我有一個(gè)這樣的數(shù)據(jù)框：import pandas as pddf = pd.DataFrame({'col1': ['abc', 'def', 'tre'], 'col2': ['foo', 'bar', 'stuff']}) col1 col20 abc foo1 def bar2 tre stuff和這樣的字典：d = {'col1': [0, 2], 'col2': [1]}字典包含要從數(shù)據(jù)框中提取的列名和值索引，以生成如下字符串：abc (0, col1)因此，每個(gè)字符串都以元素本身開頭，并在括號(hào)中顯示索引和列名。我嘗試了以下列表理解：l = [f"{df.loc[{indi}, {ci}]} ({indi}, {ci})" for ci, vali in d.items() for indi in vali]這產(chǎn)生[' col1\n0 abc (0, col1)', ' col1\n2 tre (2, col1)', ' col2\n1 bar (1, col2)']所以，幾乎沒問題，只是col1\n0需要避免的部分。如果我嘗試f"{df.loc[0, 'col1']} is great"我得到'abc is great'然而，根據(jù)需要，x = 0f"{df.loc[{x}, 'col1']} is great"我得到'0 abc\nName: col1, dtype: object is great'這怎么能解決？
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TA貢獻(xiàn)1780條經(jīng)驗(yàn) 獲得超5個(gè)贊
import pandas as pd
df = pd.DataFrame({'col1': ['abc', 'def', 'tre'],
'col2': ['foo', 'bar', 'stuff']})
d = {'col1': [0, 2], 'col2': [1]}
x = 0
[f"{df.loc[x, 'col1']} is great"
for ci, vali in d.items()
for indi in vali]
這給了你：
['abc is great', 'abc is great', 'abc is great']
這是你要找的嗎？
你也可以循環(huán)通過 x 范圍
[f"{df.loc[i, 'col1']} is great"
for ci, vali in d.items()
for indi in vali
for i in range(2)]
#output
['abc is great',
'def is great',
'abc is great',
'def is great',
'abc is great',
'def is great']
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