Question

是否有办法识别MongoDB集合中的倒数第二个对象？

我得到了最后一个：

import pandas as pd
from collections import defaultdict
import random

n_keys = 200
values_per_key = 200
n_unique_values = 200
total_rows = n_keys * values_per_key

keys = [i//values_per_key for i in range(total_rows)]
values = [random.randint(0, n_unique_values-1) for i in range(total_rows)]
data = {'key': keys, 'value': values}
df = pd.DataFrame(data)

#df = pd.DataFrame({'key': [0, 0, 1, 1, 2, 2, 2],
# 'value': ['A', 'A', 'A', 'B', 'C', 'B', 'B']})

counts = defaultdict(list)
values = df['value'].value_counts().index
keys = sorted(df['key'].value_counts().index)

for key in keys:
    for value in values:
        # the following line makes this super slow
        ind = (df['key'] == key) & (df['value'] == value)
        counts[value].append(ind.sum())

print(pd.DataFrame(counts, index=keys))

但我怎样才能得到倒数第二个呢？

感谢您的帮助

Answer 1

您可以在此使用limit和skip

collection.find({}, {sort: {Date: -1}, limit:1}).fetch();将为您提供最后一条记录，

collection.find({}, {sort: {Date: -1}, limit:1, skip:1}).fetch();为您提供倒数第二个记录

得到倒数第二个元素MongoDB

1 个答案: