将子列表中的第二项替换为dataframe的行值

Question

我有一个嵌套列表,并希望用数据帧的行值替换每个子列表的第二项.这是我的数据框和列表:

import pandas as pd
mydata = [{'id' : '12'},
          {'id' : '34'},
          {'id' : '56'},
          {'id' : '78'},]
df = pd.DataFrame(mydata)

L1 = [ ['elephant',0], ['zebra',1], ['lion',2], ['giraffe',3]  ]

期望的结果是: [ ['elephant',12], ['zebra',34], ['lion',56], ['giraffe',78] ]

这是我的代码:

for i in L1:
    for j, row in df.iterrows():
        i[1] = df["id"][j] 

哪个输出: [['elephant', '78'], ['zebra', '78'], ['lion', '78'], ['giraffe','78']]

Answer 1

使用列表推导生成第一个元素的列表,然后zip使用id col:

In[32]:
list(zip([x[0] for x in L1], df['id'].tolist()))

Out[32]: [('elephant', '12'), ('zebra', '34'), ('lion', '56'), ('giraffe', '78')]

如果你坚持列表列表,你可以将上面的内容变成一个列表:

In[35]:
L2 = list(zip([x[0] for x in L1], df['id'].tolist()))
L2

Out[35]: [('elephant', '12'), ('zebra', '34'), ('lion', '56'), ('giraffe', '78')]

In[36]:
[list(x) for x in L2]

Out[36]: [['elephant', '12'], ['zebra', '34'], ['lion', '56'], ['giraffe', '78']]

一个纯粹的熊猫方法是从你的列表中构建一个df:

In[41]:
df2 = pd.DataFrame(L1)
df2

Out[41]: 
          0  1
0  elephant  0
1     zebra  1
2      lion  2
3   giraffe  3

然后concat加入他们:

In[43]:
merged = pd.concat([df,df2], axis=1)
merged

Out[43]: 
   id         0  1
0  12  elephant  0
1  34     zebra  1
2  56      lion  2
3  78   giraffe  3

然后简单地选择感兴趣的cols并调用.values返回一个np数组,然后tolist:

In[46]:
merged[[0,'id']].values.tolist()

Out[46]: [['elephant', '12'], ['zebra', '34'], ['lion', '56'], ['giraffe', '78']]