrps', 't1pks-nrps'],
       'D': ['nrps', 'siderophore', 't1pks-nrps']}
all_val = []
for k in val:
    for v in val[k]:
        all_val.append((k,v))

df = pd.DataFrame(all_val,columns=['key','val']).set_index('key')
df_count = df.pivot_table(index='key',columns='val',aggfunc=len)

输出：

val  arylpolyene  hserlactone  hserlactone-arylpolyene  nrps  siderophore  \
key                                                                         
A            1.0          2.0                      NaN   1.0          1.0   
B            NaN          1.0                      1.0   1.0          1.0   
C            NaN          NaN                      NaN   3.0          1.0   
D            NaN          NaN                      NaN   1.0          1.0   

val  t1pks-nrps  terpene  thiopeptide  transatpks-nrps  
key                                                     
A           1.0      1.0          1.0              1.0  
B           NaN      NaN          1.0              NaN  
C           2.0      NaN          NaN              NaN  
D           1.0      NaN          NaN              NaN

Answer 3

这应该可以完成您的工作（我正在使用Python3）：

my_dict = {
            'A': ['arylpolyene', 'hserlactone', 'hserlactone', 'nrps', 'siderophore', 't1pks-nrps', 'terpene', 'thiopeptide', 'transatpks-nrps'],
            'B': ['hserlactone', 'hserlactone-arylpolyene', 'nrps', 'siderophore', 'thiopeptide'],
            'C': ['nrps', 'nrps', 'nrps', 'siderophore', 't1pks-nrps', 't1pks-nrps'],
            'D': ['nrps', 'siderophore', 't1pks-nrps']
            }

rows_list=list(my_dict.keys())
values=list(my_dict.values())
rows_size=len(rows_list)

columns_list = []

for sublist in values:
    for item in sublist:
        if item not in columns_list:
            columns_list.append(item)

columns_size = len(columns_list)

#initialize adjacent matrix
print('Initial adjacent matrix')
adjacent = [ [0]*columns_size for i in range(rows_size) ]
for row in adjacent:
    print(row)

for key, value in my_dict.items():
    for v in value:
        adjacent[rows_list.index(key)][columns_list.index(v)] += 1

print('-'*50)
print('Final adjacent matrix')
for row in adjacent:
    print(row)

在第一个循环for sublist in values:中，我创建了一个列表，其中包含您想要的值作为没有重复的列。

在adjacent = [ [0]*columns_size for i in range(rows_size) ]中，我创建了一个列表，其中的元素与字典键的数量一样多。每个元素都是一个列表，其中元素的数量与列值的数量一样。

我试图做得尽可能简单，告诉我是否有一些你不知道的事情：）

数据框存在/不存在的字典

3 个答案: