Python Pandas - 在 DataFrame 中的任何位置查找值的索引
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Pandas - find index of value anywhere in DataFrame
提问by Kemeia
I'm new to Python & Pandas.
我是 Python 和 Pandas 的新手。
I want to find the index of a certain value (let's say security_id) in my pandas dataframe, because that is where the columns start.
(There is an unknown number of rows with irrelevant data above the columns, as well as a number of empty 'columns' on the left side.)
我想security_id在我的 Pandas 数据框中找到某个值（假设）的索引，因为那是列开始的地方。（列上方有未知数量的行带有无关数据，左侧还有许多空“列”。）
As far as I see, the isinmethod only returns a boolean on whether the value exists, not its index.
据我所知，isin方法只返回一个关于值是否存在的布尔值，而不是它的索引。
How do I find the index of this value?
我如何找到这个值的索引？
采纳答案by Ujjwal
Supposing that your DataFrame looks like the following :
假设您的 DataFrame 如下所示：
0 1 2 3 4
0 a er tfr sdf 34
1 rt tyh fgd thy rer
2 1 2 3 4 5
3 6 7 8 9 10
4 dsf wew security_id name age
5 dfs bgbf 121 jason 34
6 dddp gpot 5754 mike 37
7 fpoo werwrw 342 Hyman 31
Do the following :
请执行下列操作 ：
for row in range(df.shape[0]): # df is the DataFrame
for col in range(df.shape[1]):
if df.get_value(row,col) == 'security_id':
print(row, col)
break
回答by Ravishankar Sivasubramaniam
Get the index for rows matching search term in all columns
获取所有列中匹配搜索词的行的索引
search = 'security_id'
df.loc[df.isin([search]).any(axis=1)].index.tolist()
Rows filtered for matching search term in all columns
筛选行以匹配所有列中的搜索词
search = 'search term'
df.loc[df.isin([search]).any(axis=1)]
回答by Jay
value you are looking for is not duplicated:
您正在寻找的值不会重复：
poz=matrix[matrix==minv].dropna(axis=1,how='all').dropna(how='all')
value=poz.iloc[0,0]
index=poz.index.item()
column=poz.columns.item()
you can get its index and column
你可以得到它的索引和列
duplicated:
重复：
matrix=pd.DataFrame([[1,1],[1,np.NAN]],index=['q','g'],columns=['f','h'])
matrix
Out[83]:
f h
q 1 1.0
g 1 NaN
poz=matrix[matrix==minv].dropna(axis=1,how='all').dropna(how='all')
index=poz.stack().index.tolist()
index
Out[87]: [('q', 'f'), ('q', 'h'), ('g', 'f')]
you will get a list
你会得到一个清单
回答by Peterd
A oneliner solution avoiding explicit loops...
避免显式循环的单行解决方案......
returning the entire row(s)
df.iloc[np.flatnonzero((df=='security_id').values)//df.shape[1],:]
returning row(s) and column(s)
df.iloc[ np.flatnonzero((df=='security_id').values)//df.shape[1], np.unique(np.flatnonzero((df=='security_id').values)%df.shape[1]) ]
返回整行
df.iloc[np.flatnonzero((df=='security_id').values)//df.shape[1],:]
返回行和列
df.iloc[ np.flatnonzero((df=='security_id').values)//df.shape[1], np.unique(np.flatnonzero((df=='security_id').values)%df.形状[1]) ]
回答by Adam Slack
I think this question may have been asked before here. The accepted answer is pretty comprehensive and should help you find the index of a value in a column.
我想这个问题之前可能已经问过这里了。接受的答案非常全面，应该可以帮助您找到列中值的索引。
Edit: if the column that the value exists in is not known, then you could use:
编辑：如果值所在的列未知，那么您可以使用：
for col in df.columns:
df[df[col] == 'security_id'].index.tolist()
回答by tasos bada
Function finds the positions of a value in a dataframe
函数在数据框中查找值的位置
import pandas as pd
import numpy as np
def pandasFindPositionsInDataframe(dfIn,findme):
positions = []
irow =0
while ( irow < len(dfIn.index)):
list_colPositions=dfIn.columns[dfIn.iloc[irow,:]==findme].tolist()
if list_colPositions != []:
colu_iloc = dfIn.columns.get_loc(list_colPositions[0])
positions.append([irow, colu_iloc])
irow +=1
return positions
