Question

我有这些：

df1=[('f', {'abe': 1}), ('f', {'abeli': 1}), ('f', {'mos': 1}), ('f', {'esc': 1})]

df2= [('l', {'mos': 1}), ('l', {'esc': 1})]

我想生成两个数组

L1=[1,1,1,1]

L2=[0,0,1,1]

0因为'abe'和'abeli'在第二个数组中不存在。

dict1_part = [sc[1] for sc in df1]

dict2_part = [sc[1] for sc in df2]

L1=len(dict1_part)

L2=len(dict2_part)

print(L1 , L2)

buckets=[]


for items in dict1_part:

  pl=items.keys()

  Value=items.values()

  print("pl",plant)

  print("dict2_part",dict2_part)


  if pl in dict2_part.keys():

    buckets.append(1)

  else:

    buckets.append(0)


print(buckets)

Answer 1

首先，在df1中找到每个字典的键，然后使用dict.get：

df1=[('f', {'abe': 1}), ('f', {'abeli': 1}), ('f', {'mos': 1}), ('f', {'esc': 1})]
df2= [('l', {'mos': 1}), ('l', {'esc': 1})]
df_keys= [list(b.keys())[0] for _, b in df1]
first_array = [b.get(c) for (_, b), c in zip(df1, df_keys)]
second_array = [0 if not any(x in c for _, c in df2) else [b[x] for _, b in df2 if x in b][0] for x in df_keys]

输出：

[1, 1, 1, 1]
[0, 0, 1, 1]

如何将字典的缺失值添加到数组中

1 个答案: