我有一个这种格式的文本文件。
the 0.418 0.24968 -0.41242 0.1217 0.34527 -0.044457 -0.49688 -0.17862 -0.00066023 -0.6566 0.27843 -0.14767 -0.55677 0.14658 -0.0095095 0.011658
and 0.26818 0.14346 -0.27877 0.016257 0.11384 0.69923 -0.51332 -0.47368 -0.33075 -0.13834 0.2702 0.30938 -0.45012 -0.4127 -0.09932 0.038085
我想把它读成python dict变量,以便我有
{'the': [0.418, 0.24968,..], 'and': [0.26818 0.14346,..]}
我不确定如何开始?
答案 0 :(得分:1)
您可以逐行遍历文件。然后在空格上split
,使用索引[0]
作为键,然后使用列表推导将剩余值转换为float
列表。
with open(text_file) as f:
d = dict()
for line in f:
data = line.split()
d[data[0]] = [float(i) for i in data[1:]]
print(d)
输出
{'and': [0.26818, 0.14346, -0.27877, 0.016257, 0.11384, 0.69923, -0.51332, -0.47368, -0.33075, -0.13834, 0.2702, 0.30938, -0.45012, -0.4127, -0.09932, 0.038085],
'the': [0.418, 0.24968, -0.41242, 0.1217, 0.34527, -0.044457, -0.49688, -0.17862, -0.00066023, -0.6566, 0.27843, -0.14767, -0.55677, 0.14658, -0.0095095, 0.011658]}
答案 1 :(得分:1)
这是一个有效的解决方案:
my_dict = {}
with open('your_file_name') as f:
for line in f:
elements = line.split()
my_dict[elements[0]] = [float(e) for e in elements[1:]]
请注意@CoryKramer建议的解决方案使用readlines
,它返回一个列表而不是迭代器。所以我不建议将他的解决方案用于大文件(这会占用太多内存)。
答案 2 :(得分:0)
这是一个开始吗?
import json
variable_one = "the 0.418 0.24968 -0.41242 0.1217 0.34527 -0.044457 -0.49688 -0.17862 -0.00066023 -0.6566 0.27843 -0.14767 -0.55677 0.14658 -0.0095095 0.011658"
variable_one = variable_one.split()
json_variable_one = []
json_variable_one = variable_one[0], variable_one[1:]
print(json.dumps(json_variable_one))
输出:
[“the”,[“0.418”,“0.24968”,“ - 0.41242”,“0.1217”,“0.34527”,“ - 0.044457”,“ - 0.49688”,“ - 0.17862”,“ - 0.00066023”, “-0.6566”,“0.27843”,“ - 0.14767”,“ - 0.55677”,“0.14658”,“ - 0.0095095”,“0.011658”]