Thanks. I tried a different way (you need python 3 or python 2.7 for this):<br>from collections import Counter<br>...<br>...<br>print len(data_one)<br>data_one_unique = list(set(data_one))<br><br>print len(data_one_unique)<br>
a = Counter(data_one)<br>b = Counter(data_one_unique)<br><br>c = a - b<br>print list(c.elements())<br><br><br>------------------<br>This was the output:<br><br>285<br>228<br>['DKVTIADDySDPFDAK', 'GEPEALyAAVTK', 'QHSLPSSEHLGTDGALyQVPPQPR', 'THAVSVSETDDyAEIIDEEDTYTMPSTR', 'LIEDNEyTAR', 'YMEDSTYyK', 'VyENVGLMQQQR', 'AVCSTyLQSR', 'MNHTSQAFITAASGGQPPNyER', 'ERDyAEIQDFHR', 'RTEGDyLSYR', 'NTyNQTALDIVNQFTTSQASR', 'YMEDSTyYK', 'GEPNVSyICSR', 'GEPNVSyICSR', 'SAQPSPHYMAGPSSGQIyGPGPR', 'TACTNFMMTPyVVTR', 'SDNNySTLNER', 'TVCSTyLQSR', 'SLDNNySTLNER', 'GLCTSPAEHQYFMTEyVATR', 'GLCTSPAEHQYFMTEyVATR', 'TPYEAyDPIGK', 'NLSEGNNANYTEyVATR', 'ELFDDPSyVNIQNLDK', 'LVQSPNSyFMDVK', 'VADPDHDHTGFLtEyVATR', 'TADSVFCPHyEK', 'LWLEAMDGKEPIyTLPAIISK', 'VVQEYIDAFSDyANFK', 'VEKIGEGTyGVVYK', 'AGKGESAGyMEPYEAQR', 'YVDSEGHLyTVPIR', 'KIYNGDyYR', 'LSHSSGyAQLNTYSR', 'STTNyVDFYSTK', 'IEKIGEGtyGVVYK', 'TLEPVKPPTVPNDyMTSPAR', 'IEKIGEGTyGVVYK', 'VGQGYVYEAAQTEQDEyDTPR', 'TAGTSFMMTPyVVTR', 'TAGTSFMMTPyVVTR', 'NEEENIySVPHDSTQGK', 'LCDFGSASHVADNDITPyLVSR', 'LCDFGSASHVADNDITPyLVSR', 'GPLDGSPyAQVQR', 'GPLDGSPyAQVQR', 'FLEENSSDPTyTSSLGGKIPIR', 'HAAyGGYSTPEDR', 'VADPDHDHTGFLTEyVATR', 'HLLAPGPQDIyDVPPVR', 'LTDSKEDPIyDEPEGLAPAPPR', 'HTDDEMTGyVATR', 'HTDDEMTGyVATR', 'IYQyIQSR', 'IYQyIQSR', 'VLEDDPEATyTTSGGK']<br>
<br>----------<br><br>I took the output list and made it equal to z and then subsequently i made this list unique:<br><br>z = ['DKVTIADDySDPFDAK', 'GEPEALyAAVTK', 'QHSLPSSEHLGTDGALyQVPPQPR', 'THAVSVSETDDyAEIIDEEDTYTMPSTR', 'LIEDNEyTAR', 'YMEDSTYyK', 'VyENVGLMQQQR', 'AVCSTyLQSR', 'MNHTSQAFITAASGGQPPNyER', 'ERDyAEIQDFHR', 'RTEGDyLSYR', 'NTyNQTALDIVNQFTTSQASR', 'YMEDSTyYK', 'GEPNVSyICSR', 'GEPNVSyICSR', 'SAQPSPHYMAGPSSGQIyGPGPR', 'TACTNFMMTPyVVTR', 'SDNNySTLNER', 'TVCSTyLQSR', 'SLDNNySTLNER', 'GLCTSPAEHQYFMTEyVATR', 'GLCTSPAEHQYFMTEyVATR', 'TPYEAyDPIGK', 'NLSEGNNANYTEyVATR', 'ELFDDPSyVNIQNLDK', 'LVQSPNSyFMDVK', 'VADPDHDHTGFLtEyVATR', 'TADSVFCPHyEK', 'LWLEAMDGKEPIyTLPAIISK', 'VVQEYIDAFSDyANFK', 'VEKIGEGTyGVVYK', 'AGKGESAGyMEPYEAQR', 'YVDSEGHLyTVPIR', 'KIYNGDyYR', 'LSHSSGyAQLNTYSR', 'STTNyVDFYSTK', 'IEKIGEGtyGVVYK', 'TLEPVKPPTVPNDyMTSPAR', 'IEKIGEGTyGVVYK', 'VGQGYVYEAAQTEQDEyDTPR', 'TAGTSFMMTPyVVTR', 'TAGTSFMMTPyVVTR', 'NEEENIySVPHDSTQGK', 'LCDFGSASHVADNDITPyLVSR', 'LCDFGSASHVADNDITPyLVSR', 'GPLDGSPyAQVQR', 'GPLDGSPyAQVQR', 'FLEENSSDPTyTSSLGGKIPIR', 'HAAyGGYSTPEDR', 'VADPDHDHTGFLTEyVATR', 'HLLAPGPQDIyDVPPVR', 'LTDSKEDPIyDEPEGLAPAPPR', 'HTDDEMTGyVATR', 'HTDDEMTGyVATR', 'IYQyIQSR', 'IYQyIQSR', 'VLEDDPEATyTTSGGK']<br>
>>> z_unique = list(set(z))<br>>>> len(z)<br>57<br>>>> len(z_unique)<br>50<br>>>> <br><br>It appears that if the non-unique element occurs twice than you will have only one occurence in the output but if it occurs three times then the output (print list(c.elements()) ) will have the element written twice and so on. <br>
<br><br><br><br><div class="gmail_quote">On Tue, Apr 5, 2011 at 11:01 PM, Brian Palmer <span dir="ltr"><<a href="mailto:bpalmer@gmail.com">bpalmer@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin: 0pt 0pt 0pt 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">
<br><div class="gmail_quote"><div class="im">On Tue, Apr 5, 2011 at 7:49 PM, Vikram K <span dir="ltr"><<a href="mailto:kpguy1975@gmail.com" target="_blank">kpguy1975@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin: 0pt 0pt 0pt 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">
i have this list:<br>x = ['cat','dog','dog']<br><br>i wish to identify the non-unique element in this list i.e. 'dog'. how do i do this?<br></blockquote></div><div><br>This may not be suitable depending on how big your list is, but consider<br>
<br>
x = ['cat', 'dog', 'dog']<br>
x_count = defaultdict(lambda: 0)<br>
for k in x:<br>
x[k] = x[k] + 1<br>unique_xs = [k for k in x if x[k] == 1]<br> </div></div>
</blockquote></div><br>