有如下的字典dic1={'u':-3,'v':-2,'w':-1,'x':1,'y':2,'z':3},先创建字典dict1,然后从dict1创建一个新的Series对象s3
时间: 2024-01-03 08:04:13 浏览: 29
我是AI助手,根据题目要求,可以使用以下代码:
```python
import pandas as pd
# 创建字典dict1
dict1 = {'u': -3, 'v': -2, 'w': -1, 'x': 1, 'y': 2, 'z': 3}
# 从dict1创建Series对象s3
s3 = pd.Series(dict1)
print(s3)
```
输出结果为:
```
u -3
v -2
w -1
x 1
y 2
z 3
dtype: int64
```
相关问题
key=lambda x: (-x[0],x[1])
这个问题是关于lambda表达式的用法。lambda表达式是一种匿名函数,可以用于简洁地定义一个函数。在这个问题中,key=lambda x: (-x<span class="em">1</span><span class="em">2</span><span class="em">3</span>
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[ .reference_list ]
编写一个 Python 程序,连接以下字典以创建一个新字典。例如 dic1=f1:10,2:201dic2=13:30,4:40 dic3=15:506:601期望的输出:[1:10,2: 20,3: 30,4: 40,5:50,6: 60
```python
dic1 = {1:10, 2:20}
dic2 = {3:30, 4:40}
dic3 = {5:50, 6:60}
new_dict = {}
for dic in [dic1, dic2, dic3]:
for key, value in dic.items():
new_dict[key] = value
print(new_dict)
```
输出:
```
{1: 10, 2: 20, 3: 30, 4: 40, 5: 50, 6: 60}
```
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