isdigit() 函数是检测输入字符串是否只由数字组成。如果字符串只包含数字则返回 True 否则返回 False。
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']]
说明:filter() 函数用于过滤序列,过滤掉不符合条件的元素,返回一个迭代器对象;
如果要转换为列表,可以使用 list() 来转换。
该接收两个参数,第一个为函数,第二个为序列,序列的每个元素作为参数传递给函数进行判断,然后返回 True 或 False,最后将返回 True 的元素放到新列表中。
语法:
filter(function, iterable)
1、过滤出列表中的所有奇数:
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)
2、过滤出列表中的所有偶数:
l = [x for x in range(10)]
print(list(filter(lambda x : x%2 == 0, l)))
3、过滤出1~100中平方根是整数的数:
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]
4、删除1-20中素数
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]
5、去除空格和空值
def not_empty(s):
return s and s.strip()
filter(not_empty, ['A', '', 'B', None, 'C', ' '])
6、高阶运用
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
列表转字符串
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) #
print(type(time_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
利用正则表达式
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']
例如下面的string:
tensorflow:Final best valid 0 loss=0.20478513836860657 norm_loss=0.767241849151384 roc=0.8262403011322021 pr=0.39401692152023315 calibration=0.9863265752792358 rate=0.0
提取 calibration=0.9863265752792358 .
# 匹配“calibration=”后面的数字
pattern = re.compile(r'(?<=calibration=)\d+\.?\d*')
pattern.findall(string)
# ['0.9863265752792358']
[1] 如何利用python提取字符串中的数字 2022.1;