python数据框根据其他列条件替换列中的部分字符串

Question

Dataframe点击看截图，因为我是新来的，我需要10个声望才能嵌入图片

Expected result

数据框是从 csv 文件导入的。 “类型”和“主题”是项目的属性。 “标签”是一个长字符串列，包含每个项目的混合（随机排序）标签（由“，”分隔）。基本上我需要做的是检查'Tags'列中是否有正确的主题标签（col_{theme}），如果没有，将其添加到'Tags'列。

例如：

item 8: 'Tags' 列有一个'col_t3'，它的主题是't3'。所以这是正确的，我们通过了。

item 1: 'Tags' 列中有一个 'col_t1'，但它的实际主题是 't2'，所以我需要将 'col_t1' 替换为 'col_t2' 并保持同一列中的其他标签不变

item 2 和 item 5：'Tags' 列中没有 'col_{theme}' 标签，所以我分别在它们的 'Tags' 列中添加了 add 'col_t1' 和 'col_t5'。

请帮忙！！

Answer 1

这模拟了您在屏幕截图中显示的输入：

import pandas as pd
import numpy as np

df = pd.DataFrame({"type": ["a", "c", "d", "a", "b", "a", "a", "c"], 
                  "tags": ["col_t1, col_red, large", np.nan, "col_t2, col_black, small", 
                           "col_t4, large, col_yellow", "col_gold, col_fancy,", "col_t1, thick, col_k",
                          np.nan, "col_t3, fancy, red"],
                  "theme": ["t2", "t1", "t2", "t3", "t2", "t1", np.nan, "t3"]})

df.set_index(np.arange(1, len(df)+1), inplace=True)
print df

输出：

                      tags theme type
1     col_t1, col_red, large    t2    a
2                        NaN    t1    c
3   col_t2, col_black, small    t2    d
4  col_t4, large, col_yellow    t3    a
5       col_gold, col_fancy,    t2    b
6       col_t1, thick, col_k    t1    a
7                        NaN   NaN    a
8         col_t3, fancy, red    t3    c

产生所需输出的代码：

prefix = "col_"

# Iterate over rows with non-empty theme
for row in df[df["theme"].notnull()].itertuples():

    if pd.isnull(row.tags):
        # Replace NaN in tags column with a single tag from theme column 
        df.loc[row.Index, "tags"] = prefix + row.theme
    else:
        # Extract existing tags with prefix
        inferred_tags = [t.replace(prefix, "") for t in row.tags.split(",") if prefix in t] 

        if row.theme not in inferred_tags:
            df.loc[row.Index, "tags"] = row.tags.rstrip(" ,") + ", " + prefix + row.theme     
print df

输出：

                                tags theme type
1     col_t1, col_red, large, col_t2    t2    a
2                             col_t1    t1    c
3           col_t2, col_black, small    t2    d
4  col_t4, large, col_yellow, col_t3    t3    a
5        col_gold, col_fancy, col_t2    t2    b
6               col_t1, thick, col_k    t1    a
7                                NaN   NaN    a
8                 col_t3, fancy, red    t3    c

希望这是您正在寻找的。据称itertuples() 迭代所有行比iterrows() 更快。另外，请记住，我使用 numpy 和特别是 np.nan 在您的输入中模拟 NaN，但如果您的数据来自 csv，则不需要 numpy。

--- 更新 ---

如 cmets 中所述，代码应替换与主题匹配的标签。这是更新的解决方案：

prefix = "col_"

# Find all unique themes (notnull() excludes nan from the list)
themes = df[df["theme"].notnull()]["theme"].unique()

# Add prefex to all themes for comparison with tags; convert to set 
prefixed_themes = set([prefix + t for t in themes])

# Iterate over rows with non-empty theme
for row in df[df["theme"].notnull()].itertuples():

    if pd.isnull(row.tags):
        # Replace NaN in tags column with a single tag from theme column 
        df.loc[row.Index, "tags"] = prefix + row.theme
    else:
        # Extract existing tags with prefix (do not remove prefix; remove all spaces)
        inferred_tags = row.tags.replace(" ", "").split(",")

        # Use sets to check if there is any intersection between tags and themes
        if len(set(inferred_tags).intersection(prefixed_themes)) > 0:

            # Iterate over inferred_tags to find and replace matches with themes 
            for idx, t in enumerate(inferred_tags):
                if t in prefixed_themes:
                    inferred_tags[idx] = prefix + row.theme

            df.loc[row.Index, "tags"] = ", ".join(inferred_tags) 
        else:
            # In this case, add theme to tags (no replacement)
            df.loc[row.Index, "tags"] = row.tags.rstrip(" ,") + ", " + prefix + row.theme 

print df

输出：

                                tags theme type
1             col_t2, col_red, large    t2    a
2                             col_t1    t1    c
3           col_t2, col_black, small    t2    d
4  col_t4, large, col_yellow, col_t3    t3    a
5        col_gold, col_fancy, col_t2    t2    b
6               col_t1, thick, col_k    t1    a
7                                NaN   NaN    a
8                 col_t3, fancy, red    t3    c

请注意，代码会根据主题列中存在的所有值检查标签（添加前缀）；如果某个值（如 t4）不在主题列中，则它不被视为合法主题标签，因此在处理期间不会替换第 4 项中的 col_t4。如果您需要替换所有col_t*，则需要具体说明。希望这是一个有用的解决方案，您可以从这里获得它。