python提取字符串中的数字的实现

dream = "123456"
print(dream.isdigit())  # 返回：True

dream = "123abc456"
print(dream.isdigit())  # 返回：False

dream = 'abcd'
print(dream.isdigit())  # 返回：False

# '120m' 拆为 '120'和'm'，不加key=lambda x: x.isdigit()的话，会拆成['1', '2', '0', 'm']
from itertools import groupby
str = '120m'
ret = [''.join(list(g)) for k, g in groupby(str, key=lambda x: x.isdigit())]
# 输出['120', 'm']
[k for k, g in groupby('150m', lambda x: x.isdigit())]
Out[8]: [True, False]
[k for k, g in groupby('150m')]
Out[9]: ['1', '5', '0', 'm']

[g for k, g in groupby('150m', lambda x: x.isdigit())]
Out[10]: [<itertools._grouper at 0x214eeb877f0>, <itertools._grouper at 0x214eeb87820>]

[list(g) for k, g in groupby('150m', lambda x: x.isdigit())]
Out[11]: [['1', '5', '0'], ['m']]

filter(function, iterable)

def is_odd(n):
    return n % 2 == 1
 
tmplist = filter(is_odd, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
newlist = list(tmplist)
print(newlist)

l = [x for x in range(10)]
print(list(filter(lambda x : x%2 == 0, l)))

import math
def is_sqr(x):
    return math.sqrt(x) % 1 == 0
 
tmplist = filter(is_sqr, range(1, 101))
newlist = list(tmplist)
print(newlist)

# [1, 3, 5, 7, 9]

L = range(1, 21)

def isprimer(n):
    flag = 1
    for i in range(2, n):
        if n % i == 0:
            flag = 0
    if flag == 0:
        return n

print(list(filter(isprimer, L)))
# [4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20]

def not_empty(s):
  return s and s.strip()

filter(not_empty, ['A', '', 'B', None, 'C', ' '])

def _odd_iter():
    n = 1
    while True:
        n = n + 2
        yield n
        
def _not_divisible(n): 
    return lambda x : x%n>0
 
def primes():
    yield 2
    it = _odd_iter()
    ftr = filter(_not_divisible(2), it) #1
    while True:
        n = next(ftr )        #2
        yield n                
        ftr = filter(_not_divisible(n), ftr ) #3
        
for n in primes():
    if n < 20:
        print('now:',n)
    else:
        break
输出：
now: 2
now: 3
now: 5
now: 7
now: 11
now: 13
now: 17
now: 19

number = ['12', '333', '4']
number_ = "".join(number)    # 列表转字符串
print(number_)    # 123334

a = "".join(list(filter(str.isdigit, '123ab45')))
print(a)
# 返回12345

b = list(filter(str.isdigit, '123ab45'))
print(b)
# 返回['1', '2', '3', '4', '5']

time_ = "2019年09月04日 11:00"
time_filter = filter(str.isdigit, time_)

print(time_filter)           # <filter object at 0x0000019358731BE0>
print(type(time_filter))     # <class 'filter'>
time_list = list(time_filter)       # ['2', '0', '1', '9', '0', '9', '0', '4', '1', '1', '0', '0']
time_str = "".join(time_list)       # 转为str    201909041100
time_int = int(time_str)            # 转为int    201909041100

import re
str_ = "12今天333天气4不错"
number = re.findall("\d+",str_)    # 输出结果为列表
print(number)
 
# 输出结果：['12', '333', '4']

# 匹配“calibration=”后面的数字
pattern = re.compile(r'(?<=calibration=)\d+\.?\d*')
pattern.findall(string)

# ['0.9863265752792358']