我有一个元组列表，如下所示。我必须计算有多少项的数字大于1.我到目前为止编写的代码非常慢。即使有大约10K元组，如果你看到下面的例子字符串出现两次，所以我必须得到这样的字符串。我的问题是通过迭代生成器来实现字符串计数的最佳方法是什么
列表：
b_data=[('example',123),('example-one',456),('example',987),.....]
到目前为止我的代码：
blockslst=[]
for line in b_data:
blockslst.append(line[0])
blocklstgtone=[]
for item in blockslst:
if(blockslst.count(item)>1):
blocklstgtone.append(item)
答案 0 :(得分：13)
你有正确的想法从每个元组中提取第一个项目。您可以使用列表/生成器理解使代码更简洁，如下所示。
从那时起，查找元素频率计数的最惯用方式是使用collections.Counter对象。
Counter example from collections import Counter
counts = Counter(x[0] for x in b_data)
print(counts['example'])
当然，您可以使用list.count，如果它只是一个项目，您想要查找频率计数，但在一般情况下，Counter是要走的路
Counter的优势在于它在线性（example）时间内执行所有元素（不仅仅是O(N)）的频率计数。假设您还想查询另一个元素的计数，比如说foo。这将用 -
print(counts['foo'])
如果列表中不存在'foo'，则会返回0。
如果您想找到最常见的元素，请致电counts.most_common -
print(counts.most_common(n))
其中n是您要显示的元素数。如果您想查看所有内容，请不要通过n。
要检索大多数常见元素的计数，一种有效的方法是查询most_common，然后使用itertools有效地提取计数超过1的所有元素。
from itertools import takewhile
l = [1, 1, 2, 2, 3, 3, 1, 1, 5, 4, 6, 7, 7, 8, 3, 3, 2, 1]
c = Counter(l)
list(takewhile(lambda x: x[-1] > 1, c.most_common()))
[(1, 5), (3, 4), (2, 3), (7, 2)]
（OP编辑）或者，使用列表理解来获取具有计数＆gt;的项目列表。 1 -
[item[0] for item in counts.most_common() if item[-1] > 1]
请注意，这不如itertools.takewhile解决方案有效。例如，如果您有一个带有count＆gt;的项目如果你不需要（因为most_common按降序返回频率计数），你最终会在列表中迭代一百万次和一次。由于takewhile不是这种情况，因为只要计数条件＆gt;您就停止迭代。 1变得虚假。
答案 1 :(得分：2)
第一种方法：
没有循环怎么样？
print(list(map(lambda x:x[0],b_data)).count('example'))
输出：
2
第二种方法：
您可以使用简单的dict计算，无需导入任何外部模块或不使其如此复杂：
b_data = [('example', 123), ('example-one', 456), ('example', 987)]
dict_1={}
for i in b_data:
if i[0] not in dict_1:
dict_1[i[0]]=1
else:
dict_1[i[0]]+=1
print(dict_1)
print(list(filter(lambda y:y!=None,(map(lambda x:(x,dict_1.get(x)) if dict_1.get(x)>1 else None,dict_1.keys())))))
输出：
[('example', 2)]
Test_case：
b_data = [('example', 123), ('example-one', 456), ('example', 987),('example-one', 456),('example-one', 456),('example-two', 456),('example-two', 456),('example-two', 456),('example-two', 456)]
输出：
[('example-two', 4), ('example-one', 3), ('example', 2)]
答案 2 :(得分：2)
我花了这么多时间 ayodhyankit-paul 发布相同的 - 保留生成器代码 对于测试用例和时间安排：
创建 100001 项目大约需要5秒钟，计算大约 0.3s ， 对计数进行过滤太快而无法衡量（使用datetime.now（） - 没有使用perf_counter） - 总而言之，从开始到结束大约需要少于5.1s 对您操作的数据进行计时。
我认为这与COLDSPEED s answer中的Counter类似：
item中的foreach list of tuples：
item[0]不在列表中，请dict加入count of 1 increment count 中by 1
代码：
from collections import Counter
import random
from datetime import datetime # good enough for a loong running op
dt_datagen = datetime.now()
numberOfKeys = 100000
# basis for testdata
textData = ["example", "pose", "text","someone"]
numData = [random.randint(100,1000) for _ in range(1,10)] # irrelevant
# create random testdata from above lists
tData = [(random.choice(textData)+str(a%10),random.choice(numData)) for a in range(numberOfKeys)]
tData.append(("aaa",99))
dt_dictioning = datetime.now()
# create a dict
countEm = {}
# put all your data into dict, counting them
for p in tData:
if p[0] in countEm:
countEm[p[0]] += 1
else:
countEm[p[0]] = 1
dt_filtering = datetime.now()
#comparison result-wise (commented out)
#counts = Counter(x[0] for x in tData)
#for c in sorted(counts):
# print(c, " = ", counts[c])
#print()
# output dict if count > 1
subList = [x for x in countEm if countEm[x] > 1] # without "aaa"
dt_printing = datetime.now()
for c in sorted(subList):
if (countEm[c] > 1):
print(c, " = ", countEm[c])
dt_end = datetime.now()
print( "\n\nCreating ", len(tData) , " testdataitems took:\t", (dt_dictioning-dt_datagen).total_seconds(), " seconds")
print( "Putting them into dictionary took \t", (dt_filtering-dt_dictioning).total_seconds(), " seconds")
print( "Filtering donw to those > 1 hits took \t", (dt_printing-dt_filtering).total_seconds(), " seconds")
print( "Printing all the items left took \t", (dt_end-dt_printing).total_seconds(), " seconds")
print( "\nTotal time: \t", (dt_end- dt_datagen).total_seconds(), " seconds" )
输出：
# reformatted for bevity
example0 = 2520 example1 = 2535 example2 = 2415
example3 = 2511 example4 = 2511 example5 = 2444
example6 = 2517 example7 = 2467 example8 = 2482
example9 = 2501
pose0 = 2528 pose1 = 2449 pose2 = 2520
pose3 = 2503 pose4 = 2531 pose5 = 2546
pose6 = 2511 pose7 = 2452 pose8 = 2538
pose9 = 2554
someone0 = 2498 someone1 = 2521 someone2 = 2527
someone3 = 2456 someone4 = 2399 someone5 = 2487
someone6 = 2463 someone7 = 2589 someone8 = 2404
someone9 = 2543
text0 = 2454 text1 = 2495 text2 = 2538
text3 = 2530 text4 = 2559 text5 = 2523
text6 = 2509 text7 = 2492 text8 = 2576
text9 = 2402
Creating 100001 testdataitems took: 4.728604 seconds
Putting them into dictionary took 0.273245 seconds
Filtering donw to those > 1 hits took 0.0 seconds
Printing all the items left took 0.031234 seconds
Total time: 5.033083 seconds
答案 3 :(得分：0)
让我举一个让你理解的例子。虽然这个例子与你的例子非常不同，但我发现在解决这些问题时它非常有用。
from collections import Counter
a = [
(0, "Hadoop"), (0, "Big Data"), (0, "HBase"), (0, "Java"),
(1, "Postgres"), (2, "Python"), (2, "scikit-learn"), (2, "scipy"),
(2, "numpy"), (2, "statsmodels"), (2, "pandas"), (3, "R"), (3, "Python"),
(3, "statistics"), (3, "regression"), (3, "probability"),
(4, "machine learning"), (4, "regression"), (4, "decision trees"),
(4, "libsvm"), (5, "Python"), (5, "R"), (5, "Java"), (5, "C++"),
(5, "Haskell"), (5, "programming languages"), (6, "statistics"),
(6, "probability"), (6, "mathematics"), (6, "theory"),
(7, "machine learning"), (7, "scikit-learn"), (7, "Mahout"),
(7, "neural networks"), (8, "neural networks"), (8, "deep learning"),
(8, "Big Data"), (8, "artificial intelligence"), (9, "Hadoop"),
(9, "Java"), (9, "MapReduce"), (9, "Big Data")
]
#
# 1. Lowercase everything
# 2. Split it into words.
# 3. Count the results.
dictionary = Counter(word for i, j in a for word in j.lower().split())
print(dictionary)
# print out every words if the count > 1
[print(word, count) for word, count in dictionary.most_common() if count > 1]
现在这是以上述方式解决的例子
from collections import Counter
a=[('example',123),('example-one',456),('example',987),('example2',987),('example3',987)]
dict = Counter(word for i,j in a for word in i.lower().split() )
print(dict)
[print(word ,count) for word,count in dict.most_common() if count > 1 ]