在python中如何将一位数转换为两位数的字符串？
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in python how do I convert a single digit number into a double digits string?
提问by Joe Schmoe
So say i have
所以说我有
a = 5
一 = 5
i want to print it as a string '05'
我想将它打印为字符串 '05'
采纳答案by jkerian
print "%02d"%ais the python 2 variant
print "%02d"%a是 python 2 变体
python 3 uses a somewhat more verbose formatting system:
python 3 使用了更冗长的格式系统：
"{0:0=2d}".format(a)
The relevant doc link for python2 is: http://docs.python.org/2/library/string.html#format-specification-mini-language
python2的相关文档链接是：http: //docs.python.org/2/library/string.html#format-specification-mini-language
For python3, it's http://docs.python.org/3/library/string.html#string-formatting
对于 python3，它是http://docs.python.org/3/library/string.html#string-formatting
回答by tux21b
a = 5
print '%02d' % a
# output: 05
The '%' operator is called string formattingoperator when used with a string on the left side. '%d'is the formatting code to print out an integer number (you will get a type error if the value isn't numeric). With '%2dyou can specify the length, and '%02d'can be used to set the padding character to a 0 instead of the default space.
'%' 运算符在与左侧的字符串一起使用时称为字符串格式化运算符。'%d'是打印出整数的格式代码（如果该值不是数字，则会出现类型错误）。用'%2d你可以指定长度，并且'%02d'可以用来将填充字符设置为 0 而不是默认空格。
回答by Mohammad Shahid Siddiqui
>>> a=["%02d" % x for x in range(24)]
>>> a
['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '11', '12', '13', '14', '15', '16', '17', '18', '19', '20', '21', '22', '23']
>>>
It is that simple
就是这么简单
回答by Sarvesh Chitko
In Python3, you can:
在 Python3 中，您可以：
print("%02d" % a)
回答by Alex Schwab
Branching off of Mohommad's answer:
分支穆罕默德的回答：
str_years = [x for x in range(24)]
#[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]
#Or, if you're starting with ints:
int_years = [int(x) for x in str_years]
#Formatted here
form_years = ["%02d" % x for x in int_years]
print(form_years)
#['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '11', '12', '13', '14', '15', '16', '17', '18', '19', '20', '21', '22', '23']
回答by nishant thakkar
df["col_name"].str.rjust(4,'0')#(length of string,'value') --> ValueXXX --> 0XXX
df["col_name"].str.ljust(4,'0')#(length of string,'value') --> XXXValue --> XXX0
回答by MountainWaterProducts
If you are an analyst and not a full stack guy, this might be more intuitive:
如果您是分析师而不是全栈人员，这可能更直观：
[(str('00000') + str(i))[-5:] for i in arange(100)]
breaking that down, you:
打破它，你：
start by creating a list that repeats 0's or X's, in this case, 100 long, i.e., arange(100)
add the numbers you want to the string, in this case, numbers 0-99, i.e., 'i'
keep only the right hand 5 digits, i.e., '[-5:]' for subsetting
output is numbered list, all with 5 digits
首先创建一个重复 0 或 X 的列表，在这种情况下，长度为 100，即 arange(100)
将您想要的数字添加到字符串中，在本例中为数字 0-99，即 'i'
只保留右手边的 5 位数字，即 '[-5:]' 用于子集化
输出是编号列表，全部为 5 位数字
回答by K Puri
This is a dumb solution but I was getting type errors with the other solutions above. So if all else fails, yolo:
这是一个愚蠢的解决方案，但我在使用上述其他解决方案时遇到了类型错误。所以如果所有其他方法都失败了，yolo：
images3digit = []
for i in images:
if len(i)==1:
i = '00'+i
images3digit.append(i)
elif len(i)==2:
i = '0'+i
images3digit.append(i)
elif len(i)==3:
images3digit.append(i)
