【发布时间】:2021-12-30 11:11:36
【问题描述】:
我认为我对 pandas 数据框有很好的掌握,但由于某种原因,我可以将此字典转换为数据框 - 我将发布带有常量变量的完整代码,以防有人想单独标记:
import pandas as pd
import numpy as np
patches = 26
min_sl = 2.1
max_sl = 104.1
array = np.arange(min_sl, max_sl + 1, step=1)
##Creating an array List
array_list = [i for i in array]
## Creating a dictionary
dict_l={i:"" for i in range(1,patches+1)}
## Populating the dictionary
start = 0
end = 4
for i in dict_l:
dict_l[i] = array_list[start:end]
start = start + 5
end = end + 5
这将创建以下字典
{1: [2.1, 3.1, 4.1, 5.1],
2: [7.1, 8.1, 9.1, 10.1],
3: [12.1, 13.1, 14.1, 15.1],
4: [17.1, 18.1, 19.1, 20.1],
5: [22.1, 23.1, 24.1, 25.1],
6: [27.1, 28.1, 29.1, 30.1],
7: [32.1, 33.1, 34.1, 35.1],
8: [37.1, 38.1, 39.1, 40.1],
9: [42.1, 43.1, 44.1, 45.1],
10: [47.1, 48.1, 49.1, 50.1],
11: [52.1, 53.1, 54.1, 55.1],
12: [57.1, 58.1, 59.1, 60.1],
13: [62.1, 63.1, 64.1, 65.1],
14: [67.1, 68.1, 69.1, 70.1],
15: [72.1, 73.1, 74.1, 75.1],
16: [77.1, 78.1, 79.1, 80.1],
17: [82.1, 83.1, 84.1, 85.1],
18: [87.1, 88.1, 89.1, 90.1],
19: [92.1, 93.1, 94.1, 95.1],
20: [97.1, 98.1, 99.1, 100.1],
21: [102.1, 103.1, 104.1],
22: [],
23: [],
24: [],
25: [],
26: []}
我想要的是一个简单的数据框,它将每个列表放入一行,例如索引 1 将有一个包含以下值的列表:[2.1, 3.1, 4.1, 5.1],
result = pd.DataFrame.from_dict(dict_l, orient='index')
result
如您所见,我的结果不是我想要的,因为我只想要一列
如果我没有指定函数 from_dict,我会收到一个错误,所以我想知道是否有人知道如何做到这一点..
我试图用逗号连接所有列,这在理论上是我想做的:
result['Line Name'] = result[[i for i in range(0,4,1)]].apply(lambda x: ', '.join(x[x.notnull()]),轴=1)
但是,我收到类型错误 TypeError: sequence item 0: expected str instance, float found
有谁知道如何做到这一点?
【问题讨论】:
标签: python pandas dataframe dictionary