我有一組數據，我根據月份對它們進行了分組，我想得到一個嵌套的字典，并找到每年的中位數......我該如何編寫代碼？我剛剛開始使用Python，所以我希望代碼可以清晰易懂。dictionary = {1: [ [2019, 28.7], [2019, 29.4], [2019, 29.7], [2020, 28.6], [2020, 28.7], [2020, 28.0] ], 2:[ [2018, 26.9], [2018, 28.2], [2018, 28.5], [2019, 29.9], [2019, 29.8], [2019, 28.7] ]}desired output = {1: {2019: 29.4, 2020: 28.7}, 2: {2018: 28.2, 2019: 29.8}} 非常感謝所有做出貢獻的人！我已經解決了這個問題。
2 回答
蝴蝶不菲
TA貢獻1810條經驗 獲得超4個贊
您可以使用字典理解：itertools.groupby
from itertools import groupby
from statistics import median
{k : {y : median((e[1] for e in g)) for y, g in groupby(v, key=lambda x: x[0])} for k, v in dictionary.items()}
輸出：
{1: {2019: 29.4, 2020: 28.6}, 2: {2018: 28.2, 2019: 29.8}}
上述解決方案有效，因為您的數據是按年份排序的
或者你可以使用 2 個帶有集合的循環：for
from collections import defaultdict
output = {}
for k, v in dictionary.items():
year_value = defaultdict(list)
for y, val in v:
year_value[y].append(val)
output[k] = {y: median(v) for y, v in year_value.items()}
print(output)
輸出：
{1: {2019: 29.4, 2020: 28.6}, 2: {2018: 28.2, 2019: 29.8}}
慕標5832272
TA貢獻1966條經驗 獲得超4個贊
請嘗試以下操作：
dictionary = {1: [[2019, 28.7], [2019, 29.4], [2019, 29.7], [2020, 28.6], [2020, 28.7], [2020, 28.0]],
2: [[2018, 26.9], [2018, 28.2], [2018, 28.5], [2019, 29.9], [2019, 29.8], [2019, 28.7]]}
output = {}
for obj in dictionary.keys():
temp = {}
for i in dictionary[obj]:
if i[0] not in temp:
temp[i[0]] = [i[1]]
else:
temp[i[0]].append(i[1])
output[obj] = temp
for i in output.keys():
for j in output[i].keys():
output[i][j].sort()
n = len(output[i][j])
if n % 2 == 0:
median_1 = output[i][j][n // 2]
median_2 = output[i][j][n // 2 - 1]
median = (median_1 + median_2) // 2
else:
median = output[i][j][n // 2]
output[i][j] = median
print(output)
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