[英]Python pandas replace based on partial match with list item
I have a large three-column dataframe of this form:我有一个大的三列 dataframe 这种形式:
Ref Colourref Shaperef
5 red 12 square 15
9 14 blue (circle14,2)
10 6 orange 12 18 square
12 pink1,7 [oval] [40]
14 [green] (rectsq#12,6)
...
And a long list with entries like this:还有一个长长的列表,里面有这样的条目:
li = [
'oval 60 [oval] [40]',
'(circle14,2) circ',
'square 20',
'126 18 square 921#',
]
I want to replace the entries in the Shaperef column of the df with a value from the list if the full Shaperef string matches any part of any list item.如果完整的 Shaperef 字符串与任何列表项的任何部分匹配,我想用列表中的值替换 df 的 Shaperef 列中的条目。 If there is no match, the entry is not changed.
如果没有匹配项,则不会更改条目。
Desired output:所需的 output:
Ref Colourref Shaperef
5 red 12 square 15
9 14 blue (circle14,2) circ
10 6 orange 12 126 18 square 921#
12 pink1,7 oval 60 [oval] [40]
14 [green] (rectsq#12,6)
...
So refs 9, 10, 12 are updated as there is a partial match with a list item.因此,参考 9、10、12 被更新,因为与列表项存在部分匹配。 Refs 5, 14 stay as there are.
Refs 5, 14 保持原样。
If Shaperef
and all the entries in li
are all strings you can write a function to apply over Shaperef
to convert them:如果
Shaperef
和li
中的所有条目都是字符串,则可以编写 function 以应用于Shaperef
以转换它们:
def f(row_val, seq):
for item in seq:
if row_val in item:
return item
return row_val
Then:然后:
# read in your example
import pandas as pd
from io import StringIO
s = """Ref Colourref Shaperef
5 red 12 square 15
9 14 blue (circle14,2)
10 6 orange 12 18 square
12 pink1,7 [oval] [40]
14 [green] (rectsq#12,6)
"""
li = [
"oval 60 [oval] [40]",
"(circle14,2) circ",
"square 20",
"126 18 square 921#",
]
df = pd.read_csv(StringIO(s), sep=r"\s\s+", engine="python")
# Apply the function here:
df["Shaperef"] = df["Shaperef"].apply(lambda v: f(v, li))
# Ref Colourref Shaperef
# 0 5 red 12 square 15
# 1 9 14 blue (circle14,2) circ
# 2 10 6 orange 12 126 18 square 921#
# 3 12 pink1,7 oval 60 [oval] [40]
# 4 14 [green] (rectsq#12,6)
This might not be a very quick way of doing this as it has a worst case run time of len(df) * len(li)
.这可能不是一种非常快速的方法,因为它的最坏情况运行时间为
len(df) * len(li)
。
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