Python3 在这个数组中搜索值

Question

我仍在学习python3，请就此提出建议。 我有一个很长的数组，如下所示，我如何检查任何数组元素中是否存在两个这个值（日期在第 4 个位置，字符串在数组的第二个位置）。

数组：

[
('1','200','300','500','2015-04-25 7:00:00'),
('1','200','500','500','2015-04-26 8:00:00'),
('1','200','500','500','2015-04-26 8:00:00'), # Repeated
('1','200','900','500','2015-04-27 9:00:00'),
('1','200','300','500','2015-04-28 17:00:00'),
('1','200','300','500','2015-04-28 17:00:00'), # Repeated
...
...
]

Answer 1

一些不需要使用外部库的方法是：

long_array = [
    ('1','200','300','500','2015-04-25 7:00:00'),
    ('1','200','500','500','2015-04-26 8:00:00'),
    ('1','200','500','500','2015-04-26 8:00:00'), # Repeated
    ('1','200','900','500','2015-04-27 9:00:00'),
    ('1','200','300','500','2015-04-28 17:00:00'),
    ('1','200','300','500','2015-04-28 17:00:00'), # Repeated
    # ...
]

使用一组..

values = set()
for entry in long_array:    
    value = (entry[1], entry[4])
    if (value in values): 
        print("Duplicate " + str(entry))
    else:
        values.add(value)

或使用收集计数器..

from collections import Counter

values = Counter([(entry[1], entry[4]) for entry in long_array])
for value, count in values.items():
    if count > 1:
        print(str(count) + " duplicates of " + str(value))

数组的大小在这里非常重要。这些可能会导致非常大的数组出现问题。

Answer 2

我建议使用熊猫。假设您的数组（实际上在 Python 中称为 list）称为 A，您可以加载它

import pandas as pd
df = pd.DataFrame(A)
df
   0    1    2    3                    4
0  1  200  300  500   2015-04-25 7:00:00
1  1  200  500  500   2015-04-26 8:00:00
2  1  200  500  500   2015-04-26 8:00:00
3  1  200  900  500   2015-04-27 9:00:00
4  1  200  300  500  2015-04-28 17:00:00

然后你可以像这样得到重复的行

df['Repeated'] = df.duplicated(subset=[3,4])
df

Out[463]: 
   0    1    2    3                    4 Repeated
0  1  200  300  500   2015-04-25 7:00:00    False
1  1  200  500  500   2015-04-26 8:00:00    False
2  1  200  500  500   2015-04-26 8:00:00     True
3  1  200  900  500   2015-04-27 9:00:00    False
4  1  200  300  500  2015-04-28 17:00:00    False

Answer 3

如果您想在 Python 中实际编写解决方案以进行实践，这里有一种方法：

# the indices in the tuples to be used as keys for determining repeats
# set this to whatever indices you would like (or even all of them)!
key_indices = [1, 4]

# for a given tuple tpl, construct a key consisting of the values in tpl
# that are found at the indices given in ki
def make_key(tpl, ki):
    key_elements = []
    for i in ki:
        key_elements.append(tpl[i])

    # need to return a tuple, as you cannot use a list as a key for a dict
    return tuple(key_elements)

data = [
('1','200','300','500','2015-04-25 7:00:00'),
('1','200','500','500','2015-04-26 8:00:00'),
('1','200','500','500','2015-04-26 8:00:00'), # Repeated
('1','200','900','500','2015-04-27 9:00:00'),
('1','200','300','500','2015-04-28 17:00:00'),
('1','200','300','500','2015-04-28 17:00:00') # Repeated
]

# the data structure that we'll use to remember where we've seen keys before
memory = dict()
duplicates = set()

for i in range(0, len(data)):
    # make the key for comparison
    k = make_key(data[i], key_indices)

    # find out where we've seen this before
    # if nowhere else, return an empty list
    previous_locations = memory.get(k, [])

    # note that we have now seen this key at location i
    previous_locations.append(i)

    if (len(previous_locations) > 1):
        duplicates.add(i)

    # update the dict with the new location
    memory[k] = previous_locations

print("Duplicate values found at: {}".format(list(duplicates)))


# and if you want to know which keys were duplicated where?
for k in memory.keys():
    locs = memory[k]
    if len(locs) > 1:
        print("{}: {}".format(k, locs))

输出：

Duplicate values found at: [2, 5]
('200', '2015-04-28 17:00:00'): [4, 5]
('200', '2015-04-26 8:00:00'): [1, 2]