Hello I have a dataframe that looks like this:
Index X Y Z a b c u v
0 -1.462916 -1.299055 3.015321 -0.727074 2.697548 0.822541 259 -106
1 -1.397448 -1.267354 2.935434 -0.636784 2.755215 0.836514 273 -108
2 -1.384987 -1.269575 2.937742 -0.638735 2.767929 0.836798 279 -108
3 -1.367630 -1.281251 2.959687 -0.661788 2.787479 0.834445 293 -109
4 -1.349895 -1.278706 2.951603 -0.651958 2.804420 0.836097 300 -110
... ... ... ... ... ... ... ... ...
33817 0.744647 0.701664 1.618279 1.717619 4.655669 -0.186012 1655 1169
33818 0.723408 0.674251 1.556330 1.753335 4.632607 -0.129156 1662 1168
33819 0.752354 0.693699 1.602645 1.727306 4.662931 -0.171157 1669 1168
33820 0.681987 0.622136 1.438694 1.821023 4.587733 -0.021106 1675 1167
33821 0.765870 0.691312 1.600311 1.729280 4.676410 -0.168285 1682 1166
So I have a pair of u and v values (the last two columns). So what I want to ask is, if for example I have a value of u = 279 and a value of v = -108, my output would be Index = 2, is there a way to do this?
I tried doing the code below but it only looks for one column only.
v_location = a.loc[a['v'] == -108].index[1]
v_location
[output] = 2
As you can see, this method only looks for values within one column (e.g. v column) and may give wrong index values, as you can see, index=1 and index=2 are both v == -108.
To simply put, I want my input to be values from column u and column v as a pair and would give an index value as an output.
I hope you can hekp me with this one. Thank you very much :)
CodePudding user response:
You can use multiple conditions in the expression like this df['u'].eq(279) & df['v'].eq(-108)
print(a[a['u'].eq(279) & a['v'].eq(-108)].index[0])
Prints:
2
CodePudding user response:
Try isin() function from dataframe. df.isin([-100])