Python 如何根据条件拆分元组列表？

我有一个元组列表：

my_list = [(1,'a','b','c'), (2,'d','e','f'), (3,'g','h','i'), (1,'j','k','l'), (2,'m','n','o'), (1,'p','q','r'), (2,'s','t','u')]

我需要在元组的子列表上拆分它，从第一项为“1”的元组开始

[(1,'a','b','c'), (2,'d','e','f'), (3,'g','h','i')]
[(1,'j','k','l'), (2,'m','n','o')]
[(1,'p','q','r'), (2,'s','t','u')]

---

试试这个代码。我假设在再次找到第一个字符（1）时中断。我还假设输出是一个列表

my_list = [(1,'a','b','c'), (2,'d','e','f'), (3,'g','h','i'), (1,'j','k','l'), (2,'m','n','o'), (1,'p','q','r'), (2,'s','t','u')]

ch = my_list[0][0]
all = []
st = 0
for i, t in enumerate(my_list):
    if t[0] == ch:
       if i != 0: 
           all.append(my_list[st:i])
           st = i
else:
   all.append(my_list[st:i])
   
print(all)

输出

[
[(1, 'a', 'b', 'c'), (2, 'd', 'e', 'f'), (3, 'g', 'h', 'i')], 
[(1, 'j', 'k', 'l'), (2, 'm', 'n', 'o')], 
[(1, 'p', 'q', 'r')]
]

---

试试这个代码。我假设在再次找到第一个字符（1）时中断。我还假设输出是一个列表

my_list = [(1,'a','b','c'), (2,'d','e','f'), (3,'g','h','i'), (1,'j','k','l'), (2,'m','n','o'), (1,'p','q','r'), (2,'s','t','u')]

ch = my_list[0][0]
all = []
st = 0
for i, t in enumerate(my_list):
    if t[0] == ch:
       if i != 0: 
           all.append(my_list[st:i])
           st = i
else:
   all.append(my_list[st:i])
   
print(all)

输出

[
[(1, 'a', 'b', 'c'), (2, 'd', 'e', 'f'), (3, 'g', 'h', 'i')], 
[(1, 'j', 'k', 'l'), (2, 'm', 'n', 'o')], 
[(1, 'p', 'q', 'r')]
]

---

您实际上是在计算某种“groupwhile”函数——您希望在

1

中找到的每个元组处进行拆分。这看起来非常像

itertools.groupby

，如果我们保留一点点全局状态（示例中的

one\u count

变量），我们就可以重新使用语言中内置的分组/聚合逻辑来获得所需的结果

import itertools

# The inner function is just so that one_count will be initialized only
# as many times as we want to call this rather than exactly once via
# some kind of global variable.
def gen_count():
    def _cnt(t, one_count=[0]):
        if t[0] == 1:
            one_count[0] += 1
        return one_count[0]
    return _cnt

result = [list(g[1]) for g in itertools.groupby(my_list, key=gen_count())]

一个更传统的解决方案是迭代示例并将中间输出附加到结果集

result = []
for i, *x in my_list:
    if i==1:
        result.append([(i, *x)])
    else:
        result[-1].append((i, *x))

---

您实际上是在计算某种“groupwhile”函数——您希望在

1

中找到的每个元组处进行拆分。这看起来非常像

itertools.groupby

，如果我们保留一点点全局状态（示例中的

one\u count

变量），我们就可以重新使用语言中内置的分组/聚合逻辑来获得所需的结果

import itertools

# The inner function is just so that one_count will be initialized only
# as many times as we want to call this rather than exactly once via
# some kind of global variable.
def gen_count():
    def _cnt(t, one_count=[0]):
        if t[0] == 1:
            one_count[0] += 1
        return one_count[0]
    return _cnt

result = [list(g[1]) for g in itertools.groupby(my_list, key=gen_count())]

一个更传统的解决方案是迭代示例并将中间输出附加到结果集

result = []
for i, *x in my_list:
    if i==1:
        result.append([(i, *x)])
    else:
        result[-1].append((i, *x))