diff --git a/python/Chapter 1/Question1_5/ChapQ1.5.py b/python/Chapter 1/Question1_5/ChapQ1.5.py
deleted file mode 100644
index 43eb48d1..00000000
--- a/python/Chapter 1/Question1_5/ChapQ1.5.py	
+++ /dev/null
@@ -1,69 +0,0 @@
-#Implement a method to perform basic string compression using the counts of repeated characters
-
-#aabcccccaaa would become a2blc5a3.
-
-#do nothing if this would not make the string smaller.
-
-
-def simpleCompress(compstring):
-    #Avoid using the + and += operators to accumulate a string within a loop. Since strings are immutable, this creates unnecessary temporary objects and results in quadratic rather than linear running time. Instead, add each substring to a list and ''.join the list after the loop terminates (or, write each substring to a cStringIO.StringIO buffer).
-    #Google python style guidelines http://google-styleguide.googlecode.com/svn/trunk/pyguide.html
-    outstring=[]
-    lastchar=""
-    charcount=0
-    for char in compstring:
-        if char==lastchar:
-            charcount += 1
-        else:
-            if lastchar != "":
-                outstring.append(lastchar + str(charcount))
-            charcount = 1
-        lastchar=char            
-    #final write
-    outstring.append(lastchar + str(charcount))
-    outstring="".join(outstring)
-    if len(outstring)<len(compstring):
-        return outstring
-    else:
-        return compstring
-
-#testing
-
-tocompress = "aabcccccaaa"
-
-if simpleCompress(tocompress) == "a2b1c5a3":
-    print "Test 1 Passed"
-else:
-    print "Test 1 Failed"
-
-tocompress2 = "aabca"
-if simpleCompress(tocompress2) == tocompress2:
-    print "Test 2 Passed"
-else:
-    print "Test 2 Failed"
-
-
-#Modulated version of the solution
-def countSame(string, pos):
-	#Counts number of same characters following a position
-	pos1 = pos
-	count = 0
-	while pos1 < len(string) and string[pos1] == string[pos]:
-		count = count +1
-		pos1 = pos1 + 1
-	return count
-
-
-def compress(string):
-	outstring = []
-	pos = 0
-	while pos < len(string):
-		outstring.append(string[pos] + str(countSame(string, pos)))
-		pos = pos + countSame(string, pos)
-	result = ''.join(outstring)
-	if len(result) < len(string):
-	    return result
-	else:
-	    return string
-
-print compress("aabccc   ccaaa")
diff --git a/python/Chapter 1/Question1_1/1-unique.py b/python/Chapter_1/Question1_1/1-unique.py
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diff --git a/python/Chapter_1/Question1_5/ChapQ1.5.py b/python/Chapter_1/Question1_5/ChapQ1.5.py
new file mode 100644
index 00000000..cfc31877
--- /dev/null
+++ b/python/Chapter_1/Question1_5/ChapQ1.5.py
@@ -0,0 +1,74 @@
+# Implement a method to perform basic string compression using the counts of repeated characters
+
+# aabcccccaaa would become a2blc5a3.
+
+# do nothing if this would not make the string smaller.
+
+
+def simpleCompress(compstring):
+    # Avoid using the + and += operators to accumulate a string within a loop. Since strings are immutable,
+    # this creates unnecessary temporary objects and results in quadratic rather than linear running time.
+    # Instead, add each substring to a list and ''.join the list after the loop terminates
+    # (or, write each substring to a cStringIO.StringIO buffer).
+    # Google python style guidelines http://google-styleguide.googlecode.com/svn/trunk/pyguide.html
+    outstring = []
+    lastchar = ""
+    charcount = 0
+    for char in compstring:
+        if char == lastchar:
+            charcount += 1
+        else:
+            if lastchar != "":
+                outstring.append(lastchar + str(charcount))
+            charcount = 1
+        lastchar = char
+        # final write
+    outstring.append(lastchar + str(charcount))
+    outstring = "".join(outstring)
+    if len(outstring) < len(compstring):
+        return outstring
+    else:
+        return compstring
+
+
+# testing
+
+tocompress = "aabcccccaaa"
+
+if simpleCompress(tocompress) == "a2b1c5a3":
+    print "Test 1 Passed"
+else:
+    print "Test 1 Failed"
+
+tocompress2 = "aabca"
+if simpleCompress(tocompress2) == tocompress2:
+    print "Test 2 Passed"
+else:
+    print "Test 2 Failed"
+
+
+# Modulated version of the solution
+def countSame(string, pos):
+    # Counts number of same characters following a position
+    pos1 = pos
+    count = 0
+    while pos1 < len(string) and string[pos1] == string[pos]:
+        count += 1
+        pos1 += 1
+    return count
+
+
+def compress(string):
+    outstring = []
+    pos = 0
+    while pos < len(string):
+        outstring.append(string[pos] + str(countSame(string, pos)))
+        pos = pos + countSame(string, pos)
+    result = ''.join(outstring)
+    if len(result) < len(string):
+        return result
+    else:
+        return string
+
+
+print compress("aabccc   ccaaa")
diff --git a/python/Chapter 1/Question1_5/compress.py b/python/Chapter_1/Question1_5/compress.py
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