Searching and Measuring
Once you can read pieces of a string, the next step is to ask questions about its contents: "does it contain this?", "where is it?", "how many times?", "does it start/end with�".
Membership
Use in to check if a substring appears anywhere in a string. It returns a Boolean.
123s = "Python programming" print("gram" in s) # True print("Java" not in s) # True
Finding Positions
find(sub)returns the starting index of the first match, or -1 if not found;rfind(sub)searches from the right and returns the last match index (or -1);index(sub)is likefind, but raisesValueErrorif the substring is missing.
12345s = "abracadabra" print(s.find("bra")) # 1 print(s.rfind("bra")) # 8 print(s.find("xyz")) # -1 print(s.index("xyz")) # ValueError
Counting Occurrences
count(sub) returns how many non-overlapping times the substring appears.
123s = "banana" print(s.count("a")) # 3 print(s.count("na")) # 2
Starts and Ends
To check a string's start or end, use startswith or endswith; these are clearer and safer than slicing when you only need a yes/no answer.
123fname = "report_final.pdf" print(fname.startswith("report")) # True print(fname.endswith(".pdf")) # True
Case-insensitive checks (simple and practical)
String methods are case-sensitive. For case-insensitive search, normalize both sides with .lower() (or .upper()).
123msg = "Hello, World!" print("hello" in msg.lower()) # True print(msg.lower().startswith("hello")) # True
We already used len() in the previous chapter to measure length; here we focus on content checks and locations.
1. What will this code output?
2. Which statement does not raise an error when the substring is missing?
3. Given s = "Banana", which expression returns True case-insensitively for checking the prefix "ba"?
Thanks for your feedback!
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Once you can read pieces of a string, the next step is to ask questions about its contents: "does it contain this?", "where is it?", "how many times?", "does it start/end with�".
Membership
Use in to check if a substring appears anywhere in a string. It returns a Boolean.
123s = "Python programming" print("gram" in s) # True print("Java" not in s) # True
Finding Positions
find(sub)returns the starting index of the first match, or -1 if not found;rfind(sub)searches from the right and returns the last match index (or -1);index(sub)is likefind, but raisesValueErrorif the substring is missing.
12345s = "abracadabra" print(s.find("bra")) # 1 print(s.rfind("bra")) # 8 print(s.find("xyz")) # -1 print(s.index("xyz")) # ValueError
Counting Occurrences
count(sub) returns how many non-overlapping times the substring appears.
123s = "banana" print(s.count("a")) # 3 print(s.count("na")) # 2
Starts and Ends
To check a string's start or end, use startswith or endswith; these are clearer and safer than slicing when you only need a yes/no answer.
123fname = "report_final.pdf" print(fname.startswith("report")) # True print(fname.endswith(".pdf")) # True
Case-insensitive checks (simple and practical)
String methods are case-sensitive. For case-insensitive search, normalize both sides with .lower() (or .upper()).
123msg = "Hello, World!" print("hello" in msg.lower()) # True print(msg.lower().startswith("hello")) # True
We already used len() in the previous chapter to measure length; here we focus on content checks and locations.
1. What will this code output?
2. Which statement does not raise an error when the substring is missing?
3. Given s = "Banana", which expression returns True case-insensitively for checking the prefix "ba"?
Thanks for your feedback!
