在python list中筛选包含字符的字段方式

l = [‘123a',‘456b',‘789c']

ll = [s for s in l if ‘a' in s]

lst = [“123a”, “456b”, “789c”]
lst = list(filter(lambda x: x.find(“a”) >= 0, lst))
print(lst)

# 构造筛选目标列表,确保KEY不重复
n1 = 30000
n1_set = set([random.randint(1,n1)  for n in range(n1)])
n1 = len(n1_set)
list1 = [['1108{:0>6d}27'.format(n), "".join(random.sample('ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz',10))] for n in n1_set]
 
# 构造待筛选数据列表,确保KEY不重复
n2 = 100000
n2_set = set([random.randint(1,n2)  for n in range(n2)])
n2= len(n2_set)
list2 = [['1108{:0>6d}27'.format(n), "".join(random.sample('ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz',5)),n ] for n in n2_set]

筛选目标18971条, 待处理数据63275条
遍历生成数据 耗时11.591秒 获得数据量 12024
['11080000427', 'eArVD', 4]
filter 耗时11.5秒 获得数据量 12024
['11080000427', 'eArVD', 4]
merge 耗时0.058秒 获得数据量 12024
['11080000427', 'eArVD', 4]

筛选目标189733条, 待处理数据632363条
遍历生成数据 耗时1597.4296秒 获得数据量 120180
['1108000000227', 'NkoEQ', 2]
filter 耗时1575.0432秒 获得数据量 120180
['1108000000227', 'NkoEQ', 2]
merge 耗时0.64秒 获得数据量 120180
['1108000000227', 'NkoEQ', 2]

print("筛选目标{}条, 待处理数据{}条".format(n1,n2))
 
# 直接遍历生成数据，计时
t1 = time.time()
list_temp = [n[0] for n in list1]
list3 = []
for n in list2:
    if n[0] in list_temp:
        list3.append(n)
t2 = time.time()
print("遍历生成数据 耗时{}秒".format(round(t2 - t1, 4)), "获得数据量", len(list3))
print(list3[0]) 
 
 
# 用filter筛选数据，计时
t1 = time.time()
list_temp = [n[0] for n in list1]
list3 = [n for n in filter(lambda x: x[0] in list_temp, list2)]
t2 = time.time()
print("filter 耗时{}秒".format(round(t2 - t1,4)), "获得数据量", len(list3))
print(list3[0])
 
 
# 用pd.merge 筛选数据，计时
t1 = time.time()
df1 = pd.DataFrame(list1, columns=['k1','m1'])
df2 = pd.DataFrame(list2, columns=['k1','m2','n2'])
df3 = pd.merge(df1[['k1']], df2, how='inner', on='k1')
t2 = time.time()
print("merge 耗时{}秒".format(round(t2 - t1,4)), "获得数据量", len(df3))
print(list(df3.iloc[0]))