Series对象没有split属性 - 从文本文件中读取数据
我正在读取来自.txt文件的数据，如下所示，我正在使用底部的代码提取每个括号中的第三项。
使用行data = data.iloc[0, ::4]，我每4个数据点中就删除了3个（我认为）。
但是，现在我试图保留每个数据点。所以当我删除线; data = data.iloc[0, ::4]，我收到以下错误：
"'Series' object has no attribute 'split'", 'occurred at index 0'
我不明白为什么删除该行会给我收到的错误。简而言之，我想从每个括号中提取每三个项目，并将其存储在pandas DataFrame中。但是当我尝试删除行data = data.iloc[0, ::4]时，我会抛出错误（如上所示）。
感谢您的支持！
data = pd.read_csv("C:\\Users\\Desktop\\data_test9.txt", sep="\[|\]\[|\]",engine = 'python', header = None)
data = data.iloc[0, ::4]
data = data.astype(str).apply(lambda x: x.split(',')[-1]).astype(float)
data.pop(0)
xar = range(len(data)
yar = pd.DataFrame(data)
数据：
[0.0, -0.01953125, -0.0009765625][0.0, -0.01953125, -0.0009765625][0.0, -0.01953125, -0.0009765625][0.0, -0.01953125, -0.0009765625][0.0078125, -0.912109375, -0.00390625][0.0078125, -0.912109375, -0.00390625][0.0078125, -0.912109375, -0.00390625][0.0078125, -0.912109375, -0.00390625][0.0078125, -0.912109375, -0.00390625][0.0078125, -0.912109375, -0.00390625][0.0078125, -0.912109375, -0.00390625][0.0078125, -0.912109375, -0.00390625][0.0087890625, -0.978515625, -0.00390625][0.0087890625, -0.978515625, -0.00390625][0.0087890625, -0.978515625, -0.00390625][0.0087890625, -0.978515625, -0.00390625][0.0087890625, -0.978515625, -0.00390625][0.0078125, -0.9521484375, -0.0029296875][0.0078125, -0.9521484375, -0.0029296875][0.0078125, -0.9521484375, -0.0029296875][0.0078125, -0.9521484375, -0.0029296875][0.0078125, -0.9521484375, -0.0029296875][0.0078125, -0.9521484375, -0.0029296875][0.0078125, -0.9521484375, -0.0029296875][0.0078125, -0.9521484375, -0.0029296875][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.955078125, -0.00390625][0.0078125, -0.955078125, -0.00390625][0.0078125, -0.955078125, -0.00390625][0.0078125, -0.955078125, -0.00390625][0.0078125, -0.955078125, -0.00390625][0.0078125, -0.955078125, -0.00390625][0.0078125, -0.955078125, -0.00390625][0.0078125, -0.955078125, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0048828125][0.0078125, -0.9560546875, -0.0048828125][0.0078125, -0.9560546875, -0.0048828125][0.0078125, -0.9560546875, -0.0048828125][0.0078125, -0.9560546875, -0.0048828125][0.0078125, -0.9560546875, -0.0048828125][0.0078125, -0.9560546875, -0.0048828125][0.0078125, -0.9560546875, -0.0048828125][0.0087890625, -0.9560546875, -0.0029296875][0.0087890625, -0.9560546875, -0.0029296875][0.0087890625, -0.9560546875, -0.0029296875][0.0087890625, -0.9560546875, -0.0029296875][0.0087890625, -0.9560546875, -0.0029296875][0.0087890625, -0.9560546875, -0.0029296875][0.0087890625, -0.9560546875, -0.0029296875][0.0087890625, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0078125, -0.9560546875, -0.0029296875][0.0068359375, -0.9560546875, -0.00390625][0.0068359375, -0.9560546875, -0.00390625][0.0068359375, -0.9560546875, -0.00390625][0.0068359375, -0.9560546875, -0.00390625][0.0068359375, -0.9560546875, -0.00390625][0.0068359375, -0.9560546875, -0.00390625][0.0068359375, -0.9560546875, -0.00390625][0.0068359375, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625][0.0078125, -0.9560546875, -0.00390625]
编辑1： 我的数据保存如下：
while True:
delay(DELAY_INTERVAL)
file = open("C:\\Users\\Desktop\\demo3.txt", "a")
file.writelines(str(convert(readField())))
file.flush()
file.close()
1 个答案:
答案 0 :(得分：1)
因为data是pd.DataFrame，所以当你.apply函数时，它会将pd.Series作为参数传递给函数。 pd.Series没有.split方法。当你这样做时：
data = data.iloc[0, ::4]
您正在选择第一行和每四列，从第一行开始。这*返回pd.Series。所以考虑一下：
In [16]: df
Out[16]:
a b c d e
0 -10 17 17 14 -10
1 2 10 0 4 -5
2 16 10 1 -9 -6
3 12 -6 10 5 3
4 9 10 -4 10 3
5 -8 16 4 17 0
6 9 12 4 17 -5
7 19 17 9 2 19
8 1 4 10 10 12
9 12 -1 9 -5 10
10 -1 -3 4 11 5
11 19 4 6 16 3
12 5 19 -1 1 15
13 18 2 15 0 7
14 3 10 15 6 15
15 8 12 3 -6 8
16 14 0 -3 6 2
17 -1 -5 -5 -3 16
18 -2 12 17 17 13
19 18 4 19 14 4
In [17]: df.iloc[0, ::4]
Out[17]:
a -10
e -10
Name: 0, dtype: int64
In [18]: type(df.iloc[0, ::4])
Out[18]: pandas.core.series.Series
现在，当你.apply 系列的函数时，它将系列中的每个值传递给函数，这就是你的{{1}因为每个值都是lambda，而字符串确实有 str方法。
所以，也许你想要：
.split选择每四行，并返回一个数据框。
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