The Truth value of a Series is ambiguous. Use a.Empty, a.bool(), a.item(), a.any() or a.all()

Question

I am a layperson in programming, in the stackoverflow community and I am not able to solve the following problem:

Create two lists (or if you prefer 2 columns in a New Dataframe) in which you search in a Dataframe the successor and the predecessor of the elements that are another Datatframe(reference). I made a code (I am open to new solutions) and is accusing an error message that is in the title.

antecessor = []
sucessor = []

for elemento in x[0]:
    i = 0
    while i < y[0].size:
        if elemento < y[0].iloc[i:i+1]:
            sucessor.append(y[0].iloc[i:i+1])
            antecessor.append(y[0].iloc[i-1:i])
            break
    i += 1 

Obs:

i. The elements found in the Dataframe X and Y columns are float-type

ii. The number of Dataframe X elements is less than Dataframe Y.

Thanks in advance for all your help and attention.

Sincerely yours truly, Wendel de Oliveira

x.dat (2 columns)

0.509054 0.0767 0.355404 0.0769 0.279519 0.0771 0.219150 0.0773 0.176476 0.0775 0.151162 0.0777 0.139228 0.0779 0.117106 0.0781 0.108191 0.0783 0.095156 0.0785

y. dat (4 columns)

0.0893015 1.00584 0.0725685 -1.83063 0.0938015 1.00614 0.0642253 -1.87747 0.0983015 1.00641 0.0556713 -1.9243 0.102801 1.00664 0.0469066 -1.97114 0.107301 1.00684 0.0379312 -2.01797 0.111801 1.00699 0.0287449 -2.06481 0.116301 1.00709 0.0193479 -2.11165 0.120801 1.00716 0.00974018 -2.15848 0.125301 1.00718 -7.83241e-05 -2.20532 0.129801 1.00716 -0.0101076 -2.25215 0.134301 1.00709 -0.0203476 -2.29899 0.138801 1.00697 -0.0307984 -2.34582 0.143301 1.00681 -0.0414599 -2.39266 0.147801 1.0066 -0.0523322 -2.43949 0.152301 1.00634 -0.0634153 -2.48632 0.156801 1.00603 -0.0747091 -2.53316 0.161301 1.00567 -0.0862136 -2.57999 0.165801 1.00525 -0.0979289 -2.62682 0.170301 1.00479 -0.109855 -2.67365 0.174801 1.00426 -0.121992 -2.72048 0.179301 1.00369 -0.134339 -2.76731 0.183801 1.00305 -0.146897 -2.81414 0.188301 1.00236 -0.159666 -2.86097 0.192801 1.00162 -0.172646 -2.90779 0.197301 1.00081 -0.185836 -2.95462 0.201801 0.999944 -0.199237 -3.00144 0.206301 0.999017 -0.212849 -3.04826 0.210801 0.998028 -0.226672 -3.09508 0.215301 0.996977 -0.240705 -3.14189 0.219801 0.995861 -0.254949 -3.18871 0.224301 0.994682 -0.269403 -3.23552 0.228801 0.993436 -0.284068 -3.28233 0.233301 0.992124 -0.298944 -3.32913 0.237801 0.990745 -0.31403 -3.37593 0.242301 0.989298 -0.329327 -3.42273 0.246801 0.987781 -0.344835 -3.46953 0.251301 0.986194 -0.360553 -3.51632 0.255801 0.984535 -0.376481 -3.56311 0.260301 0.982805 -0.392621 -3.6099 0.264801 0.981001 -0.40897 -3.65669 0.269301 0.979124 -0.425531 -3.70347 0.273801 0.977171 -0.442302 -3.75026 0.278301 0.975143 -0.459283 -3.79705 0.282801 0.973037 -0.476475 -3.84383 0.287301 0.970854 -0.493877 -3.89062 0.291801 0.968592 -0.51149 -3.93741 0.296301 0.96625 -0.529314 -3.98421 0.300801 0.963828 -0.547348 -4.03101 0.305301 0.961323 -0.565593 -4.07782 0.309801 0.958737 -0.584048 -4.12464 0.314301 0.956067 -0.602715 -4.17146 0.318801 0.953312 -0.621591 -4.2183 0.323301 0.950472 -0.640679 -4.26515 0.327801 0.947545 -0.659978 -4.31201 0.332301 0.944532 -0.679487 -4.35889 0.336801 0.941429 -0.699208 -4.40577 0.341301 0.938238 -0.719139 -4.45268 0.345801 0.934957 -0.739282 -4.4996 0.350301 0.931584 -0.759635 -4.54654 0.354801 0.92812 -0.7802 -4.59349 0.359301 0.924562 -0.800977 -4.64046 0.3638 0.92091 -0.821964 -4.68745 0.3683 0.917164 -0.843164 -4.73446 0.3728 0.913321 -0.864574 -4.78149 0.3773 0.909382 -0.886197 -4.82853 0.3818 0.905345 -0.908031 -4.8756 0.3863 0.901209 -0.930077 -4.92268 0.3908 0.896974 -0.952335 -4.96978 0.3953 0.892638 -0.974805 -5.0169 0.3998 0.8882 -0.997487 -5.06404 0.4043 0.88366 -1.02038 -5.1112 0.4088 0.879017 -1.04349 -5.15837 0.4133 0.874268 -1.06681 -5.20556 0.4178 0.869415 -1.09034 -5.25277 0.4223 0.864455 -1.11408 -5.29999 0.4268 0.859388 -1.13804 -5.34723 0.4313 0.854212 -1.16221 -5.39448 0.4358 0.848927 -1.18659 -5.44175 0.4403 0.843532 -1.21118 -5.48903 0.4448 0.838026 -1.23599 -5.53632 0.4493 0.832408 -1.26101 -5.58363 0.4538 0.826677 -1.28624 -5.63095 0.4583 0.820831 -1.31169 -5.67827 0.4628 0.814871 -1.33735 -5.72561 0.4673 0.808795 -1.36322 -5.77295 0.4718 0.802602 -1.3893 -5.8203 0.4763 0.796291 -1.4156 -5.86766 0.4808 0.789861 -1.44211 -5.91502 0.4853 0.783311 -1.46883 -5.96238 0.4898 0.776641 -1.49577 -6.00975 0.4943 0.769849 -1.52292 -6.05713 0.4988 0.762934 -1.55029 -6.1045 0.5033 0.755896 -1.57786 -6.15188 0.5078 0.748733 -1.60565 -6.19925 0.5123 0.741445 -1.63366 -6.24663 0.5168 0.73403 -1.66187 -6.29401 0.5213 0.726487 -1.6903 -6.34139 0.5258 0.718817 -1.71894 -6.38877 0.5303 0.711016 -1.7478 -6.43615 0.5348 0.703086 -1.77687 -6.48353 0.5393 0.695024 -1.80615 -6.53091

Answer 1

The mistake "The Truth value of a Series is ambiguous. Use a.Empty, a.bool(), a.item(), a.any() or a.all()" occurs when python cannot make a comparison because the logical test is unclear. For example, the following replicable code also returns this error:

df = pd.DataFrame({'A': [0, 1, 2, 3, 4],
                'B': [5, 6, 7, 8, 9],
                'C': ['a', 'b', 'c', 'd', 'e']})

if 3< df['A'].iloc[0:2]:
    print('ok')

Looking at your code, it seems that this error occurs in the expression elemento < y[0].iloc[i:i+1] since you are comparing a number with a list. python does not know which number to use in the comparison and returns the error.