Python 如何将文本文件加载为可以逐个元素加载的数组？

我有一个包含以下信息的文本文件：

SHRO
BJRD
ZNJK
GRMI

我想以一种可以逐行加载的方式保存它们。我的意思是，假设它的名字是“name”，我想要下面的第四列：

name[3] = "GRMI"

---

我该怎么做

您可以这样做：

with open('filename.txt') as f:
    name = f.read().split()

编辑
根据您发布的答案，我知道您只需在一行中打印姓名，因此，修改后的代码如下：

with open('filename.txt') as f:
    print f.read().split('\n')[12][:4]     # Assuming that all of the those consist of 4 characters.
                                           # That number 12 was just an example.

否则,

with open('filename.txt') as f:
    names = f.read().split('\n')
    print names[12].split()[0]            # Again, that number is an example.

---

你可以这样做：

with open('filename.txt') as f:
    name = f.read().split()

编辑
根据您发布的答案，我知道您只需在一行中打印姓名，因此，修改后的代码如下：

with open('filename.txt') as f:
    print f.read().split('\n')[12][:4]     # Assuming that all of the those consist of 4 characters.
                                           # That number 12 was just an example.

否则,

with open('filename.txt') as f:
    names = f.read().split('\n')
    print names[12].split()[0]            # Again, that number is an example.

---

你可能想要像这样的东西

with open('stationlist', 'r') as f:
    station_names = [line.split()[0] for line in f]

如果所有电台名称均为四个字符，则列表可替换为

[line[:4] for line in f]

---

你可能想要像这样的东西

with open('stationlist', 'r') as f:
    station_names = [line.split()[0] for line in f]

如果所有电台名称均为四个字符，则列表可替换为

[line[:4] for line in f]

---

逐行读取文件并在空白处拆分。显示到目前为止您已经完成的代码逐行读取文件并在空白处拆分。显示到目前为止您拥有的代码