Python实现斐波那契数列的多种写法总结
作者：lxw-pro
斐波那契数列——经典例子，永不过时！！！
1.for循环
def fibonacci1(n):
a, b = 0, 1
for i in range(n):
a, b = b, a+b
print(a)
fibonacci1(3)
或
def fib1(w):
a, b = 1, 1
for i in range(w-1):
a, b = b, a+b
return a
print(fib1(3))
[^1]刚好得出这个位置的数
2.while循环
def fibnaqi2(m):
a, b = 0, 1
i = 0
while i < m:
print(b)
a, b = b, a+b
i += 1
fibnaqi2(4)
[^1]刚好得出这个位置的数
3.使用递归
def fib2(q):
if q == 1 or q == 2:
return 1
return fib2(q-1)+fib2(q-2)
print(fib2(9))
4.递归+for循环
def fibnacci3(p):
lst = []
for i in range(p):
if i == 1 or i == 0:
lst.append(1)
else:
lst.append(lst[i-1]+lst[i-2])
print(lst)
fibnacci3(5)
5.递归+while循环
def fibnacci4(k):
lis = []
i = 0
while i<k:
if i == 0 or i == 1:
lis.append(1)
else:
lis.append(lis[i-2]+lis[i-1])
i += 1
print(lis)
fibnacci4(6)
6.递归+定义函数+for循环
def fibnacci5(o):
def fn(i):
if i < 2:
return 1
else:
return (fn(i-2)+fn(i-1))
for i in range(o):
print(fn(i))
fibnacci5(8)
7.指定列表
def fib3(e):
if e == 1:
return [1]
if e == 2:
return [1, 1]
fibs = [1, 1]
for i in range(2, e):
fibs.append(fibs[-1]+fibs[-2])
return fibs
print(fib3(12))
趣方程求解
题目描述
二次方程式 ax**2 + bx + c = 0 （a、b、c 用户提供，为实数，a ≠ 0）
# 导入 cmath(复杂数学运算) 模块
import cmath
a = float(input('输入 a: '))
b = float(input('输入 b: '))
c = float(input('输入 c: '))
# 计算
d = (b ** 2) - (4 * a * c)
# 两种求解方式
sol1 = (-b - cmath.sqrt(d)) / (2 * a)
sol2 = (-b + cmath.sqrt(d)) / (2 * a)
print('结果为 {0} 和 {1}'.format(sol1, sol2))pandas 每日一练
# -*- coding = utf-8 -*-
# @Time : 2022/7/26 21:48
# @Author : lxw_pro
# @File : pandas -8 练习.py
# @Software : PyCharm
import pandas as pd
import numpy as np
df = pd.read_excel('text5.xlsx')
print(df)
print()
程序运行结果如下：
Unnamed: 0 Unnamed: 0.1 project ... test_time date time
0 0 00:00:00 Python ... 2022-06-20 18:30:20 2022-06-20 18:30:20
1 1 1 Java ... 2022-06-18 19:40:20 2022-06-18 19:40:20
2 2 2 C ... 2022-06-08 13:33:20 2022-06-08 13:33:20
3 3 3 MySQL ... 2021-12-23 11:26:20 2021-12-23 11:26:20
4 4 4 Linux ... 2021-12-20 18:20:20 2021-12-20 18:20:20
5 5 5 Math ... 2022-07-20 16:30:20 2022-07-20 16:30:20
6 6 6 English ... 2022-06-23 15:30:20 2022-06-23 15:30:20
7 7 7 Python ... 2022-07-19 09:30:20 2022-07-19 09:30:20
[8 rows x 7 columns]
41、将test_time列设置为索引
print(df.set_index('test_time'))
print()
程序运行结果如下：
Unnamed: 0 Unnamed: 0.1 ... date time
test_time ...
2022-06-20 18:30:20 0 00:00:00 ... 2022-06-20 18:30:20
2022-06-18 19:40:20 1 1 ... 2022-06-18 19:40:20
2022-06-08 13:33:20 2 2 ... 2022-06-08 13:33:20
2021-12-23 11:26:20 3 3 ... 2021-12-23 11:26:20
2021-12-20 18:20:20 4 4 ... 2021-12-20 18:20:20
2022-07-20 16:30:20 5 5 ... 2022-07-20 16:30:20
2022-06-23 15:30:20 6 6 ... 2022-06-23 15:30:20
2022-07-19 09:30:20 7 7 ... 2022-07-19 09:30:20
[8 rows x 6 columns]
42、生成一个和df长度相同的随机数dataframe
df1 = pd.DataFrame(pd.Series(np.random.randint(1, 10, 8))) print(df1) print()
程序运行结果如下：
0
0 1
1 3
2 2
3 7
4 7
5 3
6 5
7 1
43、将上一题生成的dataframe与df合并
df = pd.concat([df, df1], axis=1) print(df) print()
程序运行结果如下：
Unnamed: 0 Unnamed: 0.1 project ... date time 0
0 0 00:00:00 Python ... 2022-06-20 18:30:20 1
1 1 1 Java ... 2022-06-18 19:40:20 3
2 2 2 C ... 2022-06-08 13:33:20 2
3 3 3 MySQL ... 2021-12-23 11:26:20 7
4 4 4 Linux ... 2021-12-20 18:20:20 7
5 5 5 Math ... 2022-07-20 16:30:20 3
6 6 6 English ... 2022-06-23 15:30:20 5
7 7 7 Python ... 2022-07-19 09:30:20 1
[8 rows x 8 columns]
44、生成新的一列new为popularity列减去之前生成随机数列
df['new'] = df['popularity'] - df[0] print(df) print()
程序运行结果如下：
Unnamed: 0 Unnamed: 0.1 project popularity ... date time 0 new
0 0 00:00:00 Python 95 ... 2022-06-20 18:30:20 1 94
1 1 1 Java 92 ... 2022-06-18 19:40:20 3 89
2 2 2 C 145 ... 2022-06-08 13:33:20 2 143
3 3 3 MySQL 141 ... 2021-12-23 11:26:20 7 134
4 4 4 Linux 84 ... 2021-12-20 18:20:20 7 77
5 5 5 Math 148 ... 2022-07-20 16:30:20 3 145
6 6 6 English 146 ... 2022-06-23 15:30:20 5 141
7 7 7 Python 149 ... 2022-07-19 09:30:20 1 148
[8 rows x 9 columns]
45、检查数据中是否含有任何缺失值
jch = df.isnull().values.any() print(jch) # 运行结果为：False print()
46、将popularity列类型转换为浮点数
fds = df['popularity'].astype(np.float64) print(fds) print()
程序运行结果如下：
0 95.0
1 92.0
2 145.0
3 141.0
4 84.0
5 148.0
6 146.0
7 149.0
Name: popularity, dtype: float64
47、计算popularity大于100的次数
cs = len(df[df['popularity'] > 100]) print(cs) # 运行结果为：5 print()
48、查看project列共有几种学历
ckj = df['project'].nunique() print(ckj) # 运行结果为：7 print()
49、查看每科出现的次数
ckc = df.project.value_counts() print(ckc) print()
程序运行结果如下：
Python 2
Java 1
C 1
MySQL 1
Linux 1
Math 1
English 1
Name: project, dtype: int64
50、提取popularity与new列的和大于136的最后3行
df1 = df[['popularity', 'new']] hh = df1.apply(np.sum, axis=1) res = df.iloc[np.where(hh > 136)[0][-3:], :] print(res)
程序运行结果如下：
Unnamed: 0 Unnamed: 0.1 project popularity ... date time 0 new
5 5 5 Math 148 ... 2022-07-20 16:30:20 3 145
6 6 6 English 146 ... 2022-06-23 15:30:20 5 141
7 7 7 Python 149 ... 2022-07-19 09:30:20 1 148
[3 rows x 9 columns]
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