将数据帧列从字符串列表转换为元组

Question

我在列表中有网址。这是数据帧的一个元素。我需要将这些字符串列表中的每一个都转换成散列类型，比如元组。我已经读过带逗号的元组（a，）在转换时保留了列表中的字符串。当应用于整个dataframe列时，我似乎无法使其工作。可能遗漏了一些简单的东西

df['url'] = tuple(df['url',])…不起作用

flatframe['url'] = flatframe['url'].apply(tuple)…工作，但不保留字符串

以下是几行数据：

index artist ranking song songurl songtext artisturl year

2280    (Lady Antebellum,)  81  [Bartender (Lady Antebellum song)]  [/wiki/Bartender_(Lady_Antebellum_song)]    "Bartender (Lady Antebellum song)"  /wiki/Lady_Antebellum   2014

2281    (Naughty Boy, Sam Smith)    82  [La La La (Naughty Boy song)]   [/wiki/La_La_La_(Naughty_Boy_song)] "La La La (Naughty Boy song)"   [/wiki/Naughty_Boy, /wiki/Sam_Smith_(singer)]   2014

2282    (Robin Thicke, T.I., Pharrell Williams) 83  [Blurred Lines] [/wiki/Blurred_Lines]   "Blurred Lines" [/wiki/Robin_Thicke, /wiki/T.I., /wiki/Pharrel...   2014

2283    (Lady Gaga, R. Kelly)   84  [Do What U Want]    [/wiki/Do_What_U_Want]  "Do What U Want"    [/wiki/Lady_Gaga, /wiki/R._Kelly]   2014

Answer 1

假设你的数据帧是这样的：

import pandas as pd
pd.set_printoptions(max_columns=10)
df = pd.DataFrame(
[[2280, ("Lady Antebellum"),  81,  ["Bartender (Lady Antebellum song)"],  ["/wiki/Bartender_(Lady_Antebellum_song)"],    "Bartender (Lady Antebellum song)",  "/wiki/Lady_Antebellum",  2014],
[2281, "(Naughty Boy, Sam Smith)",    82,  ["La La La (Naughty Boy song)"],   ["/wiki/La_La_La_(Naughty_Boy_song)"], "La La La (Naughty Boy song)",   ["/wiki/Naughty_Boy", "/wiki/Sam_Smith_(singer)"],   2014],
[2282, "(Robin Thicke, T.I., Pharrell Williams)", 83,  ["Blurred Lines"], ["/wiki/Blurred_Lines"],   "Blurred Lines", ["/wiki/Robin_Thicke", "/wiki/T.I. /wiki/Pharrel"],   2014],
[2283, "(Lady Gaga, R. Kelly)",   84,  ["Do What U Want"],    ["/wiki/Do_What_U_Want"],  "Do What U Want",    ["/wiki/Lady_Gaga", "/wiki/R._Kelly"],   2014]],
columns = ["index", "artist", "ranking", "song", "songurl", "songtext", "artisturl", "year"])

然后您可以尝试使用：

^{pr2}$

这将只对非字符串的项应用元组，然后转换为列表项，然后转换为字符串的元组项。就好像它是一个字符串，而你应用元组，它将给出一个元组，每个字符都作为条目。在

然后，您的列artisturl将显示为：

>>> df.artisturl
0                           ('/wiki/Lady_Antebellum',)
1    ('/wiki/Naughty_Boy', '/wiki/Sam_Smith_(singer)')
2    ('/wiki/Robin_Thicke', '/wiki/T.I. /wiki/Pharr...
3                ('/wiki/Lady_Gaga', '/wiki/R._Kelly')
Name: artisturl