如何获得类似于＆＃34; findall＆＃34;在python的re模块中命令？

Question

正如我们所知，＆＃34;搜索＆＃34;命令返回一个匹配对象。

mystring= 'string with numbers 990'
my_obj = re.search('(.{1}[h0])',mystring)

我希望能够打印匹配字符串列表，类似于findall方法。我尝试my_obj.groups()，但这仍然只打印第一个匹配的对象：

findall_result = re.findall('(.{1}[h0])',mystring)

>>> findall_result
['th', '90']
>>> my_obj.groups()
('th',)
>>> 

如何从匹配对象（如findall）返回所有匹配的字符串？

Answer 1

You could use mo.group(), not groups. mo.group(1) with refer
 to the first bracketed group and so on. but mo.group(0) to 
 the whole string match.

s = "somethingabcdeabcdeabcdeabcdeabcdeelseabcdeabcdeabcde"

mo = re.search(r"(abc)d(ea)", s)

print(mo.group(0))
#abcdea
print(mo.group(1))
# abc
print(mo.group(2))
# ea

 mystring= 'string with numbers 990'

mo = re.search(r'(\wg\s).*\s(\d\d)\d',mystring)

print(mo.group(0))
# ng with numbers 990
print(mo.group(1))
# ng
print(mo.group(2))
# 99

Answer 2

您可以使用finditer：

print [m.group(0) for m in re.finditer('(.{1}[h0])',mystring)]